MATHESIS ENUCLEATA: OR, THE ELEMENTS OF THE MATHEMATICKS.

By J. CHRIST. STURMIUS, Professor of Philosophy and Mathematicks in the University of Altorf.

Made English by J. R. A. M. and R. S. S.

LONDON, Printed for Robert Knaplock at the Angel, and Dan. Midwinter and Tho. Leigh at the Rose and Crown, in St. Paul's Church-yard. 1700.

The AUTHOR's PREFACE TO THE READER, Containing a SYNOPSIS of his Method.

I.

THAT the Reader may the better apprehend our design and aim, we have thought fit to premise some things concerning the Methods, both general and particular, we make use of in the following Treatise. For as heretofore a sort of a blind deference to, and superstitious Veneration of An­tiquity, and especially of Aristotle, has hindred the growth and progress of Natural Philosophy, which of late has made such considerable advances, since it has ventured to stand upon its own Bottom, to make new Additions to former Inventions, to es­say new and unknown Objects, to substitute Things instead of Names, Certainties instead of Doubts, and Experience in the room of dull Credulities; not derogating in the mean while from the deserved Praises of the Ingenious among the Ancients: So without doubt Mathematicks also, unless our Predecessors had imagined that it had long ago been brought to its utmost Perfe­ction by Euclid, Archimedes, Apollonius, and other Ingenious Ancients, would have arrived long since to a higher pitch, and by this time have surpass'd those Limits, which now we admire its arrival to.

II.

It is confess'd by all, that no Human Knowledge whatsoever can lay a more just claim to an unshaken Evidence and Certain­ty, or boast an higher necessity of its Demonstrations, or a great­er multitude of undeniable Truths, than the Mathematicks; and that those Propositions we have, found out by Archimedes, demonstrated by Euclid, Apollonius, and others, are at the same time unquestionable and altogether wonderful. But we may with Truth affirm, that most of their Propositions may either be disposed in a better order, or propounded easier, or demon­strated more evidently and directly, or taught after a more short and compendious way, now at least after they are already found out, and with a great deal of Pains demonstrated by their first Inventors; and of this Opinion are several of the best and most celebrated Mathematicians of the present Age.

III.

It is certain Euclid has demonstrated several Propositions, (as Prop. 2, 3, 20, 30, lib. 1. 2, 5, 6, 10, 15, 28, 29, lib. 3. &c.) whose Truth to any attentive Person appears from the very terms, more clearly and certainly than the truth of Ax­iom 13. lib. 1. which his Interpreters dare not admit without a Demonstration. And tho' those superfluous Demonstrations derogate nothing from the certitude of the thing, yet by an un­necessary increase of the number of Propositions, and (which fre­quently follows thence) an inverting the order of things, they breed Tediousness and Confusion.

IV.

There are none, unless those who are bigotted to Antiquity, but must own that the Elements of Euclid are destitute of a just and orderly Disposition of things. For to omit, that in the first Book there are h [...]ndled several sorts of Subjects, and a great variety of Properties demonstrated of them promiscuously without any respect to similitude or conveniency; there is this never to be [Page] excused, that, after he has in the first Book deduced and demonstrated some particular Affections of Magnitude, he proceeds, in the second, to those things which are universal and common to any quantity; then in the third and fourth, he contemplates the Circle and the Properties of Figures inscrib'd in it, or circumscribed about it; in the fifth again he treats of the universal Doctrine of Reasons and Proportions; and yet not so universally, but in the seventh again he is obliged to demonstrate the same of Numbers particularly, which might have been done for all Quantities whatsover, by one general Demonstra­tion.

V.

Then as for the method of Demonstrating used by the Anci­ents; it is true that it nicely regarded the certainty of its Con­clusions, nor would it admit any thing into its Demonstrations, which was not either a first Principle and so self evident, (cal­led by them an Axiom) or might not be supposed, beyond all Con­tradiction, possible to be effected (and on that account named a Postulate) or, thirdly, an arbitrary Denomination of the thing proposed which needed no Demonstration, (and was called a De­finition or Explication of a Term;) or, lastly, which had not been evidently demonstrated before: Yet I believe none will deny, but that this Method would have been more deservedly esteem'd, if with the certainty of its Conclusions it had joined a greater Easiness, Brevity, and Evidence, which is wanting in most of the Demonstrations of the Ancients; who thought it enough, firmly and infallibly to establish the truth of their Theorems, and extort the Assent of their Readers; little regarding by what Ambages, by how many circumambulatory Propositions, and almost whole Volumes, it was done; that thereby they might be forced to acknowledge the thing to be so, while how it came to be so, or from what intrinsick Cause o [...] Condition of the Subject requiring it, such and such an Attribute agreed to it, remained in the mean while obscure, or altogether unknown.

VI.

Hence they made such frequent use of Apagogical Demonstra­tions, or deductive ad absurdum & impossibile, which ought not to be done, but where no ostensive Demonstration can be had, or for illustrating negative Propositions rather than demostrat­ing them; for the method of Deduction ad impossibile, does not so much demonstrate the Truth it self directly, as the conse­quent Absurdity of the opposite Supposition; whence it follows very indirectly, (tho' most certainly) that the Proposition is true, while in the mean time the original Reason of its Truth remains altogether hid and in the dark.

VII.

But that we may not seem unjustly to reject the particular Method of the Ancients, made use of by Euclid, as in lib. 12. Prop. 2, 10. &c. and by others, but especially by Archimedes, who peculiarly addicted himself to it, whence it has been by some called the Archimedean Method, and by Renaldinus the Method per Explosum excessum atque defectum; besides its Deduction doubly ad absurdum, whereon it always relies, as e. g. it infers the equality of two Magnitudes A and B by a far fetch round about way, by shewing, that if B be supposed greater or less than A, from either Position there would follow an absurdity; and thence as it were begging the Equality by a new Inference, which tho' it may pass free from Suspicion, yet it neither ought, nor can be admitted, without this Limita­tion: In comparing those things whose Natures are capable of Equality, if it can be demonstrated that the one is neither great­er nor less than the other, we may thence justly infer their E­quality.

VIII.

The Learned are now generally agreed, that, besides that Syn­thetick Method, whereby the Ancients either ostensively deduced their Problems and Theorems from evident and common Princi­ples, [Page] or apagogically demonstrated them by Deductions ad absur­dum, they also made use of a certain sort of Analysis, whereby they found out those Theorems and Problems; and which, to raise the greater Admiration in their Readers, they afterwards studiously conceal'd and kept to themselves: Which Method is undoubtedly preferable to the other, as not only demonstrating the certainty of the Propositions so found, but at the same time shewing the invention of them too; and this is that Method that Vieta, Harriot, and Des Cartes, and their Followers have not only brought to light in this last Age, but to a great degree of Perfection too, and whereof Carolus Renaldinus, in that vast Work he has intituled Ars Analytica Mathematum, has given us a large Treatise.

IX.

There has appear'd moreover of late another particular Me­thod Dr. Wallis is of Opinion this is nothing but an Im­provement of the Ancients Method of Ex­haustions. invented by Bonaventura Cavallerius, which is called the Method of indivisibles, where­by the most difficult and abstruse Problems of Ge­ometry are found out and Demonstrated with an incredible ease, which is the above-mentioned Renaldinus's Opinion of it, lib. 1. Resol. & Comp. p. 239. which, to demonstrate the Equality or Proportions of Figures and Bodies that may be compared with one another, goes to work after a way which seems to be more natural than any other, by supposing plane Fi­gures to consist of innumerable lines, and solids of innumerable Plans (called their indivisible Parts or Elements, because the Lines are conceived without latitude, and the Plans without any thickness,) and relying on this self-evident Axiom, That if all the Indivisibles of one Magnitude collectively taken, be equal or proportional to all the correspondent Indivisibles of another, or taken separately each to each, then also those Magnitudes will be equal or proportional among themselves. Which Infe­rence can be guilty of no Fallacy, nor liable to any Error, as long as those Elements are taken and conceived in that sense their Authors design them; which is sufficiently demonstrated by Renaldinus, lib. 1. Compos. & Resol. p. 245. towards the bot­tom, [Page] and at the beginning of p. 306. and also by Honoratus Fa­ber, in his Synopsis, p. 24. and Dr. Barrow in Lect. Geom. p. 24. and the following, against Tacquet and the other Ad­versaries of this method.

X.

There is another Method a-kin to this, which may be proper­ly named Generative, very much followed by Faber in his Sy­nopsis, and Barrow in his Lect. Geom. whose Author Renal­dinus tells us was Guldinus, lib. Cit. p. 253. shewing at large its Rules and Foundations in the following pages, viz. The rise of Lines from certain motions of points; of Plane and Curvili­near Surfaces, from the determinate progress or rotation of a given Line, and of Solids by the various motions of various Surfaces; the Productions whereof are so represented to the Imagination, that the intrinsick nature of the magnitudes thence arising may become known, and their Properties and Affections may, from their Natures thus known, be easily and briefly deduced.

XI.

Near a-kin to this Method of Cavellerius is that other of In­finite Progressions, wherein having found a certain Progression of like Parts circumscrib'd about, or inscrib'd in any given mag­nitude, which may be continued by Bisection ad infinitum; and then at length (by vertue of the Doctrine of Exhaustions, foun­ded on Prop. 1. lib. 10. Eucl.) will terminate in the magni­tude it self, I say, wherein the sum of those infinite Terms, col­lected by Rules on purpose for the addition of those Progressions, and consequently the quantity or proportion of the proposed Mag­nitude, to any other given one, may be expressed or defined. But this termination of Figures infinitely circumscribed or inscri­bed in a Circle, not pleasing Renaldinus (altho' his Dissention seems only to consist in Words,) he exhibits another Method like it, (which he peculia [...]ly calls his own) built on twelve funda­mental Theorems, and illustrated by several Examples, lib. 1. de Resol. & Comp. p 277 & seqq.

XII.

Of late also, the most ingenious Mr. Isaac Newton, to de­monstrate his Philosophiae Naturalis Principia Mathematica, lib. 1. sect. 1. premises some Lemma's of his method of Rati­ones primae & ultimae, or evanescent quantities, thereby to a­void the tediousness of deducing long and perplext apagogical De­monstrations after the way of the Ancients. For finding that his Demonstrations might be very much contracted by the method of Indivisibles, and knowing at the same time, that that me­thod was scrupled by some, and thought not very Geometrical, he rather chose to found his Method on the sums and proportions of quantities which he calls Evanescent, which performs the same as the Method of Indivisibles, and may be more safely used, which he inculcates in these very words, and others such like, in Schol. of Lemma. 11. And also answers se­veral objections which might seem to make against it.

XIII.

But it would be in vain for us to attempt, in this place, to explain all and each of those various methods at length; having only proposed to our selves, to demonstrate the chief and principal Theorems and Inventions of the Mathematicks, and to use some­times one of them, and sometimes another, (having first Demon­strated their Foundations) according as we shall judge this or that of them, fittest to Demonstrate the thing in hand, and so shew the reasons, and use of each of them in the process of this discourse. And altho' H. Faber in his Synopsis, p. 8. Insinuates, that Analytick terms ought not to be made use of in Geometrical Demonstrations, because that Algebraick method seems to be too difficult for young beginners; yet we are of the quite contrary opi­nion (nay we can scarce doubt but that that Ingenious man would also agree with us herein, if he saw the way we make use of those foundations of Algebra, which is only of the most simple and ge­neral principles of it) especially in this case, where the said me­thod is by little and little instill'd with the Demonstrations themselves; and the literal Computations taught from their first [Page] Principles, than which nothing is more easie: And this is that which we design to do, and so use the learner by degrees to this sort of Demonstration, thereby to prepare him the bet­ter for the Analytick Geometry of the Moderns, which is the highest apex of the Mathematicks. But we had rather our Reader should himself find, than we trouble our selves any further to tell him here, how compendiously we Demonstrate the Propositions of Geometry, by the help of these Analytick notes, without the tedious Concatenation of a long Chain of Consequences, which would be otherwise unavoidable.

XIV.

After this way we design to go through the following Scheme. 1. We shall deduce several propositions of Euclid, Archi­medes, and Apollonius from our definitions, and the genera­tions of Magnitudes therein proposed; as Corollaries necessarily flowing from them, and confirm'd only by an immediate and simple consequence. 2. We shall demonstrate their chief Theo­rems (for the sake of which they were forced to Demon­strate several others before hand, the knowledge whereof for their own sake was not so necessary or valuable) without a­ny long series of antecedent Propositions, or Foreign princi­ples, from a few direct and intrinsick Principles of their own. Whence 3. It will follow, that after this Method we shall propose things, and treat of them; in a more natural Or­der, and first of all deliver those which are most universal and common to all quantities, and then descend to those which in a more special manner regard Magnitude, and distribute and dispose all according to certain general distinct Classes of the things to be treated of, and their affections. Hence also 4. We deduce from those universal Theorems, by way of Corollary, the Precepts of vulgar Arithmetick, and specious Computation, which afterwards we make use of in particular Demonstra­tions after a very short and compendious way; and, for this very reason, some Learned men of the present Age are of opi­nion, that the Ancients often fell into that tedious and intri­cate prolixity in their Demonstrations, because they would not acknowledge the great affinity there was between Arithme­tick [Page] and Geometry, taking particular care not to introduce the Terms and Operations of Arithmetick into Geometry; tho' at the same time they never scrupled to transfer the names of Plan, Square, Cube, and such like to numbers. 5. Lastly, Having first Demonstrated the first and Fundamental Theo­rems of Elementary Geometry, we may safely build on them the Praxes of all kinds of Mathematical Arts, that are most useful and requisite to several Exigencies of human Life, as, first, Trigonometry both Plain and Spherical, the Constructi­on and use of the Tables of Sines and Tangents. 2. The Construction of Logarithms, and a compendious application of them to Trigonometry: And in the 3. and last place, the fundamental Precepts of Algebra, or the Analytic Art; by the help whereof the learner may at length arrive to the higher and more recluse parts of Geometry, and become master thereof: Not to mention several Geometrical and A­rithmetical Problems, which we have all along derived from several of our Theorems, by way of Corollary, which it may be some other time, may make an Appendix of this work.

XV.

And thus when we shall have Demonstrated not only the chief Theorems of the Ancient Mathematicians, omitting the unnecessary crowd of those that are only Subsidiary, but also have demonstratively deduc'd the fundamental Precepts of the most necessary and useful Arts that flow from them, and that are abstracted from matter, as of Arithmetick, Trigono­metry, and Algebra; I hope none will doubt but that in this little Volum we have exhibited, as it were the Nucleus or Kernel of the pure and genuine Mathematicks (for those other Sciences and Arts which go by the name of mixt Ma­thematick, are most of them parts of natural Philosophy, from the application of Mathematicks to the Phaenomena of nature) and so may justly bear the name of Mathesis Enucleata.

XVI.

Nor are we ignorant, nor shall we conceal what several Learned men have both proposed and already done, for remov­ing those difficulties and blemishes of the Ancient Mathematical Methods we have just now mentioned. The late Admonish­ments of the anonymous Authour of L'Art de Penser, no less ingeniously than modestly delivered, Part. 4. Chap 9. 10. of the said Treatise, are sufficiently known; as also the lau­dable endeavours of A. Tacquet and Honoratus Faber and several others above mentioned for contracting, new order­ing, and more easily and directly Demonstrating the chief Ge­ometrical Inventions of the Ancients. There are moreover ex­tant of a certain anonymous Author, Elementa Geometrica novo ordine ac methodo feré Demonstrata, Printed at London about 26. Years ago. There are also F. Ignatius Gaston Par­dies Elemens de La Geometrie, &c. Translated into Latin after the third Edition, by the Famous Schmidtius Professor at Geneva: As also of F. Mich. Mourgues's, of the Society of J. Nouveaux Elemens de Geometrie, abreges par des Metho­des particulieres en moins de Cinquante propositions, &c. There are also several other Essays of reducing the Mathema­ticks into a better Order and Method, the titles whereof we have only as yet seen; and even while these papers were in the Press, there happen'd into our hands a Treatise of F. Lamy's Entituled Les Elemens de Geometrie, ou de la mesure des Corps, &c. Printed at Paris in 1685; so that we may only seem to some to do what has been done already, in endeavour­ing to shew our Reader a new and shorter way to the Ma­thematicks.

XVII.

But as none can blame Jacobus le Maire, because, after the happy discovery of the Magellanick Passage from the Atlantick into the Pacifick Sea, he would needs yet endeavour to find ano­ther shorter, which he accordingly did; nor can they be blamed, who now a days consult about finding one from these parts of [Page] the World, by the North to the East India's. Thus also, in an affair of that moment, that one or a few are not sufficient to bring it to Perfection, if any one who comes after, not only in­vited, but also assisted by the ingenious Endeavours of those who have gone before him, shall undertake to add to their Inventi­ons, to help on the business by his advices, and shew what things are capable of a further Polish, and the method how to perform it, doubtless such an one ought not to be blamed, nor accused of arrogance, unless at the same time he endeavours to depretiate the essays of others, and cry up his own as the only va­luable; Which how far it is from our design, the work it self will abundantly shew. Moreover as the senses of men are differently affected by different Objects, and their Palates have different Relishes of the same thing, according to different Pre­parations of them: So the same truth takes and insinuates it self more easily with one proposed and demonstrated after this way, more with another after that way; and we are so much the more likely to suit the different genii of different Persons, by how many more and different methods and ways we shew them, leading to the same end, of which every one may take that which he likes best.

XVIII.

We therefore Publish, by the Divine assistance, these our En­deavours also, after so many other ingenious and elaborate ones in the same kind; nor can we doubt the approbation of some of our Readers. This at least we can experimentally affirm: That not a few of those to whom these our thoughts were partly pub­lickly read in Lectures, and partly privately taught (for they were only design'd for Learners) were not a little taken with the Concise brevity and facility of the Demonstrations; so that we may reasonably hope to be acceptable to those, to whom ei­ther time, or sufficient force of genius is wanting, to run over the vast Volumes of the Ancient Mathematicians, and compre­hend their prolix Demonstrations, and long series's of far fetch'd Consequences; and as for those who have both leisure and geni­us to do so, this may serve for an Encouragement towards it; that after they have gone through the chief truths and propo­sitions [Page] they contain, Demonstrated in a more easie and shorter way, they may so much the more confidently adventure upon those celebrated and ingenious Treatises, from the reading whereof they were before deterr'd by the length and almost insuperable diffi­culty of their tedious and perplext Demonstrations.

XIX.

Being now about to enter upon the matter it self, we will on­ly further hint these few things. 1. Since our whole design is for the advantage of young Students (which ought to be a Professors chief Care and Study) we must not omit the Expli­cation of the most simple terms; especially since we design to deduce some Corollaries immediately from them, which hereto­fore have unnecessarily increas'd the number of Propositions and Demonstrations. 2. To encumber our work as little as we can with words we have made use, especially in our Analytick Cal­culus, of some Symbols, as = for Equality, as also of □ and ▭ for Square and Rectangle, and of √ the common Radicall sign for the square Root, with the Line √¯ on the top for connecting of quantities together, the Root where­of is jointly taken, √ 3 for the Cube Root, √√ for the Biquadratick Root.

XX.

That the Reader may at one view see the Contents of the following Treatise, we have thought fit to present him here with it, by way of Synopsis. It is divided in two Books.

I.

The first whereof contains the chief and most select Proposi­tions of Euc [...]id's Elements, of Archimedes's Treatises of the Sphere and Cylinder, as also of the dimension of the Circle, &c. Wherein that which these [...]uthors have demonstrated by a long and tedious Series of Consequences, and for the most part in­directly, we have here endeavoured to Demonstrate directly, and so that the Demonstration of each Proposition depends, ei­ther [Page] on no other, and so is evident by its own light, or on a very few of the antecedent ones.

II.

After the same way in the second Book we treat of the Co­nick Sections, and Demonstrate the chief properties of the Co­noid, Spheroid, Cycloid, Conchoid and Spiral Lines, which are extant either in Apollonius or Archimedes and others, and what they have Demonstrated by long and tedious process's we have here exhibited in a short and easie Compendium. And that,

III.

In such a method, as does not so much require intent and severe thinking, as a bare and easie inspection, and application of the Principles of specious Algebra, and method of Indivisi­bles. which yet,

IV.

We don't barely suppose, and remit our Readers to other Books to learn (which would be too troublesome) but in the Process of the Work it self, they are gradually, and as occasion presents, de­rived frrm their Original Fountain, and first Principles.

V

By the same way also the most useful and necessary Mathema­tical Praxes are laid down under the names of Corallaries and Scholia, the Construction of the Tables of Sines and Tangents taught, the Original and use of the Logarithms Demonstrated, and the Precepts both of Plain and Spherical Trigonometry deduced from their first Principles, &c.

VI.

The Praxis also of pure Arithmetick, and of common Deci­mal, and (which is seldom used) of Tetractical, as also the [Page] Doctrine of Surds, are derived from their first Original Whereunto,

VII.

As a Complement of the whole Work, we have added an Intro­duction to the Specious Analysis, or new Geometry of the Moderns, particularly according to the Method of Des Cartes, but much facilitated by later Inventions, and Comprizing the Precepts of the Art in six or seven Pages, but illustrated with above forty Examples in the different degrees of Equations.

What the Reader's opinion will be of these our Endeavours, design'd only for the use of young Students, time must teach us. The Author himself at least, amongst his other performances, al­lows these the first place.

ADVERTISEMENT.
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Writing, Arithmetick, Book-keeping, Algebra, Geometry, Measuring, Surveying, Gauging, Astronomy, Geography, Navigation and Dialing.

By Robt. Arnold.

Persons Taught abroad.

Elementa Arithmeticae Numerosae & Speciosae. In usum Juventutis Oxoniae è Theatro Sheldoniano. Prostant Venales Londini apud Dan Midwinter & Tho. Leigh ad Insigne Rosae Coronatae in Caemeterio Divi Pauli.

…

Mathesis Enucleata: OR, The Elements of the Mathematicks. Book I.

Explaining the First Principles of the Mathematicks; among which are (in the first place) Definitions, and some Consectaries that flow from them.

CHAP. I. Containing the Definitions or Explications of the Terms which relate to the Object of Mathematicks.

DEFINITION I.

MAthematicks is the Science or Knowledge of Quan­tity, and of Beings, as far as they are subject to it, or measurable; and may justly claim the Name of Ʋniversal, while it is employ'd in Demonstra­ting those Properties which are common to all or most Quan­tities: But when it descends to the different Species of Quantity, and is busied in contemplating the Affections belonging par­ticularly to this or that Quantity, it is distinguished by va­rious Names, and distributed into various Parts, according to the various diversities of the Objects.

DEFINITION II.

QUantity may be defined in General, whatever is capable of any sort of Estimate or Mensuration, as immediately the Habitudes and Qualities of Things, as e. g. the multitude of Stars in the Heaven, or of Souldiers in an Army, the length of a Rope, or Way, the weight of a Stone, the swiftness or slowness of Motion, the Price of Commodities, &c. but medi­ately, the very things themselves wherein those Estimable Qua­lities are inherent. Whence with the ingenious Weigelius we may not incongruously reduce them all to these four Kinds or Genders, viz. 1. to Quanta Naturalia, Natural Quantities, or such as Nature has furnish'd us with, as Matter with its Exten­sion and Parts, the Powers and Forces of Natural Bodies, as Gravity, Motion, Place, Light, Opacity, Perspicuity, Heat, Cold, &c. 2. to Moral Quanta or Quantities, depending for the most part on the Manners of Men, and arbitrarious De­terminations of the Will; as for Example, the Values and Price of Things, the Dignity and Power of Persons, the Good or Evil of Actions, Merits and Demerits, Rewards and Punish­ments, &c. 3. to Quanta Notionalia, arising from the Notions and Operations of the Understanding, as e. g. the amplitude or narrowness of our Conceptions, universality or particularity, &c. in Logick; the length or brevity of Syllables, Accent, Tone, &c. in Grammar: And lastly, to Quanta Transcendentia, Tran­scendent Quantities, such as are obvious in Moral, Notional, and Natural Beings; as Duration, i. e. the Continuation of the Existence of any Being; which in Physicks especially is named Time, and may be conceived as a Line, &c. To these you may moreover add Ʋnity, Multitude or Number, Necessity and Contingency.

DEFINITION III.

NƲmber (whereon we shall make some special Remarks) if it be taken in the Concrete, is nothing else than an Aggregate or Multitude of any sort of Beings; taken abstract­edly, it is, as Euclid calls it, [...], a multitude, or (as they call it) Quotity of Ʋnities, on the one hand Number, i. e. many are opposed to one; and in that sense Unity is not a Num­ber: [Page 3] On the other hand Unity may be esteem'd a Number, since it is no less (if I may be allow'd that term) some Quo­tity than two or three. But as we denote or signifie particular Things, when we speak of them Universally, by the Letters of the Alphabet, A, B, C, (a, b, c,) &c. as universal Signs or Symbols of them; so for distinctly and compendiously Expres­sing the innumerable Variety of Numbers, Men have found out various Notes, the most natural whereof, are Points disposed in particular extended Orders, as . . . to denote Three,

[9 dots arranged in 3 rows and 3 columns]

to denote Nine, &c. But that way which is most commodious for Practise, is by the common Notation, or Cyphers, 1, 2, 3, 4, 5, 6, 7, 8, 9. the invention whereof, as we have it by Vulgar Tradition, is owing to the Arabians. By a very few of these we express any number tho never so great, by a won­derful, tho now adays familiar, Artifice; the first Inventor of them having Establish'd this as an arbitrary Law, that the first of them shall signifie unity or one; the 2d two, &c. as often as they stand alone; but placed in a row with others, or on the left hand of one or more 0, or noughts, (which of themselves stand for nothing, but fill up empty places) if in the second, be­fore a nought, they denote Tens; if in the 3d. Hundreds; in the 4th Thousands; in the 5th Myriads or Tens of Thousands; in the 7th so many Thousands of Thousands, or Millions; in the 8th Tens of Millions, &c. and so onwards, increasing al­ways in decuple Proportion, by Tens, Hundreds, Thousands, &c.

COROLLARY I.

HEnce you have a way of expressing or writing any Sum by these Notes, which you may hear expressed in Words; as if we were to express in Notes the year of our Lord, One Thousand Six Hundred Ninety and Nine, it is manifest, that according to the method above described, by placing 9 on the right hand in the first place, and nine again in the second to­wards the left, six in the third, and one or unity in the fourth, the business will be done. Thus it will be easie to any one with a little attention, to express any Number whatsoever by these Notes; (as suppose that which Swenterus proposes, in Delic-Physico-Math. [Page 4] Part. 1. Probl. 75.) Eleven Thousand, Eleven Hundred, and Eleven.

COROLLARY II.

HEnce you have also the Foundation and Reason of the Rule of Numeration in Arithmetick, or expressing any Numbers or Sum in Words, which you see written in Cyphers; which for greater ease may be done thus, viz. beginning from the first Figure towards the right hand, over every fourth Figure note a Point, (including always that which was last pointed) and at every second Punctation or Point, draw short strokes thus, one over the 2d, two over the fourth, &c. the first denoting Millions, the second Millions of Millions, or Bimillions, the third Trimillions, &c. and the Intercepted Points the Thousands, in their kinds, &c.

SCHOLION.

HEre I cannot omit, on this occasion, what the foremen­ti ned Weigelius has hinted about another way of Nume­ration, and which Dr. Wallis mentions, Oper Mathemat. Part 1. p. 25. & 66. shewing there a way (and illustrating it by Ex­amples) of Numeration, and of Expressing the Figures; which proceeds thus; whereas now adays in numbring we ascend from uni y or 1 to ten (the reason whereof; after which Aristotle makes a prolix Inquiry, Probl. 3. Sect. 15. was taken without doubt from the denary Number of our Fingers) if from unity we proceed only to four, (which Aristotle in the same place tells us some of the Thracians used to do of old,) and thence retur­ning back again to Unity, we should proceed again after the same way; we might after that way obtain a vastly more simple and easie Arithmetick, than we have now adays. Which, even hence we may conclude; because for Multiplication and Divi­sion there would need no other Table (or Pythagorick Abacus) than this easie and short one:

  • 1.1.1 once one is one;
  • 2.2.10 i. e. twice two are four;
  • 2.3.12 twice three are four and two.
  • 3.3.21 thrice three are twice four and one
Pag. 5.

Fig. 1.

2

3

4

5

6

7

8

9

10

11

12

13

And altho it is pity that we can't hope now a-days to substi­stute this vastly easier way of Computing, in room of the other now in use, because the other is universally receiv'd, and most sorts of Measures and other Quantities are fitted and accommo­dated to the decuple Proportion; yet it ought not to be alto­gether neglected in Mathematicks, which might receive very great advantage hereby, especially in Trigonometry, if the in­genious Invention of the Logarithms had not already supply'd its use therein. The whole Foundation of this Tetractys or Quaternary Arithmetick, is placed only in these three Notes, 1, 2, 3; [...]o that any one of them alone, or in the first place, should denote Units, in the second place, Tetrads, or so many Fours (or Quaternions,) in the third place so many Sixteens, in the fourth place so many times Four Sixteens, or 64.s. &c. al­ways proceeding in a Quadruple Proportion. For which way of Numeration there might be found out terms as commodious as those we now use, and which are thereby grown Familiar to us, as one, ten, twenty, a hundred, a thousand, &c. which will be evident by what follows:

One, Ʋnum 1
One.
Ten, Decem 10
Quatuor, Tetras, a Quaternion or Four.
Twenty, Viginti 20
a Biquaternion
Thirty, Triginta 30
a Triquaternion
Hundred, Centum 100
a Tetraquaternion
Thousand, Mille 1000
a Quartan
Ten Thousand, 10000 &c.
a Tetraquartan, &c.
DEFINITION IV.

A Magnitude is whatever is conceived to be Extended or Continuous, or has parts one without another, and con­tained within some common Term or Terms: wherein that is called a Point which is conceived (as indivisible, or) to have no Parts, and so no Magnitude, but is notwithstanding the be­ginning or first Principle of all Magnitude.

DEFINITION V.

IF we conceive a Point (A) ( Fig. 1.) to be moved towards B, by this motion it will leave a trace, or describe the Magni­tude [Page 6] AB of one only Dimension, that is Length without Lati­tude, or which at least we are to conceive so, and is called a Line: If that Line AB be conceiv'd again so to be moved, as that its extreme Points AB shall describe other Lines BC and AD, it will describe by that Motion the Magnitude AB CD, or (to denote it more compendiously by the Diagonal Letters) AC or BD, having both length and breadth, but without any depth or thickness, or at least so to be conceived, and this is called Superficies or Surface: Lastly, if this Surface AC be con­ceived so to move, e. g. upwards or downwards, that its opposite Points A and C again describe other Lines AF and CH, and consequently each of its Lines other Surfaces, &c. by this Mo­tion there will be formed a Magnitude of three Dimensions, which we call a Solid or Body, which we will also denote by the two diametrically opposite Letters AH and DG. But as this Motion of the Point, Line, or Surface, may be various, so there will be produced by them various sorts of Lines, Sur­faces, and Solids: But these Productions stop here, and proceed no further; for the Motion of a Body can only produce ano­ther Body greater than the first, but no more new Dimensions.

CONSECTARYS.

I. POints therefore being moved thro' equal Intervals in the same or a like way or trace ( e. g. in a streight or the shortest trace) describe equal Lines; and

II. The same or equal Lines moved thro' the same Right-lined or Curvilinear Paths, describe equal Surfaces; and

III. Equal Surfaces moved according to the same Methods and Conditions describe equal Solids: which, if rightly under­stood, are the first certain and infallible Foundations of the Method of Indivisibles. But here you must take care to distin­guish between the way which the Line it self describes, and that which its Ends or extreme Points describe: For altho e. g. the Point a (Fig. 24.) moves along in a more oblique way than A, and so describes a longer Line a c; yet the Line a b describes by a parallel Motion, an equal space with the Line A B, (viz.) the same which the whole Line A b, whereof they are parts, would describe. See Faber's Synopsis, p. m. 13.

DEFINITION VI.

BUT that we may a little further prosecute this Genesis of Magnitudes (as very much conducing to understand their Nature and Properties) if the Point A moves to B the shortest way, it describes the Right Line AB; but if in any (one) other it will describe the Curve or Compounded Line ACB: From whence, with F. Morgues, we may infer these

CONSECTARYS.

I. THat two Right Lines Eucl. Ax. 14. beginning from the same Point A, and ending in the same Point B, will necessarily coincide, nor can they comprehend or inclose Space; for if they did, one must deviate, and so would cease to be a Right Line.

II. In a Space comprehended by three Right Lines AB, BC, CA, Eucl. Ax. 14. any two taken together, must needs be greater than any one alone. Moreover we may add this before hand;

III. In a Circle a Right Line drawn from A to B ( Fig. 3) will fall within the Circle, because the Curve Line ADB de­scribed, as we shall hereafter shew, being longer than a Right Line, must necessarily fall beyond it, or on the outside of it. And lastly,

IV. A Tangent, or Line (b) which does not cut or enter into the Circle, touches it only in one Point.

Moreover if a Right Line AB ( Fig. 4.) move on another Right Line BC, remaining in the same Position to it, it will generate a Plan Surface, to which a Right Line being any way applied, will touch it with all its point, as Faber rightly describes it; if a Right Line be moved on a Curve, or a Curve on a Right Line, &c. they will generate a Curve Surface, call'd Gib­bous, or Convex without, and Concave within.

DEFINITION VII.

IF a Right Line be fixed at one of its ends A, and the other be moved round ( Fig. 5.) it will describe in this Motion a [Page 8] Circular Plane, or a Circle; and by the motion of its end or extreme Point B, Eucl. l. 1. Prop. 20. the Periphery or Circumference of that Circle BEF. Eucl. Prop. 2. lib. 3. The fixed Point A is called the Center of that Circle; the Lines AB, AC, &c. its Radii or Semi-Diame­ters; all of which are equal one to another. Any Right Line BC drawn from one part of the Circumference thro' the Center to another, is called the Diameter, and divides the Circle into two Semicircles BECB and BFCB. The Circumference of a Circle, whether great or small, is divided into 360 equal parts called Degrees, and each Degree into 60 Minutes, &c. From this Geniture of the Circle presupposed, there evidently fol­low these

CONSECTARYS.

I. THat 2 Circles which cut one another cannot have the same common Center; for if they had, the Radii ED and EA drawn from the common Center E ( Fig. 6.) would be equal to the common Radius EB that is the part to the whole.

II. Nor can two Circles touching one another within side, have one and the same Center, for the same reason.

III. Of Lines falling from any given Point without the ( Fig. 7.) Circle, & 6 of the same Book. and Eucl. 8. l. 3. passing thro' the Pe­riphery to the opposite Concave part of it, that which passes through the Center of it, is the longest, viz. AB; and of the other that which is nearest to it is longer than that which is more remote: But on the contrary of those which fall on the Convex Periphery, that which tends towards the Center, as Ab, is the least, and the rest gradually greater, and there can be but two, as AE and AF, or Ae and Af, equal: All which will appear very evi­dent by drawing other Circles from the Center A thro' B, D, E, and b, d, e. Or thus; having drawn two other Circles, from the Radii AB and Ab, if we conceive the Radii Ab and Cb to move towards the right hand, their ends will always recede further from one another; the same is also evident of the Ra­dii AB and CB, moved also to the right together.

IV. Moreover ( Fig. 8.) of all the Lines drawn within the Circle Eucl. 5 l. 3 the Diameter is the greatest, and the rest gradually less, by how much the more remote they are from the Center, &c. [Page 9] Which will be very evident to any one who contemplates a Circle inscribed in a Square, as also the Genesis of Curvity it self; as also many other ways which I shall now omit; or to mention one more thus; because the two Radii CA and CB being mo­ved, in order to meet together, necessarily approach nearer to one another in their extreme Points.

DEFINITION VIII.

THE Aperture or opening of two Lines ( Fig. 9.) AB, AD, &c. that are both fixed at one end at A, and the other ends opened or removed farther and farther from one another, is called an Angle, Prop. 15. lib. 3. and usually de­noted by 3 Letters, D, A, B, (whereof that which denotes the Angular Point, always stands in the middle,) and measured by the Arch of a Circle BD, or a certain num­ber of Degrees which it intercepts. The greatest Aperture of all BAC is when the 2 Legs of the Angle AB and AC make one Right Line, and is measured by a Semicircle, or 180 De­grees. The mean or middle Aperture BAE or CAE, when one Leg EA is erected on the other AB or AC at Right An­gles, so that it inclines neither one way nor the other, (thence called a Perpendicular) is named a Right Angle, whose measure is consequently a Quadrant (or quarter part) of a Circle or 90 Gr. Wherefore a Semicircle is the measure of two Right An­gles: An Aperture or Angle BAD less than a Right Angle (and so measured by less than 90 degrees) is called an Acute Angle; and that which is greater than a Right Angle, as DAC (and so consisting of more than 90 degrees) is called an Obtuse An­gle. Whence me may now draw these

COROLLARYS.

I. TWO or more Contiguous Angles Eucl. 13. lib. I. with the Coroll. con­stituted on the same Right Line BC, and at the same Point A (as DAB, and DAC or DAB, DAE and EAC) make two Right An­gles, as filling the Semicircle; and consequently,

II. All the Angles that can be constituted about the Point A (as filling the whole Circle) are equal to 4 Right ones: As [Page 10] also on the other side Euclid. l. 1. Prop. 14. if two Right Lines AB and AD meet on the same point A of another Right Line AC, and make the Contiguous Angles equal to 2 Right ones, that is, if they fill a Semicircle, BC will necessarily be the Diameter of a Circle, and consequently a Right line.

III. If one of the Contiguous Angles BAE be a Right one, the other CAE will be so also.

IV. If two Right Lines AB, CD, cut one another in E, the 4 Angles they make will be equal to 4 Right ones.

V. And as it is evident at first sight ( Fig. 10.) that any Circle having one half (or Semicircle) folded on the other, at the Diameter ECD, the two Semicircles EHD, and EID, must needs agree, or every where coincide one with the other; so if the Angle ACD be supposed equal to the Angle BCD, that is, the Arch AD to the Arch BD, having one Leg CK or CL common; the others AC and BC being supposed before equal,

1. The Bases BL and AL, KB and KA, will be also equal; for these will coincide too, and therefore the Angles also.

2. The Line AB being bisected in K, the two Angles Eucl. l. 1. 4 & 3. lib. III. 3. at K will also coincide and be equal, and con­sequently Right Angles: and contrarywise,

3. The Angles at the Base of equal lib. 1. 5. Legs, CAB, CBA, and also those below the Legs, the Legs being produced to F and G, are equal.

4. Consequently the Spaces ACL and BCL, ACK and BCK are equal to one another.

5. The Contiguous Angles AED and BED insisting on equal Arches AD and BD are equal, and è contra; as also those that are not Contiguous, if their Vertex's are equidistant from E, &c.

6. It is hence also manifest, that a Perpendicular erected on the middle of any Line AB, inscribed in any Circle, passes through its Center, by what we have just now said; and if you likewise erect Perpendiculars on the middle of any 2 Lines, ab and bm (Fig. 11.) connecting any 2 Arches, or any 3 Points, a, b, m, that are not placed all in the same Right Line, those 2 Per­pendiculars ke, no, will determine (by their Intersection) the Center of a Circle that shall pass through these 3 Points.

DEFINITION IX.

IF one Right Line DE cut or pass thro' another AB ( Fig. 12) the opposite Angles at the top or intersection ACD and ECB are called Vertical; as also the other two ACE and DCB: Whence follow these

COROLLARYS.

I. THat the Vertical Angles are always Eucl. lib. 1. 15. Equal; for both ACD and ECB with the third, ACE, which is common to both, fill or are equal to a Semicircle; as likewise both ACE and DCB with the third ECB, which is common.

II. Contrarywise, if at The same Prop. Schol. 1. the Point C of the Right Line DE, the 2 opposite Lines AC and CB make the Vertical An­gles x and z equal, then will AC and CB make one Right Line; for, since x and o make a Semicircle, and z and x are equal, by Hypoth. o and z will also make or fill a Semicircle, whose Diameter will be ACB.

III. By the same Argument it will appear, that of 4 Lines Schol. 2. proceeding from the same Point so as to make the opposite Vertical Angles equal, the 2 oppo­site ones AC and CB, as also DC and CE, will make each but one Right Line; for since all the 4 Angles together make a whole Circle, or 4 Right Angles, and the sum of x and o is equal (by Hypoth.) to the sum of o and z, it follows, that both the one and the other will make Se­micircles, whose Diameter will be AB and DE, and so Right Lines.

DEFINITION X.

IN any Circle, a Right Line, as D.G, that subtends any Arch of it DGB, is called the Chord of that Arch ( Fig. 13.) BF (a part cut off from the Semidiameter BC passing thro' the middle of the Chord) is called the Sagitta or Intercepted Ax, but most commonly the Versed Sine; and DF let fall from the other ex­tremity of the given Arch BD, on the Semidiameter at Right Angles, is called the Right Sine of that Arch BD, or of the Angle [Page 12] BCD; also DI is called the Right Sine of the Complement (or for brevity sake, Sine Compl.) of that Arch DH, or Angle DCH, &c. but the greatest of all the Right Sines HC let fall from the other extremity (or end) of the Quadrant (which is indeed the same as the Semidiameter of the Circle) is called the whole Sine or Radius; lastly, BE is called the Tangent of the Arch BD or Angle BCD, and CE its Secant: Whence Mathe­maticians, for the sake of Trigonometrical Calculations, have divided the whole Sine or Radius of the Circle into 1000, 10000, 100000, 1000 000, 10000 0000, &c. parts, thence to make a proportionable Estimate of the number of Parts in the Sine, Tangent, or Secant of any Arch, &c. as may be seen in the Tables of Sines, Tangents, and Secants. From these Suppositions and Explications of the Terms, we shall now in­fer from this Definition the following

COROLLARYS.

I. IN equal Circles (and so much more Among the rest Vid. Eucl. 28 & 29 lib. 3, and also part­ly the 26 and 27. in one and the same) as the Radii or Semidia­meters BC and bc are equal, so also it is evident, that the Right Sines DF & Df, of equal Arches BD and bd, or equal Angles BCD and bcd, also the Tangents BE and be, and Secants CE and ce, and Subtenses or Chords DG and dg, also the Sagittae or intercepted Axes BF and bf, of double the Arches DBG and dbg, &c. will be equal, and so consist of an equal number of Parts of the whole Sine or Radius, &c. which both is evident from what we have said before, and may be further evinced, if one Circle be con­ceived to be put on the other, and the Radius BC on the Ra­dius bc, that so they may coincide, by reason of the equality of the Arches BD and bd; and so of all the rest. And è contra,

II. In unequal Circles, the Sines, Tangents, &c. of equal Angles BCD or bcd (Fig. 14.) or similar Arches, or Arches of an equal number of Degrees, BD and bd, will be also simi­lar or like, &c. i. e. the Sine df consists of as many parts of its Radius bC, as the Sine DF does of its Radius BC, &c. e. g. if the Radius BC be double of the Radius bc, each thousandth part of the one, will be double of each thou­sandth [Page]

Pag. 12.

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[Page] [Page 13] part of the other, but they are alike 1000 in each; be­cause the degrees in the Circumference of the little one, parti­cularly in the Arch bd are but half as big as those in the Arch BD, and yet equal in number in both. Thus also if the Sine DF contains 700 of the 1000 parts of its Radius BC, df will also contain 700 of the 1000 parts of its smaller Radius bc, and in like manner the Chords DG and dg, and the Tan­gents BE and be, &c. contain a like number of parts, each of its own Radius.

SCHOLION.

IT may not be amiss here to note by the by (altho it may seem more proper to be taught after the Doctrin of Pro­portions) that if, v. g. the degrees of a greater Circle be each of them respectively double, or triple, or quadruple, &c. of the degrees of a less Circle, according as the Radius of the one is double or triple to the Radius of the other, then, at least as far as Mechanical Practice can require, you may find the Arch of a greater Circle equal to the whole Periphery of a less, viz. if you take reciprocally that part of the greater Periphery, which shall be as the Radius of the less to the Radius of the greater, or as one degree of the less Periphery to one degree of the greater. e. g. if the less Radius bc be half the greater BC, and so also the Periphery, and each of the degrees of the one, be one half of the Periphery, and of each of the degrees of the other, one half of the greater Periphery will reciprocally be equal to the whole less Periphery, or 180 degrees of the one to 360 of the other, &c.

2. The same (at least in this case where the Radius cb is double of the Radius CB) may be done also Geometrically by the same reason. Having described Circles on each Radius, suppose the Radius CB ( Fig. 15.) so to move with an equable motion about its Center c, as to take or move the Radius of the greater Circle cb along with it, and coming, v. g. to I. stops that also at 1, and going forward, to II. stops that again at 2, &c. Hence it will be manifest to any attentive Reader, that when the less Radius CB shall have described the Semicircum­ference B. II. III. the greater Radius cb having moved to 3, will have described precisely a quarter of its circumference; [Page 14] and if still the less Radius C. IV. moves on to the right hand, and continues to carry the greater c 4 along with it onwards the same way, it will necessarily follow, that in the same mo­ment as the Radius C. IV. (together with c 4.) shall come to its first situation in B, having described a whole Circle, the op­posite Radius c 4 will be come to 5, and have described half its Circle, having moved all along with an equal Motion. Hence it is evident, that the whole least Circle answers exact­ly to half the greater, and half of the first to a quarter of the last; as also the Quadrant B. II. to the Octant (or 8th. part) b 2, &c. whence any Arch being given, as B. I. in the least Circle, if you draw thro' I the Radius of the greater Circle c 1. you'l cut off an Arch b 1. equal to the given Arch in mag­nitude, but only half in the number of degrees.

3. Hence follows naturally that celebrated Proposition of Euclid, that the Angle at the Center BC. I. or BC. II. is double of the corresponding Angle at the Periphery bc. 1. or bc. 2, &c. which in this case is manifest, and in the other 2 ( Fig. 16.) of the wholes or remainders DCD and DPD it is also Eucl. p. 20. l. 3. certain; which is true also of the parts BCD and BPD to be added or subtracted by the first Case.

Eucl. p. 9. l. 1.4. Hence we have a new way of bisecting any given Angle CDE, or Arch CE ( Fig. 17.) viz. if you make C B equal to the Leg DC, and from this, as Radius, describe an Arch BF equal to the Arch CE, and draw DF.

And with the same facility we might obtain the Tri­section, if the greater Radius being triple to the less, was thus carried along by an equable Motion, as we have shewn how to do already in a double Radius; and this at first sight may seem very probable.

But whether the triple Radius be immediately carried round by the simple Radius C B, or by means of the double Radius cb, neither the one nor the other will cause an equable Motion. For in the latter Case, while the Radius cb describes the quadrant Bb, the Radius de will not describe so much as a Quadrant; but while cb with the same velocity describes the other Quadrant bf, the Radius de will come to g, de­scribing an Arch as much greater than a Quadrant as the for­mer [Page 15] was less. In the former case on the contrary, the Radius C B moved on to D beyond a Quadrant, while de was carried from B to e, but if the Radius CD moving on, should again carry de along with it, the one would describe the same Arch eB, while the other would describe one less than before.

5. Hence the Angle at the Center ACE ( F g. 19.) upon the Arch AE, is equal to the Angle ADB at the Periphery, upon double that Arch AB.

6. Hence the Angle ADB in the Semicircle (Num. 1.) Eucl. 31. lib. 2. is a Right Angle, in a Segment less than a Semicircle (Num. 2.) is an obtuse Angle, and in a greater (Num. 3) an Acute one, because the Angle at the Center ACE upon the half Arch, is equal to the Angle ADB pr. praeced. 5. and is a Right Angle in the first Case, Obtuse in the second, and Acute in the third.

7. Hence Angles in the same Segment, or Eucl. 26 and 27. lib. 3. on equal Segments of equal Circles, or on the same or equal Arches, are all equal and è contra.

DEFINITION XI.

WHen 2 or more Lines AB and CD are so continued as to keep always the same distance from one another (whose Genesis may be conceived to proceed from the uniform Motion of 2 Points A and C, always keeping the same distance from each other) they are called Parallels: But as it evidently fol­lows from this Definition, that Eucl. lib. 1. 30. those Lines which are Pa­rallel to one third, are parallel to one another (since adding or subtracting equal Intervals to or from other equal ones, the sums or remainders must needs be equal;) so if the Parallels are Right Lines and cut transversly (or slopingly a-cross) by another Right Line EF, you'l have these

COROLLARYS.

I. THE Angles Eucl. lib. 1. 29. which we call Alternate ones, GHK and HGI ( Fig. 21.) are equal by Corollary I. Definition X. since the distances GK and HI, which are the Right Sines of the said Angles, are supposed equal.

[Page 16]II. The External Angle EGA is also equal to the Internal opposite Angle GHK, by Consect. 1. Definit. 9. because that External Angle EGA is equal to the alternate Vertical one HGI.

III. The same Internal Angle GHK, with the other internal opposite one on the same side AGH (as well as the External one EGA, by Coroll. 1. Definit. 8.) are equal to two Right ones.

IV. On the contrary, If any Right Line EF lib. 1. Prop. 27 & 28 cutting 2 others AB and CD transversly, makes the alternate Angles GHK and HGI equal, their Right Sines, by Consect. 1. Definit. 10. will be equal, and consequently the Lines AB and CD parallel: and the same will follow, if the External Angle be supposed equal to the Internal, or the 2 Internal ones on the same side equal to 2 Right ones; since from either Hypothesis the former will immediately follow.

V. From whence it appears more than one way (b) That the 3 Internal Angles of any Triangle (e. g. H, G, K, which will serve for all) taken together, lib. 1. Prop. 32, are equal to two Right ones, and the External one GHD is equal to the two Internal opposite ones. For we might either conclude with Euclid, that 1, 2, 3, together make 2 Right ones, by Consect. 1. Definit. 8. but 2=II and 3=III pr. 1 and 2 of this, therefore I, II, III=2 Right ones; or with others, 1, II, 4 are = 2 R. but 1=I and 4=III pr. 1st of this. There­fore, &c. or more briefly with F. Pardies, 1=I pr. 1st of this, but 1, II, III, together = to 2 Right ones, by the 3d of this; therefore I, II, III=2 R. Q. E D.

DEFINITION XII.

IF a Right Line AB ( Fig. 22.) be conceived to move from the top of a plain Angle CAD with a motion always paral­lel to its self, so that at one end A it shall always touch the Leg AC, and all along cut the Leg AD, while at length being come to F, it shall only touch that Leg with its other end B, and so fall at length wholly within the Angle CAD: It will describe by this motion within the Legs CAD the Triangular Figure EAF, and without them the Triangular Figure BAF; its parts within them a f continually increasing, and the other without [Page 17] fb continually decreasing; but with all its Parts, or the whole Line, it will describe the Quadrangular Figure AEFB: Conse­quently if the other Leg AD of the given Angle CAD ( Fig. 23.) or any part of it AB, be moved along the other Leg remain­ing parallel to it self, it will also describe a Quadrilateral Fi­gure, which will be also equilateral, if the Line describing it AB, be equal to the Line AE according to which it is directed; but if either of the Lines, as AD be greater than the other, the opposite Sides will be only equal; for the describent or describ­ing Line is always necessarily equal to its self, and the Points A, B, D, moved with an equal Motion, describe also in the same time equal Lines AE, BF, DG. From these Geneses of Quadrangles and Triangles we have the following

CONSECTARIES.

I. THese Quadrilateral Figures are also Parallelograms, i. e. they have their opposite Sides Parallel; Schol. Prop. 34. lib. I. because the Line that describes them is supposed to re­main always parallel to its self, and the Points A and D, or A and B, to be always equidistant.

II. Because the 2 Internal opposite Angles The first part of the same Proposit. A and E, and also E and F, &c. are equal to 2 Right ones, by Consect. 3. Definit. 11. if one Angle v. g. that at A be a Right one, all the others must neces­sarily be so too [in which case the quadrilateral and equilateral Figure AF is called a Square, and the other AG an Oblong, or Rectangle:] if there be no Right Angle, the opposite Angles transversly or cross-ways, a and f, or a and g are equal, because both the one and the other, with the third (e) make 2 Right ones [in which case the quadrilateral Equi­lateral af is called a Rhombus, but the other ag a Rhomboid.

III. The Transversal (or Diagonal) Line the latter part of the same Prop. in any Parallelogram, divides it into two equal Triangles AEF and FAB; for all the Lines and Angles on each side are equal, and as the descri­bent (Line) AB moved thro' the Angle EAF upon the Line AE described the Triangle AEF; so the Line EF, equal to the former, moved after the same way, thro' the Angle AFB also equal to the former Angle, upon the equal Line FB, must [Page 18] necessarily describe an equal Triangle; or, in short, all the In­divisibles af, or their whole increasing Series, are necessarily equal to the like number of Indivisibles fb, increasing recipro­cally after the same way.

IV. All Parallelograms that are between the same Parallels AB and CF ( Fig. 24.) i. e. having the same Altitude lib. 1. Prop. 35 & 36 and the same or equal Bases, as CD or CD and cd, are equal among themselves; for they may be conceived to be described by the equal Lines AB and ab equally moved thro' the same or equal Intervals of the Parallel Lines; so that all or each of the Indivisibles or Elements AB will necessarily be equal to all and each of the Indivisibles ab; for they all along answer one to the other both in num­ber and magnitude.

SCHOLIUM.

HEre you have a Specimen of the Method of Indivisibles, introduced first by Bonaventura Cavallerius, and since much facilitated; and altho these Indivisibles placed one by another, or as it were laid upon an heap, cannot compose any Mag­nitude, yet by an imaginary Motion they may measure it, and as it were, after a negative way, demonstrate the Equality of two Magnitudes compared together, viz. if we conceive a cer­tain number of such Elements in any given Magnitude, and thence conclude that in another consisting of the like Elements, ordered or ranked after the same way, there can be neither more nor less in number than in the first; thence follows their Equality, &c.

V. Hence therefore it is also Evident, that Triangles upon the same and equal Bases as CD and cd, and placed between the same Parallels, are necessarily equal, because they are the half of equal Parallelograms AD and ad, by the 3d Consectary of this Definition.

VI. F. Mourgues ingeniously concludes from hence, viz. be­cause the 2 Internal opposite Angles lib. 1. Prop. 37 & 38, & 41. lib. I. Prop. 32. otherwise. on the same side in any Parallelogram, are equal to two Right ones, and so all together equal to four; that therefore the three Angles of any Triangle ABC ( Fig. 25. which may always be compleated into a [Page]

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[Page] [Page 19] Parallelogram) are equal to half of those four, or to two Right ones. This may be yet more briefly conceived thus; the Angles b+ c+ d (the sum of the two inferiour ones) = to two Right ones; but a = alternate d: therefore b+ c+ a = to two Right ones, Q. E. D.

VII. Because it is manifest in Rectangular Parallelograms, if the Altitude AB, and Base BC ( Fig. 26.) measured and divided by the same common Measure, be conceived to be (multiplied or) drawn one cross the other, that the This is ma­nifest from its Genesis; for the five parts of the Line AB by its motion along the part of the Base BE, de­scribes 5 little Squares, and moving along the following part, describes five more, &c. Area AC, thereby described, will be divided into as many little square Measures or Area's, as the number of their Sides multiplied together would produce Units; therefore the A­rea of any other Parallelogram will be after the like manner produced, if the Base be multiplied by the Perpendicular heigth, equally as if it were a Rectangle of the same Base and Alti­tude.

VIII. Consequently also you may have the Area of any Triangle, by Consectary 3 and 5. if the Base be multiplied by half the Perpendicular heigth; or, the whole Base being multiplied by the heighth, if you take the half of the product.

DEFINITION XIII.

BUT as there are various Species of Triangles, while first with relation to their Sides, one is called Equilateral, as ABC ( Fig. 27.) because all its Sides are equal; another Equi­crural or Isosceles, as DEF, because it has two equal Sides DE and EF, while its Base DF may be either longer or shorter; and a third is called Scalenum, as GHI, because it has all its Sides un­equal; then again in respect to their Angles, one is called Rectangled, as a, b, c, because it has one Right Angle at a;

Another Obtusangled, as d, e, f, because it has one obtuse An­gle at d; a third is called Acuteangled, as g, h, i, because all its three Angles are Acute: So each of these kinds has its pecu­liar properties, which we shall partly hereafter demonstrate in their proper places, and partly deduce here as

CONSECTARYS.

I. ALL Equilateral Triangles are also Equiangular, and con­sequently Acutangled; for, having found a Center for three Points, and the Periphery A, B, C, (see Fig. 28.) by Consect. 6. Definit. 8. the three Arches AB, BC, and AC, answering to equal Chords; and consequently the three Angles at the Center O are equal, by Consect. 1. of Definit. 10. and therefore the three Angles at the Periphery also, as being half of the other, by the 3d Consectary of the same Definition. Each Angle therefore is one third part of two Right ones, by Consect. 6. Definit. 12. two thirds of one Right Angle, i. e. 60 degrees, and consequently Acute.

II. It follows also by the same Reason in an Isosceles Triangle, that the Angles at the Base opposed to equal Sides are equal, and lib. I. Prop. 5. the same otherwise demonstrated. consequently Acute; for having circum­scribed a Circle about it, equal Arches will cor­respond to the equal Chords DE and EF, and equal Angles at the Center DOE and FOE will correspond to them, and equal ones at the Peri­phery DFE, and FDE to these again. And it is evident that each of these are less than a Right Angle h. e. an Acute one, because all three are equal to two Right ones. Wherefore if the third is a Right Angle, the other two at the Base will ne­cessarily be half Right ones.

SCHOLIUM.

WE will here (a) shew by way of Anticipation, the truth of the Pythagorick Theorem, esteemed worth an He­catomb: Which hereafter we will demonstrate after other dif­ferent ways; viz. In a Right Angled Triangle BAC (Fig. 29.) the Square of the greatest Side opposite to the Right Angle, is equal to the Squares of the other two Sides taken together. For having de­scribed the Squares of the other two Sides, AC dE, DE ab (taking ED=AB) and the Square of the greatest BC cb, it will be evident, that the parts X and Z are common to each, and that the two other Triangles in the greatest Square BAC and BDb, are equal to the two Triangles bac and Cdc which [Page 21] remain in the two Squares of the lesser Sides; and so the whole truth of the Proposition will be evident, while these two things are undoubtedly true: 1. That the Side of the greatest Square Bb will necessarily concur with the Extremity of the less Db, and the other Side of the greatest Square Cc with its Extremity c, will precisely touch the Continuation of the Sides of the two least Squares dEa; as you'l see them both expressed in the Fi­gure. 2. The said two Triangles are every way equal; for the Angles at C with the intermediate one at Z, make two Right ones, therefore they are equal; but the Side CA is equal to the Side cd, and CB to Cc, and the Angles at A and d Right ones. Wherefore if we conceive the Triangle ABC to to be turned about C, as a Center to the right hand, it will exactly agree with the Triangle Cdc, and the Point B will ne­cessarily fall on the continued Line d E, as agreeing with the Line AB. Hence it is now evident, that Ca= BD, and be­cause ba is also = bD, and the Angles at a and D Right ones. Where, if we conceive the Triangle bac to be moved about b as a Center, untill ba coincides with bD, and ac with DB, bc will also necessarily coincide with bB Q. E. D.

To this Demonstration of Van Schooten's, which we have thus illustrated and abbreviated, we will add another of our own, more like Euclids, but somewhat easier, which is this: Having drawn the Lines (as the other Figure 29 directs) the ▵ ACD being on the same Base AC with the Square AI, and between the same Parallels, is necessarily one half of it, but it is also half of the Parallelogram CF being on the same Base with it, viz. DC; therefore this Parallelogram = ▭ AI. In like manner ▵ ABE is half the ▭ AL, and also half the Parallelogram BF, therefore BF=▭ AL: therefore CF + BF that is the ▭ of BD = to the two ▭ ▭ AI + AL. Q. E. D. For because the Side BE occurs to, or meets the Side LK, and the Side CD the Side IH continued, it yet more apparently follows; because the An­gles a and b, and also c and d, are manifestly equal, as making both ways, with the Intermediate x or z, Right Angles. There­fore the ▵ BAC being turned on the Center B and laid on BLE will exactly agree with it, and turned on the Center C and laid on CID, will agree with that also, &c.

DEFINITION XIV.

ALL Rectilinear or Right Lined Figures that have more than three or four Sides (to the latter sort of which, there remains to be added another Species besides Pa­rallelograms, call'd Trapeziums, whose Angles and Sides are un­equal, as K, L, M, N, Fig. 30.) are called by one common Name Poligons, or Many-sided and Many-Angled Figures, and par­ticularly according to the Number of their Sides and Angles, Pen­tagons, Hexagons, Heptagons, &c. All whereof, as also Trapezia, being resolvible into Triangles by Diagonal Lines, (as may be seen in the 31 and foregoing Fig.) you have these

CONSECTARYS.

I. YOU have the Area of any Polygon by resolving it into Triangles, and then adding the Area's of each Tri­angle found by Consect. of Definition 12 into one Sum.

II. The Area of the Trapezium KLMN (in the first of the Fig. 30) whose two opposite Sides, at least KL and MN, are Parallel, may be had more compendiously, if the Sum of the Sides be multiplied by half the common heigth KO.

SCHOLIUM.

HEnce we have the foundation of Epipedometry or Masuring of Figures that stand on the same Base, and Ichnography; in the Practise whereof this deserves to be taken special Notice of, that to work so much the more Compendiously, you ought to divide your Figure into Triangles, so that ( Fig. 31.) 2 of their Perpendiculars may (as conveniently can be) fall on one and the same Base. For thus you'l have but one Base to measure, and 2 Perpendiculars to find the Area of both: But for Ichnography, the distance of the Perpendiculars from the nearest end of the Base must be taken; which we shall supersede in this Place and Discourse more largely on hereafter.

2. This resolution of a Polygon into Triangles may be perform'd by assuming a point any where about the middle, and making the sides of the Polygon the Bases of so many [Page 23] Triangles; (see the 2d Figure mark'd 31) wherein it is evi­dent; 1 That all the Angles of any Polygon are equal to twice so many right ones, excepting 4, as the Polygon has sides; for it will be resolv'd into as many Triangles as it has sides, and each of these has its Angles equal to 2 right ones. Subtract­ing therefore all the Angles about the Point M (which always make 4 right ones by Cons. 2. Def. 8.) there remain the rest which make the Angles of the Polygon. 2 All the external Angles of any right lined Figure (e, e, e, &c.) are always equal to 4 right ones; for any one of them with its Contiguous internal Angle is equal to 2 right ones pr. Consect. 1 of the said Def. and so altogether equal to twice so many right ones as there are Sides or internal Angles of the Figure. But all the inter­nal Ones make also twice so many right Ones, excepting 4 therefore the external Ones make those 4.

DEFINITIOF XV.

AMong all these plain Figures those are call'd Regular whose Angles and Sides are all equal, as among trilateral Figures the Equilateral Triangle, among Quadrilateral ones, the Square, and in other kinds, several Species which are not particulariz'd by Names; but all others in whose Angles or Sides there is any inequality, are call'd Irregular: Tho' some of these also, and all the other may be inscrib'd in a Circle. Whence you have these

CONSECTARYS.

I. THE Areas of the Regular Figures may be obtained yet easier, if having found their Center (by Consect. 6. De­finit. 8.) you draw from thence the Right Lines CB, CA, &c. (Fig. 32.) till there be form'd as many Triangles ACB, ACF, &c. as the Figure has Sides; for since all these Triangles have their Bases AB, BF, as so many Chords, and their Altitudes CD, CG, as so many parts of intercepted Axes DE and GH, and also equal pr. Consect. 1. Definit. 10. and so by Consect. 5. Definit. 10. are equal among themselves; one of their Area's being found and multiplied by the number of Sides, or half the Altitude by the Sum of all the Sides, you'l have the Area of the [Page 24] whole Polygon: For it is manifest from what we have already said, and very elegantly Demonstrated by F. Pardies, That any Regular Polygon inscribed in or circumscribed to a Circle, is equal to the Triangle Aza, one Legg whereof is equal to the Perpendicular heighth let fall from the Center upon any Side, and the other to the whole Pe­riphery of the Polygon. Now if the Triangles into which the Poly­gon is resolved, do all stand on the same Right Line Aa, ( Fig. 32) and are all equal and of the same heighth, to which the Per­pendicular AZ is equal, it will necessarily follow, that each pair of Triangles ABZ and ABC, BZF and BCF, &c. are equal among themselves, pr. Consect. 5. Definit. 12. and consequently the Sum of all the former will be equal to the Sum of all the latter, that is, the Triangle Aza to the Polygon given.

II. Since Regular Figures inscrib'd in a Circle, by bi­secting their Arches AB, BF, &c. may be easily conceived to be changed into others of double the number of Sides, (as a Pentagon into a Decagon, &c.) and that ad Infinitum; a Circle may be justly esteemed a Polygon of infinite Sides, or consisting of an infinite Number of equal Triangles, whose common Al­titude is the Semidiameter of the Circle: So that the Area of any Circle is equal to a Right Angled Triangle (as AZa) one of whose Sides AZ is equal to its Archimedes of the Dimen­sion of the Circle, Prop. 1. Semidia­meter, and the other Aa to its whole Circum­ference.

SCHOLIUM.

IT may not be amiss to note these few things here, concer­ning the Inscription of Regular Figures in a Circle.

I. Having described a Circle on any Semidiameter AC, Euclid. Prop. 15. lib. 4. ( Fig. 33. N. 1.) that Semidiameter being placed in the Circumference, will precisely cut off one sixth part of it, and so become the Side of a Re­gular Hexagon: and so the Triangle ABC will be an Equilateral one, and consequently the An­gle ACB and the Arch AB 60 Degrees, by Cons. 1. Definit. 13.

II. Hence a Right Line AD, omitting one point of the di­vision B, and drawn Eucl. 2d Coroll. of the same. to the next D, gives you the Side of a Regular Triangle inscrib'd in the Circle, and subtends twice 60, i. e. 120 Degrees.

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[Page 25]III. If the Diameters of the Circle AD and DE ( N. 2.) cut one another at Right Angles in the Center C, the Right Lines AB, BD, &c. will be the Sides of an inscribed Square ABDE: For Eucl. 6. lib. 4. the Sides AB, BD, &c. are the equal Chords of Quadrants, or Quadrantal Arches, and the Angles ABD, BDE, &c. will be all Right ones, as being Angles in a Semicircle ( per Schol. 6. Definit. 10.) composed each of two half Right ones, by Consect. 2. Definit. 13.

IV. Euclid very ingeniously shews us how to Inscribe a Re­gular Pentagon also, Lib. 4. Prop. 10 & 11. and also a Quin­decagon (or Polygon of 15 Sides) Prop. 16. But though the first is too far fetch'd to be shewn here, yet (supposing that) the second will easily and briefly follow:

In a given Circle from the same point A ( N. 3.) inscribe a Regular Pentagon AEFGHA, and also a Regular Triangle ABC; then will BF be the Side of the Quindecagon, or 15 Sided Figure. For the two Arches AE and EF make together 144 Degrees, and AB 120: (a) Therefore the difference BF will be 24, which is the 15th. part of the Circumference.

V. The Invention of Renaldinus would be very happy, if it could be rightly Demonstrated; (as he supposes it to be in his Book of the Circle) which gives an Universal Rule of dividing the Periphery of the Circle into any number of equal Parts re­quired, in his 2d Book De Resol. & Comp. Mathem. p. 367. which in short is this: Upon the Diameter of a given Circle AB Fig. 34.) make an Equilateral Triangle ABD, and having divided the Diameter AB into as many equal Parts, as you design there shall be Sides of the Polygon to be Inscribed, and omitting two, e. g. from B to A, draw thro' the beginning of the third from D, a Right Line, to the opposite Concave Circumference, and thence another Right Line to the end of the Diameter B, which the two parts you omitted shall touch thus, e. g. for the Triangle, having divided AB into three equal parts, if omitting the two B2, thro' this beginning of the 3d you draw the Right Line DIII, and thence the Right Line III.B, which will be the Side of the Triangle; and so IV.B will be the Side of the Square, VB the Side of the Pentagon, &c.

N. B. The Demonstration of these ( Renaldinus adds, p. 368.) we have several ways prosecuted in our Treatise of the Circle: Some of the most noted Antient Geometricians, have spent a great deal of pains in the Investigation and Effection of this Problem, and several of the Moderns have lost both time and pains therein: Whence, we hope, without the imputation of Vain Glory, we may have somewhat obliged Posterity in this point.

DEFINITION VI.

IF the Plane of any Parallelogram AC ( Fig. 25.) be concei­ved to move along a Right Line AE, or another Plane AF downwards, remaining always Parallel to its self; there will be generated after this way a Solid having six opposite Planes Pa­rallel, two whereof, at least, will be equal to one another, whence it is called a Parallelepiped; and particularly a Cube or Hexaedrum, if the Parallelogram ABCD that describes it be a Square, and the Line along which it is moved, AE, equal to the Side of that Square, and Perpendicular to the describing Plane, and consequently all the six Parallel Planes comprehend­ing this Solid, equal to one another. But if the describing or Plane Describent ( Fig. 36.) be a Triangle or Polygon, the So­lid is call'd a Prism, if a Circle, it is called a Cylinder. Now from the Genesis of these Solids you have the following

CONSECTARYS.

I. IF the Planes or Parallelograms Describent Eucl. l. 11. p. 29, 30, 31 ABCD and abcd (Fig. 37.) are equal, and their Lines of Motion AE and ae also equal; the Solids thereby de­scribed, viz. Parallelepipeds, Cylinders, and Prisms, (which will therefore have their Bases and heigths equal) will be equal among themselves; be­cause the describent Indivisibles of the one, will exactly answer, both in number and position, to those of the other, as we have already shewn in Parallelograms; Consequently therefore,

II. Any Parallelepiped Eucl. l. 11. p. 28. may be divided by a Diagonal Plane BDHF (or a Plane passing thro' its Diagonals) into two equal Prisms; for by [Page 27] Consect. 3. Definit. 12. the Triangles ABD and BCD, are equal, and are supposed to be moved by an equal Motion thro' equal spaces.

III. And since it is evident, even by this Genesis of them, that in Right Angled Cubes and Parallelepipeds, if the Base ABCD ( Fig. 38.) being divided into little square Area's, be multiplied by the heigth AE, divided by a like measure for length, after this way you may conceive as many equal little Cubes to be generated in the whole Solid, as is the number of the little Area's of the Base multiplied by the number of Divi­sions of the side AE; you may moreover obtain the Solidity of any other Parallelepipeds, that are not Right Angled ones, by multiplying their Bases and Perpendicular Heigths together.

IV. Moreover since every Triangular Prism is the half of a Paral­lelepiped, and any Multangular Prism may be resolved into as ma­ny Triangular ones, as its Base contains Triangles; you may obtain the Solidity (or Solid Contents) either of the one or the other, if you multiply the Triangular, or Multangular Base of them into their Perpendicular Heigth.

V. After the same manner you may likewise have the Soli­dity of a Cylinder, which may be considered as an Infinite Angled Prism, just as the Circle is as an Infinit-Angled Po­lygon.

DEFINITION XVII.

IF any Triangle ABC ( Fig. 39. N. 1.) be conceived to move with one of its Plane Angles C, from the Vertex or top of a Solid Angle (determined by two Planes aAb and cAa joined together in the common Line Aa) with a motion always parallel to it self; so that its extreme Angular Point A shall al­ways remain in the Line Aa, but with its Sides AB and AC shall all along raze on the two Angular Planes, till at length it falls wholly within the Solid Angle: by this its motion it will describe within the Solid Angle, the Figure we call Pyramidal, whose Base will be the Triangle abc, and its Vertex A will also describe without it another Quadrangular Pyramid, whose common Vertex will be the same A, but the Base the Quadrangle Cb, described by the Side of the moveable Triangle BC: The first Py­ramid it will describe with its Triangular Parts, [...] continually in­creasing [Page 28] from the point A, and ending in the Triangle abc; but the latter Pyramid will be described by the remaining parts, continually decreasing downwards from the whole ABC, and the Quadrangular Trapeiza [...] at length ending in the Right Line bc: So that in the mean while the said Triangle with its whole space describes, according to what we have said before, the Triangular Prism composed of those two Pyramids. From this Geeesis of Pyramids you'l have the following

CONSECTARYS.

I. OF what sort soever the Describing Triangles are ABC, ABC ( N. 2.) fo they are equal; and whatever the Solid Angles are, comprehended under the Planes abA, and acA, abA and acA, so they are accommodated to the Plane Angles A and A, and such that, &c.

DEFINITION XVIII.

THere may be exhibited another easier Genesis of Cones and Pyramids, but it respects only the Dimension of the Surface, and not of the Solidity of them, viz. If you have a fix'd point A that is not in the Angular Plane BCDEF ( Fig. 41.) and a Rght Line AF let fall from that point to any Angle of the Plane, be con­ceived to move round the sides BC, CD, &c. This Plane by its motion will describe as many Triangles ABC, CAD, &c. as the Angular Plane has Sides. And these Triangles all meet­ing at the point A, make that Solid which we call a Pyramid. Now if instead of the Angular Plane there be supposed a Cir­cular one, (or an Angular one of Infinite Sides) the Solid thence produc'd is called a Cone, whose Surface is equal to Infinite Tri­angles, constituted on the Base BCDE, and whose Solidity would consequently equal an Infinite Angled Pyramid of the same heigth. And after the same manner by the motion of the Line AF, remaining always Parallel to it self about Parallelo­grams or Triangular Planes, will be generated Parallelipipeds, Prisms, and Cylinders. But as one Pyramid will be produced more upright than another, according as the point A stands more over the middle of the Plane BCD, &c. (Fig. 42.) or [Page]

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Schemes for rai-sing the five Regular Bodys

[Page] [Page 29] respects it more obliquely: So in particular a Cone is called a right Cone, when the line AO being let fall to the Center of the circular Plane (otherwise call'd the Axis of the Cone) constitutes on all sides right Angles with it; but it is call'd an Oblique or Sca­lene Cone, when the Ax stands obliquely on the Base. Which distinction may also be easily understood when apply'd to Cylin­ders, tho' a right Cone and Cylinder may also be conceiv'd to be generated after another way, as if for the one a Triangle and for the other a right angled Parallelogram AOB and LAOB be be conceiv'd to be moved round a line AO considered as immoveable (whence it is call'd an Ax;) and also a truncated Cone may be formed if a right angled Trapezium, Archimedes lib. I de Con. and Cylin. prop. 7. and 8. 2 of whose Sides are parallel be moved, &c. And as we have deduc'd the So­lidity of these Bodies from the foregoing Genesis, so their external Surfaces, as also of Prisms and Parallelepipeds may easily be found from the present Genesis, by any one who attentively considers the following

COROLLARYS.

I. SInce the whole external Surface, except the Base, of any Pyramid is nothing but a System of as many Triangles ABC, CAD &c. as the Pyramid has Sides; if the Area's of those Triangles separately found by Consect. 8. Def. 12. be added into one Sum, you'l have the Superficial Area of the whole Pyramid.

II. If a Pyramid be cut with a Plane b, c, d, e, parallel to its Base BCDE ( Fig. 41.) The Surface of that truncated Py­ramid comprehended between the prallel Planes may be obtain'd if having found the Surface of the Pyramid A bcde cut off from the rest by Consect. 1. you subtract it from the Surface of the whole Pyramid.

II. The external Surface of a right Pyramid that stands on a regular Polygon Base is equal to a Triangle, whose Altitude is equal to the Altitude of one of the Triangles which compose it, and its Base to the whole Circumference of the Base of the Pyramid.

IV. Therefore the Surface of a right Cone, by what we have already said, is equal to a Triangle whose heighth is the [Page 30] side of the Cone, and the Base equal to the Circumference of the Base of the Cone.

V. The Surface of a truncated right Cone, or Pyramid is equal to a Trapezium which has 2 parallel Sides the lowest of which is equal to the Perifery of the Base, and the other to the Perifery of the Top or upper Part, and the heigth, to the inter­cepted Part.

VI. The Surface of a right Cylinder or Prism is equal to a Parallelogram which has the same height with them, and for its Base a right line equal to the Perifery of that Cylinder or Prism.

DEFINITION XIX.

IF a semicircular Plane ADB ( Fig. 43. N. 1.) be conceived to move round its Diameter AB which is fixt, as an Axis, by this Motion it will describe a Sphere, and with its Semicircum­ference the Surface of that Sphere; every part whereof is equally distant from the middle Point of that Axis C (which is there­fore call'd the Center of that Sphere.) Now if ( N. 2.) this semicircular Ambitus Archimedes lib. I. de Cono & Cyl. Prop. 22. Coroll. & Prop. 23. be conceived to be divi­ded before that revolution, first into 2 Quadrant [...] AD, and BD, and then each of those again into as many parts equal in Number and Magnitude as you please, and having drawn the Chords AF, FE, ED, &c. let the Polygon AFEDGHB In­scribed in the Semicircle be conceived together with it to be turn'd about the Axis AB; then will A 1 F, and B 4 F de­scribe 2 Cones about the Diameters F f, and H h; and the Tra­pezia about the Axes 1, 2, 2 C, C 3, and 34, will describe so many truncated Cones, and the lines AF, FE, &c. so many Conical Superficies, by the Antecedent Def. and so the whole Po­lygonal Plane AFEDGHB a Conical Body inscribed in the Sphere, and contain'd under only Conical Surfaces. And as any attentive Person may easily perceive such a Body to be less than the ambient Sphere, and its whole Surface less than the Surface o [...] the ambient Sphere; so he may as easily trace these following

CONSECTAYS.

I. IF the Arches AF, FE, &c. be further bisected, and a Polygonal Figure of double the number of Sides inscri­bed in the Semicircle, and conceived to be moved round after the way we have already shewn, the Pseudoconical Body hence arising, will approach nearer and nearer to the Solidity of the Sphere, and the Surface of the one to the Surface of the other, and hence (if we continue this Bisection, or conceive it to be continued ad Infinitum) you may infer.

II. That a Sphere may be look'd upon as much a Pseudoco­nical Body, consisting of infinite Sides, and it's Surface will be equal to the infinite Conical Surfaces of that Body; which we will take further notice of below.

DEFINITION XX.

IF the Diameter AB of the Semicircular Plane ADB ( Fig. 43. N. 3.) be conceived to be divided into equal Parts (as here the Semidiameter AC into 3.) and if the circumscribing Pa­rallelograms CE, 2 E, 1 G on the transverse Parallels CD, 2 e, 1 f be conceived together with the Semicircle it self to revolve about the fixed Ax AB; it is evident that there will be formed from the Semicircle a Sphere as before, and from the Circum­scribed Parallelograms, so many circumscribed Cylinders of e­qual heighth: but if all the Altitudes or Heighths of these are bisected or divided into two, and so make the number of circum­scribing Parallelograms double, there will be formed (by mo­ving them round as before) double the Number of Cylinders of half the heighth, but which yet being taken together, approach much nearer the solidity and roundness of the Sphere, than the for­mer, which were fewer in Number ( viz. the six latter Parallelo­grams approach nearer to the Plane of the Circle than the three former) and thus if that bisection of the Altitudes be conceived to be continued ad infinitum, the innumerable Number of those in­finitely little Cylinders will coincide with the Sphere it self. More­over if you conceive any Polyedrous or Multilateral Figure to be circumscribed about the Sphere (which we here endeavour to de­lineate by the Polygon ABCD N.4. circumscribed about the Cir­cle) [Page 32] and the solid Angles thereof to be cut by other Planes ab, which shall touch the Sphere; it is manifest there will thence a­rise another Polyedrous Figure, the Solidity whereof will approach nearer to the Solidity of the Sphere, and its Surface to the Sphe­rical Surface than the former, and if the Angles of this be a­gain in like manner cut off, there will still arise another new So­lid, and new Surface approaching yet nearer to the Solidity and Surface of the Sphere than the former, &c. and so after an infi­nite Process they will coincide with the Sphere and its Surface themselves. Whence flow these

COROLLARYS.

I. THE Sphere may be considered as a Polyedrous Figure, or as consisting of innumerable Bases, i. e. composed of an innumerable Number of Pyramids, all whose Vertex's meet in the Center, and so whose common heigth is the Semidia­meter of the Sphere, and the sum of all the Bases equal to the Superficies of the Sphere.

II. If you can find a Proportion between a Cylinder of the same heigth with any Sphere, and whose Base is equal to the greatest Circle of that Sphere, and innumerable Cylinders cir­cumscribed about it, as we have just now shewn; then you may also obtain the Proportion between the said circumscrib'd Cylinder and the inscrib'd Sphere: Which to have here hinted may be of service hereafter in its proper place.

DEFINITION XXI.

THere remain those Bodies to be consider'd which are call'd Regular, which correspond to the Regular Plane Figures; and as those consist of equal Lines and Angles, so these likewise are comprehended under Regular and Equal Planes meeting in equal solid Angles; and as those may be Inscribed and Circum­scribed about a Circle, so may the latter likewise in and about a Sphere. But whereas there are infinite Species of Regular Plane Figures, there are only five of Regular Solids; the first whereof is contained under four Equal and Equilateral Trian­gles, whence it is nam'd a Tetraedrum; the second is terminated by six equal Squares, and thence is call'd Hexaedrum, and other­wise [Page 33] a Cube; the third being comprehended under eight Equal and Equilateral Triangles, is call'd an Octaëdrum; the fourth is contained under twelve Regular and Equal Pentagons, and so is nam'd a Dodecaëdrum; the fifth, lastly, is contained under twenty Regular and Equal Triangles, and is thence nominated an Icosa­ëdrum. Besides these five sorts of Regular Bodies there can be no other; for from the concourse of three Equilateral Triangles arises the Solid Angle of a Tetraëdrum, from four the Solid An­gle of an Octaëdrum, from five the Solid Angle of an Icosaëdrum; from the concourse of four Squares you have the Solid Angle of an Hexaëdrum; from that of three Pentagons you have the Solid Angle of a Dodecaëdrum; and in all this Collection of Plane Angles, the Sum does not arise so high as to four Right ones. But four Squares, or three Hexagons meeting in one Point, make precisely four Right Angles, and so by Consect. 2. Definit 8. would constitute a Plane Surface, and not a Solid Angle. Much less therefore could three Heptagons or Octagons, or four Pentagons meet in a Solid Angle, to form a new Regular Body; for those added together would be greater than four Right An­gles. But now, for the Measures of these five Regular Bodies, take the three following

CONSECTARYS.

I. SInce a Tetraëdrum is nothing else but a Triangular Py­ramid, and an Octaëdrum a double Quadrangular one, their Dimension is the same as of the Pyramids in Schol. of De­finit. 17.

II. The Solidity of an Hexaëdrum may be had from Consect. 3. Definit. 13.

III. A Dodecaëdrum consists of twelve Quinquangular Py­ramids, and an Icosaëdrum of twenty Triangular ones, all the Vertex's or tops whereof meet in the Center of a Sphere that is conceived to circumscribe the respective Solids, and consequent­ly they have their Altitudes and Bases equal: Wherefore ha­ving found the Solidity of one of those Pyramids, and multi­plied it by the number of Bases (in the one Solid 12, in the o­ther 20) you have the Solidity of the whole respective Solids.

DEFINITION XXII.

BEsides these Definitions of the Regular Bodies, we may al­so form like Idea's of them from their Genesis, which particularly Honoratus Fabri has given us a short and ingenious System of, in his Synopsis Geometrica, p. 149. and the follow­ing.

I. Suppose an Equilateral Triangle ABD to be inscrib'd in a Circle ( Fig. 44. N. 1.) whose Center is C, whence having con­ceived the Radii CA, CB, CD, to be drawn, imagine them to be lifted up together with the common Center C, so that the point C ascending Perpendicularly, at length you'l have the Line [...] EA, EB, ED, equal to the Lines AB, BD, DA, After this way there will be generated, or made a Space consisting of fou [...] Equal and Equilateral Triangles, which is call'd a Tetraëdrum Hence we shall by and by easily demonstrate, the quantity o [...] the Elevation CE, and the Proportion of the Diameter of th [...] Sphere EF to be Circumscribed to the remaining part CF and so the reason of the Euclidean Genesis proposed lib. 13 Prop. 13.

II. Much like this, but somewat easier to be conceived, is th [...] Genesis of the Octaëdrum, where by a mental raising of the Cen­ter C ( Fig. 44. N. 2.) of the Square ABDE inscribed in th [...] Circle, together with the Semidiameters CA, CB, CD, CE until being more and more extended they at length become th [...] Lines AF, BF, DF, EF, all equal among themselves, and [...] the side of the Square AB or BD; and its manifest, that by th [...] like extension conceived to be made downwards to G, the [...] will be formed eight equal and regular Triangles, which w [...] all concur in the two opposite Points F and G. We migh [...] also deduce another Genesis of the Octaëdrum from a certai [...] Section of a Sphere, and also give the like of a Hexaëdrum o [...] Cube: but we have already given the easiest, of the one, vi [...] that which is also common to Parallelepipeds; and that of th [...] other just now given is sufficient to our purpose.

CHAP. II. Containing the Explication of those terms, which relate to the affections of the Objects of the Mathematicks.

DEFINITION XXIII.

EVery Magnitude is said to be either Finite if it has any bounds or terms of its Quantity; or Infinite if it has none, or at least Indefinite if those bounds are not determined, or at least not considered as so; as Euclid often supposes an Infinite Line, or ra­ther perhaps, an Indefinite one, i. e. considered without any re­lation to its bounds or Ends: By a like distinction, and in reality the same with the former, all quantity is either Measurable, or such that some Measure or other repeated some number of Times, either exactly measures and so equals it, (which Euclid and other Geometricians emphatically or particularly call Measur­ing) or else is greater; or on the other side Immense, whose Amplitude or Extension no Finite Measure whatsoever, or how many times soever repeated, can ever equal: In the first Case, on the one Hand, the Measure ( viz. which exactly measures any quantity) is called by Euclid an aliquot Part lib. 5. Def. 1. or simply a Part of the thing measured: as e. g. the Length of one Foot is an aliquot Part of a Length or Line of 10 Foot. In the latter Case the Mea­ [...]ure (which does not exactly measure any Quan­tity) is called an Aliquant Part, as a line of 3 or 4 Foot is an Aliquant part of a Line of 10 Foot. Now therefore, omitting [...]hat perplext Question, whether or not there may be an infinite Magnitude, we shall here, respecting what is to our purpose, deduce the following

CONSECTARY.

EVery Measure, or part strictly so taken, is to the thing Measured, or its whole, as Unity to a whole number, for that (which is one) repeated a certain number of times, is sup­posed exactly to measure the other.

DEFINITION XXIV.

IF the same Measure measures 2 different quantities (whether the one can exactly Measure the other or not) those Quantities are said to be absolutely Commensurable; but if they can have no common Measure; they are called Incommensurable▪ Notwithstanding which they both retain one to the other a certain relation of Quantity, which is call'd Reason or Proportion, a [...] we shall further shew hereafter. In the mean while we hav [...] hence, as an infallible Rule to try whether Quantities can admit o [...] a Common Measure or not, this

CONSECTARY.

THose Quantities are Commensurable, whereof Euclid lib. 10. Prop. 5, 6, 7, 8. one [...] to the other, either as Unity to an whole Number, or a [...] one whole Number to another, for either one o [...] them is the Measure of the other, as also of i [...] self, and then it is to that other as Unity to [...] Number by the Consect. of the preced. or els [...] they admit of some third Quantity for a common Measure which will be to either of them separately as Unity to some Number: therefore they are one to another as Number to Num­ber.

DEFINITION XXV.

IF 2 Quantities of the same Kind, considered as Measures on [...] of the other, being applyed one to the other, exactly agree or are exactly equal every way, (as e. g. 2 Squares on the sam [...] common Side, or two Triangles whose Lines Angles and Space exactly agree and conicide) or at least may be equally mea­sured by a common Measure applyed to both) as e. g. a Square and an Oblong, or a Rhombus, or Triangle, each of whose Area's were 20 square Inches, altho' they do not agree in Line [...] and Angles; the first may be called Simply Equal, and the othe [...] totally equal, or equal as to their wholes: But if one be greater and the other less, they are Ʋnequal, and that which exceeds i [...] called the greater, and that which is deficient the less, and tha [...] [Page 37] part by which the less is exceeded by the greater, in respect to the greater is call'd Excess, in respect to the less Defect, and by a common Name they are call'd the Difference. All which as they are plain and easy, so they afford us a great many self-evident Truths, which are used to be call'd Axioms, as these and the like

CONSECTARYS.

I. THe whole is greater than its Part, whether it be an Ali­quot or aliquant Part.

II. Those Quantities which are equal to a third are equal be­twixt themselves.

III. That which is greater or less than one of the equal Quantities is also greater or less than the other.

IV. Those Quantities which, being applyed one to the o­ther, or placed one upon the other, either really or mentally, a­gree; may be esteemed as totally equal: And on the Contrary,

V. Those Quantities which are totally equal will agree, &c. To which might be added several others which we have already made use of and supposed as such in the preceding Definitions.

DEFINITION XXVI.

THere are moreover Addition, Subtraction, Multiplication and Division, which are common affections of all Quantities as well as of Numbers. Addition is the Collection of several Quantities (for the most part of one kind) into one total or Sum; which is either done so, that the whole (which is com­monly called the Sum or Aggregate) obtains a new Name, or else by a bare connexion of the Quantities to be added by the Copu­lative and, or the usual Sign + ( i. e. plus or more) as for Ex­ample 2 Numbers . . . and . . . . (suppose 3 and 4) added together make the Sum . . . . . . . ( i. e. 7, or which is the same thing 3+4;) and this Line— added to this other—gives the Sum— which is nothing but the 2 Lines joyn'd, or taken together. But now if we would treat of these Lines, or any other 2 Quantities to be added, more ge­nerally; by calling the first a (a) and the latter (b) we may fitly write their Sum a+ b.

SCHOLIUM.

HAving thus explained the Term of Addition, these and the like Axioms emerge of themselves: If to equal Quantities you add Equal the Sum will be Equal; but if to Equal you add unequ [...] the Aggregate will be unequal, &c. Moreover it may not be amiss to admonish the Tyro of these 2 things. 1. In Addition may be see [...] the vast usefulness of that very Ingenious tho' familiar Invention mentioned in Definit. 3. for hereby we may collect into one Su [...] not only Tens, and Hundreds, but Thousands, Millions, My riads, as tho' they were only Units; which we will Illustrate by an Example.

DEFINITION XXVII.

SUbtraction is the taking one Quantity from another (of th [...] same kind;) which is so performed that either the remainde obtains a new Name, or by a bare separation of the Subtrahen [...] by the privative Particle less, or the usual Sign − which stand for it, as e. g. . . . or three being subtracted from . . . . . . ▪ or 7, the remainder or difference is . . . . or 4 and this Lin [...] — Subtracted from that — leaves — Now if we would signify this more generally either of the [...] Lines, or the Number above, or any 2 Quantities whatsoeve [...] that are to be Subtracted one from the other, by naming th [...] first (a) and the latter (b) we shall have the remainder a — [...] Herein are evident these and the like Axioms: If from equ [...] Quantities you Subtract Equal ones, the Remainders or Differences [...] be equal. Here it will be worth while to take notice of, from this and the preced. Definit. the following

CONSECTARYS.

I. IF a negative Quantity be added to it self considered a positive (as − 3 to + 3 or − a to + a) the Sum wi [...] be [...] for to add a Privation or Negative is the same thing a [...] to Subtract a Positive, wherefore to join a Negative and Pos [...] ­ [...]ve together, is to make the one to destroy the other.

[Page 39]II. If a negative be subtracted from its positive (− a from + a) the remainder will be double of that positive (+2 a) for to subtract or take away a privation or negative, is to add that very thing, the privation of which you take away; for really that which in words is called the addition of a Privation, is in reality a Subtraction, and a subtraction of it, is really an ad­dition; and what is here call'd a Remainder, is indeed a Sum or Aggregate; and what is there call'd a Sum, is truly a Remain­der. Thus,

III. If the positive Quantity (+ a) be taken from the pri­vative one (− a) the remainder is double the privative one (−2 a) since, taking away a positive one, there necessarily arises a new Privation which will double that you had before. Hence,

IV. You have the Original of the Vulgar Rules in Literal Addition and Subtraction: If the Signs of the unequal Quantities are different, in the room of Addition you must subtract, and in room of Subtraction add, and to the sum or remainder, prefix the Sign in the first place of the greatest, in the next of that from which you Subtract: but if the Signs are both the same, and the greatest quan­tity to be subtracted from the less, you must, on the contrary, subtract according to the natural Way, the least from the greater, and prefix the contrary Sign to the remainder: Which Rules you may see Illustrated in the following Examples:

Addition Subtraction.  
4 b−2 a from 2 a+ b from 3 a+2 b
3 b+5 a Subst. a− b Subst. 2 a+3 b
7 b+3 a R. a+2 b R. a− b
NOTE.

☞ Instead of the Authors 4th Consect. as far as it relates to Subtraction, which may seem a little perplext, take this ge­neral Rule for Subtraction in Species, viz. Change all the Signs of the lower Line, or Subtrahend, and then add the Quantities, and you have the true Remainder.

SCHOLIUM.

IN this Literal Subtraction, we have not that conveniency which the invention of Vulgar Notes supplies us with, that from the next foregoing Note we may borrow Ʋnity, which in the following Series goes for 10, &c. This is done in Te­tractycal Subtraction only with this difference, that an Unite here borrowed goes only for 4. That the easiness of this O­peration may appear, we will add one Example, wherein from this number, — you are to subtract this,

1232002310232
321012321223
310323323003

Whereever therefore the inferiour Note is greater than the superiour one, the facility is much greater here than in com­mon Subtraction, because never a greater number than 3 is to be subtracted out of a greater, than 4 and 2: but if the in­feriour number be greater than the superiour, you borrow unity from the left hand, which is equivalent to 4; the rest is perform'd as in common Subtraction.

DEFINITION XXVIII.

MƲltiplication, generally Speaking, is nothing else but a Complex or manifold Addition of the same quantity, wherein that which is produced is peculiarly call'd the Product, and those quantities by which it is produced, are called the Multiplicand and the Multiplier: The first denotes the Quantity which is to be multiplied, or added so many times to its self; and the other the Number by which it is to be multiplied, or determins how many times it is to be added to it self. The same terms are applyed moreover to Lines and other Quan­tities. But here are two things to be chiefly noted; 1. That the Multiplication of one number by another, or of a Line by a Line, may be considered as having a double Event; for the Product may be either of the same or a different kind, as, e. g. when . . . . 4 is multiplied by 3 . . . the product may be considered either as a Line, thus, . . . . . . . . . . . . or as [Page 41] a Plane Surface in this Form,

[12 dots arranged in three rows and four columns]

Whence it is also named a Plane Number, and the product is conceived to be formed by the motion of an erect Line AB, consisting of 3 equal parts, along another BC, consisting of 4 equal parts, and conceived as lying along. So also the Multiplication of Lines ( e. g. of the Line A — B by the Line B—C) may be conceived to be so performed, that the Product also shall be a Line, e. g. C—D (concerning the usefulness of which Multiplication in Geometry, we shall have occasion to speak more hereafter;) or so, that the Pro­duct shall be a Plane or Surface, arising from the motion of the erect Line AB, along AC, conceiv'd as lying along; as we have already shewn. But as for the most part these Planes so produced are called Rectangles, if the Lines that form them are unequal; but if they are equal they are call'd Squares, (otherwise the Powers of the given Quantities;) and in this case the Lines that form them are called Square Roots; so also if those Planes are multiplied again into a third Quantity (as either a Line or a Number) there will arise Solids, and parti­cularly if that third Quantity be the Root of the Square, the Product is called a Cube, &c. The other thing to be noted is, That both these ways of Multiplying either Numbers or Lines, are expressed by a very compendious, tho arbitrary way, of Notation, viz. by a bare Juxtaposition of the Letters which denote such and such Species of Quantities, as, e. g. if for the forementioned Number or Line AB we put a, and for BC b, the Product will be ab; or if the Efficients are equal, as a and a the Square thence produced, will be aa or a (powerof2); and if this Square be further multiplied by its Root a, then the Cube thence produced will be aaa or a (powerof3), &c. Which being premised, you have these following

CONSECTARIES.

I. IF a Positive Quantity be multiplied by a Positive one, the Product will be also Positive; since to multiply is to repeat the Quantity according as the Multiplier directs: Where­fore to multiply by a Positive Quantity, is to repeat the Quan­tities positively; as on the other side, to multiply by [Page 42] a Privative, is so many times to repeat the Privation of that Thing: Which we shall shew further hereafter.

II. Equal Quantities ( a and a) multiplied by the same (b), or contrariwise, will give equal Products ( ab and ab or ba).

III. The same Quantity (z) multiplied by the whole Quan­tity ( a+ b+ c) or by Eucl. lib. 2. prop. 1. all its parts separately, will give equal Products. Also

IV. The whole ( a+ b) whether it be mul­tiplied by lib. 2. prop. 2. it self, or by its parts separately, will give equal Products.

SCHOLIUM I.

THe Vulgar Praxis of Numeral Multiplication, is founded on these two last Consectarys, as e. g. to multiply 126 by 3; you first multiply 6 by 3, then 2, i. e. 20 by 3, then 1, i. e. 100 by the same, and then add each of those partial Products into one Sum: In like manner being to multiply 348 by 23, you first multiply each Note of the Multiplicand by the first of the Multiplier (3) and then by the second (2) ( i. e. 20) &c. which is to be done likewise after the same man­ner in Tetractical Multiplication; only in this latter, which is more easie, you have nothing to reserve in your mind, but all is immediately writ down, (which might also be done in Vulgar Multiplication) as may be seen by this Example un­derneath, as also the great easiness of this sort of Multiplica­tion, beyond the common way, because there is no need of any longer Table than that we have shewn page 7.

[...]

SCHOLIUM II.

It is manifest from what we have said,

I. IF the Base of a Parallelogram be called (b) and its Alti­tude a, its Area may be expressed by the Product ab, by Cons. 7. Definit. 12.

II. If the Base of a ▵ be b or eb, and its Altitude a its Area will be half ab or half eab, by Consectary 8. of the same De­finition.

III. If the Base of a Prism or Parallelepiped or Pyramid be half ab or ab, and its Altitude c, the solid Contents of that Prism will be half abc, and of the Parallelepiped abc, by Con­sect. 3 & 4. Def. 16. and of the Pyramid ⅙ abc, by Cons. 3. Def. 17.

DEFINITION XXIX.

DIvision, in general, is a manifold or complicated Sub­traction of one quantity (which is called the Divisor) out of another (which is called the Dividend) whose multiplicity, or how many times the one is contained in the other, is shewn by another quantity arising from that Division, which is there­fore called the Quote or Quotient. Here also the Divisor is of the same kind with the Dividend, or of a different kind, e. g. of the same kind if the product . . . . . . . . . . . . (12) be divided by (3) whence you'l have the Quotient . . . . (4) or dividing the aforementioned Line CD by the Line AB you'l again have the Line BC; but of a different kind, if the plane number a­bove found

[12 dots arranged in 3 rows and 4 columns]

or the Rectangle ABCD be divided by a Retroduction, or a moving backwards again the erect Side AB, by whose motion the Rectangle was first formed, that so the Line BC may remain alone again. But both these kinds of Division as they have their peculiar Difficulties in Arithme­tick and Geometry, which we shall further elucidate in their proper places; so they may be universally and very easily per­formed in Species (or by Letters) which will be sufficient to our present purpose; or by a bare separation of the Divisor from the Dividend, if it be actually therein included; or by [Page 44] placing the Divisor underneath the Dividend with a Line be­tween. Thus if ab be to be divided by (b) the Quotient will be a; if by a, the Quotient will be (b); but if a or ab be to be divided by c which Letter since it is not found in the Dividend, cannot be taken out of it) the Quotients are a / c and ab / c i. e. a or ab divided by c, after the same manner as if 2 were to be divided by 3; which Divisor, since it is not contained in the Dividend, is usually placed un­der it, by a separating Line thus, ⅔, 2 divided by 3.

SCHOLION.

HOW difficult Common Division is, especially of a large Dividend by a large Divisor, is sufficiently known: but how easily it is performed by Tetractical Arithmetick, we will barely bring one Example to shew. If the Product found in Schol. 1. of the preceding Definition, 1200 203 22 be again to be divided by its Multiplier 133, it may be performed after the usual way, but with much more ease, as the following Opera­tion will shew; or according to a particular way of Weigelius, by writing down the Divisor, and its double and triple, in a piece of paper by it self, after this way:

123 312 1101
Divisor, Double, Triple.

and then moving that piece of Paper to the Dividend, note, which of those three Numbers comes nearest to the first Fi­gures of the Dividend; for that barely subtracted gives the Re­mainder, and will denote the Quotient to be writ down in its proper place; as the operation itself will shew better than any words can.

[...]

Thus after Weigelius's way: [...]

DEFINITION XXX.

EXtraction of Roots is a Species of Division, wherein the Quo­tient is the Root of the given Square or Cube, &c. But the Divisor is not given, neither is it all along the same (as it is in Division) but must be perpetually found, and they are se­veral. And as the Squares of Simple Numbers 1, 2, 3, &c. viz. 1, 4, 9, 16, &c. and their Cubes 1, 8, 27, 64, &c. may be had immediately out of a Multiplication Table, as also their Roots, without any further trouble; and likewise in Spe­cies, as the Roots of the Square aa or a (powerof2), or of the Cube aaa or a (powerof3), are without doubt (a); so if the Square Root be to be extracted out of de, or the Cube Root out of fgm (because the letters are different, and no one can be taken for the Root) the Square Root is commonly noted by this Sign √ de, the Cube Root by this √C, or 3√ fgm, &c. as also in Numbers that are not perfect Squares (as e. g. 2, 3, 5, 6, 7, 8, 10, 11, 15, 17, 19, &c.) we can no otherwise express the Square Roots, then after this manner √2, 7, √19, &c. and in those that are not perfectly Cubical (as all between 1, 8, 27, 64, &c.) we can only express their Cube Roots after some such manner, √c. 7, or 3√7. √c. 61, or 3√61 &c. Which forms of Roots in specious Computation, we call Surd Quantities, in Vulgar Arithmetick Surd Numbers, i. e. such as cannot be perfectly expressed by any Numbers; altho we have Rules at hand to determine their Values nearer and nearer ad Infinitum.

These Rules accommodated to Square and Cube Numbers, &c. which otherwise are more difficult to be comprehended, appear plain and easie to him, who multiplies a Root expressed [Page 46] by 2 Letters (called therefore commonly a Binomial) first Qua­dratically, then Cubically, &c. For he will have as

CONSECTARYS.

I. THE Square of any assumed Root, as also, Prop. 4. lib. 2. Eucl. and at the same time a general Rule for Ex­tracting the Square Root, all expressed in these few Notes:

aa+2 ab+ bb.

II. The Cube of the same Root, a New Theorem, and at the same time a Rule for Extracting any Cube Root, con­tained in this Theorem:

a (powerof3)+3 aab+3 abb+ bbb.

SCHOLIUM I.

WHich that we may more plainly shew, especially as far as it relates to the Rules of Extraction, consider, 1. That the Root of the Square aa+2 ab+ bb is already known (for we assumed for the Root the Quantity a+ b) so that now we are to see which way this Root is to be obtain'd out of that Square by Division. It will presently appear, that the first Note of the Root a, will come out of the first part of the Square aa, and the other part b must be obtain'd out of the remainder 2 ab+ bb; and so as there are 2 Notes of the Root, the Square must be distinguish'd as it were into 2 Classes, each of which gives a particular note of the Root. Then it is manifest, that the first Note of the Root (a) may be obtain'd out of the Square aa by a simple Extraction. Now it is e­vident, if I would have by Division the other Note of the Root, the next following part of the remaining Classis must be divided by 2 a, the double of the Quotient just now found, and that nothing should remain after this Division (for now we have the whole Root a+ b) you must not only subtract the Product of the Divisor and this new Quotient, but also the Square of this new Quotient: Which is the Vulgar Method and Rule for the Extraction of Square Roots taught in common Arithmetick.

Likewise if you would extract the Root of the above-men­tioned Cube, which we already know, having formed it from a+ b, it is manifest, that the first Note of the Root a will come out of the first part of the Cube a (powerof3), and the other b, must be obtain'd out of the remainder 3 abb+ bbb, and so, as there are two Notes of the Root, the Cube must be distinguish'd, as it were into two Classes, each of which will give a particular Note of the Root. Now it is manifest, the first Note of the Root a is obtained by simple Extraction of the Root out of the Cube aaa. It is moreover evident, if I would have by Division the other Note of the Root b, the next remaining part must be di­vided by 3 aa (the triple Square of the precedent Quotient, or thrice the precedent Quotient multiplied by it self) and, that nothing should remain after this division (for now we have the whole Cube Root a+ b) you must not only subtract from the remaining Dividend the Product of the Divisor, and the new Quotient (3 aab) but also the Product of the Square of the new Quotient, and thrice the precedent Quotient (3 abb) and more­over the Cube of that new Quotient b (powerof3): Which is the Method of extracting Cube Roots in Vulgar Arithmetick.

SCHOLIUM II.

FRom what we have said you have also the Reason of ano­ther rule in Arithmetick which teaches how to approach continually nearer and nearer to the Square and Cube Roots of numbers that are not exact Squares and Cubes; viz. by adding to the given Number perpetually new Classes and Cyphers or o's, two at a time, to the Square, and three to the Cube, and so continue on the operation as before; which will add Deci­mal Parts to the Integrals before found; and the next opera­tion (if you add a second Classe of Cyphers) will exhibit Cen­tesimal Parts, and so on ad Infinitum. For Example, If I would have the Square Root of 2 pretty near, I can assign no nearer whole Number than 1. But by adding a new Classe of 2 Cy­phers, i. e. multiplying the given Number by 100 (whereby the Root is multiplied by 10) you'l have 14, nearly the Root of 200, that is, 14/10 or 1 4/10 much nearer the Root of 2 than the former; and thus you may always come nearer and nearer ad Infinitum, but never to an exact Root. For if you could have [Page 48] the exact Root of 2, or 3, or 5, &c. in any Fraction what­soever, that Fraction must be of such sort, that its Numerator and Denominator being squared, the Fraction thence arising must exactly equal 2 or 3, or 5, &c. that is, its Numera­tor must be exactly double, or trible, or Quadruple, &c. of of the Denominator; which can never be, because both are Squares, and in a Series of Squares no such thing can happen. Hence you have these

CONSECTARYS.

III. THat it is a certain mark of Incommensurability, if on [...] quantity is 1, and other the √2, or √3, or √5, &c▪

IV. That these sorts of Quantities are notwithstanding Com­mensurable in their Powers, i. e. their Squares are as 1 and 2 or as a number to a number.

V. Those Quantities which are to one another, as 1 an [...] √√2, or as √2 and the √√3 are incommensurable in Powe [...] also. Which being rightly understood, you may easily compre­hend several Eucl. l. 10 Prop. 9, 10, 11 12, 13 &c. Propositions of lib. 10 Eucl. especially aft [...] some few things premised concerning Reason and Proportion.

SCHOLIUM. III.

FRom what we have shewn may easily be concluded, th [...] to any proposed Quantity whatsoever, which Euclid cal [...] lib. 10. Def. 5, 6, [...] 8, 9, 10, 11. Rational, and for which we may always put I, there may be several others both commensurable and incommensurable, and that either simply or in power so; those which are commensurable to a Rational given Quantity, either Simply or only in Power (which, e. g. are to it, as 2, 3, 4, &c. ½, ⅓, ¼, &c. or as √2, √3, √4, √½, √⅓, √¼, &c.) are called also Rational: but those which are Incommensurable both ways ( i. e. both simply and in power) as (√√ [...] √√⅓, &c.) are called Irrational. In like manner the Squar [...] of a given Rational Quantity (as I) is called Rational, an [...] Quantities commensurable to it (as 2, 3, 4, 5, &c. ½, ⅓, [...] &c. □) are called also Rational; but incommensurable on [...] (√√2, √√5, &c.) Irrational, and the Sides and Roots of the [...] more Irrational.

DEFINITION XXXI.

ANy Quantities whatsoever of the same kind, whether com­mensurable or incommensurable, equal or unequal, ad­mit of a twofold respect or relation of their magnitude, one whereof, when only the difference or excess of one above ano­ther is respected (as 10 which is 3 more than 7) is called an Arithmetical Reason or Respect; the other, wherein respect is ra­ther had to the Amplitude, whereby one is contained once, or a certain number of times in the other (as 3 is contained thrice in 10 and ⅓ part more) is called Geometrical Reason, or by way of Emphasis, only Reason, and by others Proportion; and this Rea­son or Proportion, if the less is exactly contained a certain num­ber of times in the greater (as 3 in 6, or 4 in 12) is generally called, on the part of the greatest term, Multiple, and on part of the less, Submultiple, and particularly in the first Example double, when 6 is taken in respect to 3, and subduple when 3 is taken in respect to 6; in the other triple and subtriple, &c. If the less be contained in the greater once or more times, unity only remaining over and above (as 3 in 4 and 4 in 9) the Reason or Proportion is called Superparticular and Subsuperparticular, and is noted by the terms Sesqui & Subsesqui, joyning the ordinal Name of the lesser Term; as the the Reason of 4 to 3 is called Ses­quitertian, and contrariwise Subsesquitertian; the Reason of 9 to 4 is called double Sesquiquartan, and contrariwise Subduple Subsequi­quartan, &c. If lastly, the less be contained in the greater once, or a certain number of times, several units remaining over and above, it is commonly called Superpartient Reason and is expres­sed by the word Super or Subsuper, joyned with the adverbial Name of the remaining Parts, and the ordinal Name of the lesser Term; thus, e. g. the Reason of 7 to 4, is cal­led Supertriquartan, 12 to 5 double Superbiquintan, &c. but when the Quotient arises by the division of the greater term by the less, and is commonly expressed in the same words, it is also commonly called by the name of Reason (e. g. 2 is the name of the Reason of 6 to 3, 2¼ of 9 to 4, or contrariwise, &c.) as also the quotient arising by division of the Consequent by the Antecedent (as ½ in the first case, 4/9 in the latter) by which name the antecedent Term of the Reason being multiplied, pro­duces its Consequent; which is evident by naming any Reason [Page 50] e or i, or o, &c. Thus if the antecedent Term be called a or b &c. the Consequent may be rightly call'd ea or eb, oa or ob, &c. and because in an Arithmetical Relation we only respect the excess of the first above the following, or of the following a­bove the foregoing (which may be called x or z) if the ante­cedent (which may be called a or b) be less, the consequent may properly be called a+ x or b+ z; but if it be greater, the other will be a− x or b− z.

CONSECTARYS.

I. WE may hence readily infer, that if the Diameter of any Circle be called a the Circumference may be called ea, (for whatever the proportion is between them, i [...] may be expressed by the Letter e) and the Area, according to Consectary 2. Definition 15, will be ¼ eaa.

II. If for the Base of any Cylinder or Cone you put ¼ eaa and the Altitude (b) the Solidity of that Cylinder may be rightly expressed by ¼ eaab, by Consect. 5. Definit. 16, and of the Cone by ½ eaab, by Consect. 4. Definit. 17.

DEFINITION XXXII.

AS the Identity (or sameness) of several Geometrical Rea­sons used to be called Geometrical Proportionality, or em­phatically Proportion; so the similitude (or likeness) of severa [...] Arithmetical Reasons, is deservedly call'd Arithmetical Proportiona­lity, or by a particular Name Progression; and consequently those Progressionals, or Arithmetical Proportionals, which exceed one ano­ther by the same difference, either uninterruptedly or continually as 2, 5, 8, 11, 14, &c. ascending, or 30, 28, 26, 24, 22, 20, &c. descending; or interruptedly, as 2 and 5, 7 and 10, 11 and 14, &c. ascending; or 30 and 26, 24 and 20, 1 [...] and 13, &c. descending: For which, and all other in what cas [...] soever, we may universally put this (or such like) continua [...] Progression, v. g. a, a+ x, a+2 x, a+3x, &c. ascending; o [...] a, a− x, a−2x, a−3x, &c. descending, but in an interrupted Progression, v. g. b and b+ z and c and c− z, d and d− z, &c. descending. Whence you have this

CONSECTARY.

ANY Difference being given, the following Terms of me Progression, continually proceeding from the first assumed or given one, may be found; as also several Antecedents that interruptedly follow the given or assumed ones, viz. by adding or subtracting the given Difference to or from the former Terms to find the latter.

DEFINITION XXXIII.

IN like manner, since Reasons are said to be the same, which have the same Denomination of Reason, those quantities will be proportional which continually ascend by the same de­nomination of Reason, as 2, 4, 8, 16, 32, 64, &c. or descend, [...]s 81, 27, 9, 3, 1. there by the Denomination of the Reason 2, here 3; or that ascend interruptedly, as 2, 4; 3, [...]: 5, 10, &c. or descend, as 40, 10; 28, 7; 20, 5; 8, [...], &c. Whence you have these

CONSECTAYS.

HAving two Terms given, or only one with the Deno­mination of the Reason ( e. g. the Term 2 with the Denomination of the Reason 3, or universally the first Term a with the Denomination of the Reason e) it will be easie to find [...]s many more Terms of the Geometrical Progression or Pro­ [...]ortion as you please, viz. by always multiplying the Antece­ [...]ent by the Denomination of the Reason, that you may have 2, [...], 18, 54, &c. or a, ea, e (powerof2) a, e (powerof3) a, &c. in continued, or 2 and [...], 4 and 12, 5 and 15, &c. and aea, beb, ded, &c. in [...]iscontinued or interrupted Proportion.

Thus having rightly understood what we have said in this [...]3 and 31 Definition, there will follow these Corollarys as so ma­ [...]y Axioms.

II. That equal Quantities have the same proportion to the [Page 52] same Quantity Eucl. lib. Prop. 7. and the same has the like to equal Quan­tities.

III. But a greater quantity has a greater Reason to the same prop. 8. than a less, and the same has a greater proportion to a less Quantity than to a greater.

IV. On the contrary, those that have the prop. 9. same pro­portion to the same quantity, and that likewise the same to them are equal.

V. But that which bears a prop. 10 greater proportion to the same is greater; but that to which the same bears a greate [...] proportion is less.

VI. Proportions equal to one third 16. are also equal amon [...] themselves, &c.

DEFINITION XXXIV.

HEre remain two things to be taken notice of; First th [...] If any whole (quantiy) be so divided into two equ [...] parts Eucl. De­finit. 3. lib. 6 that the whole, the greater part an [...] the less are in a continual proportion; th [...] (whole) is said to be cut in extreme and me [...] Reason. 2. In a continual Series of that kind [...] Proportionals ( e. g. 2. 4. 8. 16. 32, &c. or a, [...] e (powerof2) a, e (powerof3) a, e (powerof4) a, &c.) the Reason of the first Ter [...] to the third Eucl. De­finit. 10. l. 5. (2 to 8, or a to e (powerof2) a) is pa [...] ticularly called Duplicate, and to the 4th ( [...] or e (powerof3) a) Triplicate, &c. of that Reason which the same first Te [...] has to its second, or any other antecedent of that Series to [...] Consequent: But generally these Duplicate and Triplicate Re [...] sons, &c. as others also of the first Term to the third or four [...] of Proportions continually cohering together, (whether the are the same as in the foregoing Examples, or different as [...] these, 2, 4, 6, 18, or a, ea, eia, eioa, &c. viz. if the nam [...] of the first Reason be e, of the second i, [...] the third o, &c.) I say, the Reasons of the fir [...] Term (2 or a) to the third (6 or eia) [...] to the 4th (18 or eioa) are said to be compoun [...] ed of the continual intermediate Reasons.

Now from our general Example, what Eucl [...] says, is manifest,

CONSECTARY I.

THat the denomination of a compounded Reason arises from the Multiplication of the denominations of the given Simple Eucl. l. 6. Def. 5. Reasons; as the denomination of the reason com­pounded of both ( viz. a to eia) is produced by multiplying the denomination of the first Reason e by the denomination of the second Reason i, and the denomination of the Reason com­pounded of the three ( viz. a to eioa) is produced by the deno­mination of the first Reason e, multiplied by the denomina­tion of the second Reason i; and the Product of these by the denomination of the third Reason o, &c.

CONSECTARY II.

SO that it is very easie after this way, having never so many Reasons given, whether continued (as 2 to 3, 3 to 6, or [...]a, ea, eia,) or interrupted or discrete (as 2 to 3, and 5 to 10, or a to ea, and b to i b) to express their compounded Reason: [...]n the first case it easily obtain'd by the bare omission of the in­termediate Term or Terms (2 to 6, or a to eia;) and in the other by multiplying first of all the Names of the com­pounding Reasons among themselves (1 ½ and 2, e. and i.) and by the Product (3 or ei) as the name of the Reason compound­ing the first Term (2 or a) that you may have the o her 6 or eia) or (if any one had rather do so in this latter case) by turning the discrete or interrupted Reasons into continued ones, by making as 5 to 10 in the second Reason, so is the Consequent of the first 3 to 6, or as b to ib, so ea to eia,) and then by re­ferring the first 2 to the third 6, or the first a to the third eia, &c. In a word therefore, any Duplicate Reason may be appositely expressed by a to e (powerof2) a, and Triplicate by a to e (powerof3) a, the one immediately discernible by a double, the other by a triple Multiplication into itself; as you may also commodiously, and denote others compounded, e. g. of 2 by a to eia, of 3 by a to eioa, &c.

SCHOLIUM.

WE will here advertise the Reader, that tho the Names [...] duplicate & triplicate Reasons, &c. are chiefly appropriate to Geometrical Proportionality, yet the Moderns have also accom­modated them to Arithmetical also; as e. g. That Arithmetic [...] Progression is called Duplicate, whose Terms are the Squares [...] Numbers Arithmetically Proportional ( e. g. 1, 4, 9, 16, 25 &c.) and Triplicate, whose Terms are Cubes, (&c. as 1, [...] 27, 64, &c.

DEFINITION XXXV.

AND now at length we may understand what Magnitud [...] Geometers particularly call like, or similar. Whereas [...] General one number may be said to be like another, one rig [...] Line to another, one obtuse Angle to another, a Triangle [...] a Triangle, and the like; but an Acute Angle is not like [...] Obtuse one, nor a Triangle like a Parallelogram, or a rig [...] Line like a Curve one; or a Square like an Oblong, &c. Yet [...] mong those Figures which may after that rate in general be sa [...] to be like, there is notwithstanding a great deal of dissimi [...] tude; therefore in a strict Sense we call only those Right Li [...] ed Figures similar or like (α) which have each of their Angl [...] respectively equal to each of the other (as A and A) B and B C and C, &c. Fig. 48.) and the Sides about those equal A [...] gles Proportional, viz. as BA to AC, so BA to AC, &c. ( [...] and among Solid Figures those are said to be Similar, each o [...] whose Planes are respectively Similar one to the other, and equ [...] in number on both sides; as, e. g. the Plane AC is similar to th [...] Plane AC, and CG to CG, &c. and six in number on bo [...] sides.

Pag. 54.

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48

49

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53

Book I.

Section II. Containing several Propositions demonstrated from the foregoing Foundations.

CHAP. I. Of the Composition and Division of Quantities.

Proposition I.

THE Sum and Difference of two unequal Quantities ad­ded together, make double of the greatest.

Demonstration.

Suppose a be the greatest, b the least, then will their Sum be

  a+ b
And their Difference a− b
Their Sum 2 a,

by Consectary 1. Definition 27, Q. E. D.

CONSECTARY.

HEnce by a bare Subsumption Eucl. lib. 6. Definit. 1. you have the truth of Consect. 1. Definit. 8. that 2 unequal contiguous Angles on the same Right Line, ACD and ACE ( Fig. 49. lib. II. Def. 9.) i. e. if we call the Right Angle BCD or BCE (a) and the dif­ference between the one and the other (b) a+ b and a− b, make 2 a, i. e. are equal to 2 right ones. Eucl. 13. l. 1.

Proposition II.

IF the Difference of two unequal Quantities be subtracted from their Sum, the Remainder will be double of the least.

Demonstration.
If from the Aggregate or Sum a+ b
You subtract the Difference a− b
The Remainder will be 0+2 b

by Con­sectary 2. Definition 27. Q. E. D.

Proposition III.

BUt if the Sum or Aggregate be subtracteed from the Dif­ference, the remainder is so much less than nothing, as is the double of the last Quantity.

Demonstration.
For if from the Difference a− b
You subtract the Sum or Aggregate a+ b
The Remainder will be — 0−2 b

by Con­sectary 3 of the aforesaid Definition. Q. E. D.

Proposition IV.

IF a Positive Quantity be multiplied by a Negative one, or contrariwise, the Product will be a Negative Quantity.

Exposition.

If a− b is to be multiplied by c; it is certain, that a multiplied by c, makes ac a Positive Quantity, by Consect. 1. Definit. 28. Moreover b by the same c (a Negative by a Positive) will make − bc; and so the whole Product of a− b by + c, will be ac− bc.

Demonstration.

Suppose a− b= e; therefore ec will be = to the Product of a− b by c: and since a− b is = e by the Hypoth. adding on both sides h, you'l have a= e+ b. by Schol. Definit. 26. and multiplying both sides by c, ac= ec+ bc, by Consect. 2. Definit. 28. and by further subtracting from each side bc, you'l have ac− bc= ec, that is, to the Product of a− b by c. Q. E D.

CONSECTARY.

SInce ac− bc is the Product of a− b by c, it is manifest also, that if ac− bc be divided by c, you'l have a− b for the Quotient; and so always a Positive Quantity (as ac) di­vided by a Positive one, c, will give a Positive Quotient; but a Negative Quantity − bc divided by a Positive one, will give a Negative Quotient.

Proposition V.

IF a Negative Quantity be multiplied by a Negative one, the Product will be Positive.

Exposition.

Suppose a− b is to be multiplied by − c; it is certain, that [...] multiplied by − c will give the Negative Quantity − ac, by Prop. 4. but − b multiplied by the same − c will produce + bc, and so the whole Product will be − ac+ bc.

A Demonstration like the former.

Suppose a− b= e, then will − ec = the Product of a− b by − c: and since a− b is = e, adding b on both sides you'l [...]ave a= e+ b, by Schol. Definit. 26. and multiplying both sides by − c, you'l have ac=− ec− bc, by Prop. 4. and Consect. [...]. Definit. 28. and by adding bc on both sides, you'l have [Page 58] − ac+ bc=− ec, i. e. to the Product of a− b by − c Q. E. D.

CONSECTARYS.

I. SInce therefore a− c+ bc arises from a− b by − c, it is manifest, that if − ac+ bc be divided again by − c, you will again have a− b, and consequently a Negative Quantity divided by Negative, will give a Positive Quotient but a Positive Quantity + bc divided by a Negative one, wil [...] give a negative Quotient − b.

II. We have therefore the Foundation and Demonstration o [...] the Rules of Specious Computation, in the multiplication an [...] division of Compounded Quantities, viz. that the same Sig [...] multiplied together (as + by + or − by −) give + but different (as + by − or − by +) give the Sign − Which Rules the following Examples will Illustrate, as als [...] several other we shall meet with in the following Chapter.

Multiplication.

[...]

Division.

[...]

CAAP. II. Of the Powers of QUANTITIES.

Containing (after a compendious Way) most part of the 2d Book of Euclid; and the Appendix of Clavius to Lib. 9. Prop. 14.

Proposition VI.

IF any whole Quantity be divided into two parts Eucl. lib. 2. prop. 3. A. C. also the third. the Rectangle contained under the whole, and one of its parts, is equal to the Square of the same part, and the Rectangle contain'd under both the parts.

Demonstration.
Let a+ b represent the whole a+ b  
  b one part of it, or a the other.
  ab+ bb the Rectangle, aa+ ab the Rectangle.

(See Fig. 50.) Q. E. D.

Proposition VII.

IF a whole Quantity be divided into two parts Eucl. Prop. 4. lib. 2. the Square of the whole is equal to the Squares of both those parts and 2 Rectangles contained under them.

Demonstration.

This is evident from the preceding, and may moreover thus appear further.

Let the Parts be a and b, then will the whole be [...]

Which if you multiply by it self [...]

You have the Square [...]

(See Fig. 51. N. 1.) Q.E.D.

CONSECTARYS.

I. HEnce you have the Original Rule for Extracting of Square Roots, as we have shewn after Definition 30. and here have further Illustrated in Scheme N o 2.

II. Hence it naturally follows, that the Square of double any Side is Quadruple of the Square of that Side taken singly.

III. Hence also you have the addition of surd Numbers, or in general of surd Quantities, by help of the following Rule (supposing in the mean while their Multiplication:) Suppose these 2 Surds √8 and √18, or more generally √75 aa and √27 aa, are to be added together; first add their Squares 8 and 18, &c. then double their Rectangle (√144) that is, multiply it by the √4, and then the double of this √576, i. e. having ex­tracted the Square Root, (24) and added it to the Sum of the first Squares (26) the Root of the whole Summ (50) viz. √50, is the Sum of the two surd Quantities first proposed.

SCHOLIUM.

BUT if it happens that the Root of the double Product can­not be expressed by a Rational Number (as, when the proposed Quantities are Surds, as √3 and √7, to whose Squares 3+7, i. e. 10, you must add the double Product of √7 by √3, i. e. √84, which cannot be expressed by a Rational Num­ber) then that double Product must be joined under a Surd Form, or Radical Sign, to the Sum of the Squares (thus, viz. 10+√84) and to this whole Aggregate prefix another Radi­cal Sign, thus, [...]; or also you may only simply join the Surd Quantities proposed by the Sign + thus, √3+√7. Here also you may note, that the two Surd Quantities proposed [Page 61] in the first case of Consectary 2. are called Communicants; in the other case of this Scholium, Non-Communicants: For in this case each quantity under the Radical Sign may be divided by some Square, and have the same Quotient ( e. g. 8 and 18, may be divided the first by 4, the other by 9, and the Quotient of both will be 2; likewise 75 aa and 27 aa may be divided, the one by 25 aa, the other by 9 aa, the Quotient of both being 3; and then if the Quotient on both sides be left under the Radical Sign, and the Root of the dividing Square set before it, the same quantities will be rightly expressed under this form: 2√2 and 3√2, also 5 a√3 and 3 a√3; and then the addition is easie, viz. only collecting or adding together the Quantities prefixt to the Radical Sign; so that the Sums will be of the one 5√2 and of the other 8 a√3, which are indeed the same we have shewn in Consect. 2. For if contrarywise we square the Quantities that stand without, or are prefixt to the Radical Sign, and then set those Squares (25 and 64 aa) under the Ra­dical Sign, multiplying by the Number prefixt to it, you'l have for the one √50, for the other √192 aa (Consect. after Schol-Prop. 22.)

Proposition VIII.

IF any whole Quantity ( viz. Line or Number) be divided Eucl. & Clav. 5. into two equal parts, and two unequal ones, the Rectangle of the unequal ones, toge­ther with the Square of (the intermediate part or) the difference of the equal part from the unequal one, is equal the Square of the half.

An Universal Demonstration.

Suppose the parts to be a and a, and the whole 2 a; let one of the unequal Parts be b, the other will be 2 a− b, and the dif­ference between the equal and unequal part a− b.

The equal ones [...]

Rectangle [...]

Difference [...]

The Sum will be aa (the other parts destroying one another) Q.E.D. ( Vid. Fig. 52.)

Proposition IX.

IF to any whole Quantity divided into two equal parts Eucl. & Clav. 6. you add another Quantity of the same kind, the Rect­angle or Product made of the whole and the part added, mul­tiplied by that part added, together with th [...] square of the half, will be equal to the Square o [...] the Quantity compounded of that half, and th [...] part added.

Demonstration.

Let the whole be called 2 a, the part added b, then th [...] quantity compounded of the whole and the part added will b [...] 2 a+ b; and that compounded of the half and the part added a+ b.

The Quantity compounded of the whole, and the part added is, [...] the half a Comp. [...]

Multip. by the part added [...]

2 ab+ bb□ of the half aa=□ aa+2 ab+ bb

( Vid. Fig. 53.) Q. E. D.

Proposition X.

IF a Quantity be divided any how into Eucl. & Clav. prop. 7. l. 2 two parts, the Square of the whole, together with that of one of its parts, is equal to two Rectangles contained under the whole and the first part, together with the Square of the other part.

Pag. 63.

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60

The Universal Demonstration.

Let a be one part and b the other, the whole a+ b

a+ b the whole. The whole a+ b
a the first part *   a+ b
aa+ ab     aa+ ab
2     ab+ bb
2 aa+2 ab the double rectangle □ of the whole aa+2 ab+ bb
add bb the □ of the other part * add aa
Sum 2 aa+2 ab+ bb = to the Sum .... 2 aa+2 ab+ bb

( Vid. Fig. 54. N o 1.) Q.E.D.

CONSECTARY.

HEnce you have the Subtraction of Surd Numbers, or more generally of Surd Quantities, by help of the following Rule.

Add the Squares of the given Roots according to Consect. 3. Prop. 7. and from their Sum subtract the double Rectangle of their Roots; the Root of the Remainder will be the difference sought of the given Quantities.

As, if the √8(BC) is to be subtracted from √50AC ( Fig. 54, N o 1.) you must add 50, i. e. the whole Square AD) and 8, ( i. e. the other Square superadded DE,) and the Sum will be 58, equal to the two Rectangles AF and FE+□GH, by [...]his Prop. I find therefore those two Rectangles by multiplying √50 by √8, and then the Product √400 by 2 or √4, there­by to obtain the double Rectangle √1600, i. e. (having actu­ally Extracted the Root) 40. This double Rectangle therefore [...]or 40, being subtracted out of the superiour Sum, the remain­der 18 will be the □ GH, and so its Root ( viz. √18) gives [...]he required Difference between the given Surd Quantities.

SCHOLION.

BUT this Subtraction may be performed yet a shorter way, if each quantity under its Radical can be divided by some [Page 64] square, so that the same Quotient may come out on both sides that is, if the Surd Quantities are Communicants, as e. g. √5 [...] (the number 50 being divided by 25) is equal to 5√2 a [...] √8 to 2√2; for then the numbers prefixt to the Radical Sig [...] being subtracted from one another ( viz. 2√2 from 5√2) yo [...] have immediately the remainder or difference 3√2, i. e. √1 [...] But if the proposed Quantities are not Communicants (as if th [...] √3 is to be subtracted from √7) the remainder may be brief [...] expressed by means of the Sign − thus, [...], or accordin [...] to the foregoing Consectary, thus, [...].

Proposition XI.

IF any Quantity be divided into two parts, Eucl. & Clav. prop. 8. the Quadru [...] Rectangle contained under the whole and one of its parts, toget [...] with the Square of its other parts, will be equal to the Square of [...] Quantity compounded of the whole and the other part.

Demonstration.
Suppose a+ b the whole.
  b one part.
  ab+ bb the Rectangle of these two.
mult. by 4  
  4 ab+4 bb the Quadruple Rectangle.
Add aa the Square of the other part.
Sum aa+4 ab+ bb  

The Quantity compounded of the whole and the first part [...]

Square of the Compound Quantity Q. E [...]

( Vid. Fig. 55.)

Proposition XII.

IF any Quantity be divided into two equal parts Eucl & Clav. prop. 9. and into two other unequal ones, the Squares of the unequal parts taken together will be double the Square of half the quantity, and the Square of the difference, viz. of the equal and unequal part, taken to­gether.

Demonstration.

Suppose the equal parts to be a and a, the difference (b) the greater of the unequal Parts to be a+ b, the less a− b.

The greater part [...] The less [...] Half [...] Difference [...]
Sum of these 2 aa+2 abb Sum aa+ bb

Q. E. D. ( Vid. Fig. 56.)

Proposition XIII.

IF to any whole Quantity Eucl. & Clav. x. divided into two equal parts there be added another Quantity of the same kind, the Square of the Quantity compounded of the whole, and the quantity added, together with the square of the quantity added, will be double the square of the half the quantity, and the square of the Sum of the half and the part added taken together.

Demonstration.

Suppose the whole to be 2 a, the half parts a and a, the quantity added b; then the quantity compounded of the whole and the quantity added, will be 2 a+ b, and that of the half and the quantity added a+ b.

Comp. of the whole [...] and quantity added

Sum

Half [...]

Qu. compouded of hal [...] [...] and qu. added,

Sum 2 aa+2 ab+ [...] Manifestly the half of the former Sum. Q.E.D.

CHAP. III. Of Progression, or Arithmetical Proportionals.

Proposition XIV.

IF there are 3 Quantities in continued Progression, or [...] Arithmetical continued Proportion, the Sum of the Extrem [...] is double of the middle Term.

Demonstration.

Such are e. g. a, a+ x, a+2 x ascending, or a, a− x, a−2 x descending.

By Definition 32. the Sum of the Extremes in the first 2 a+2 x, in the latter 2 a−2 x; in both manifestly double the middle Term Q.E.D.

Proposition XV.

IF there are 4 of these Continued Proportionals, the Sum [...] the Extreme Terms is equal to the Sum of the me [...] Terms.

Demonstration.

Such are e. g. a, a+ x, a+2 x, a+3 x, &c. ascending, or a, a− x, a−2 x, a−3 x, &c descending; in the one the Sum of the Extremes is 2 a+3 x, in the other 2 a−3 x; and also of the means 2 a+3 x and 2 a−3 x Q. E D.

Proposition XVI.

IF there are never so many of these continued Proportionals, the Sum of the Extremes is always equal to the Sum of any 2 other of them, equally remote from the Ex [...]remes, or also double of the middle Term, if the number of the Terms is odd.

Demonstration.

Suppose a, a+ x, a+2 x, a+3 x, a+4 x, a+5 x, a+6 x, &c. or a, a− x, a−2 x, a−3 x, a−4 x, a−5 x, a−6 x, and the Sum of the Extremes, as also of any 2 equally remote from the Extremes, and the double of the middle Term is in the first Series 2 a+6 x, in the latter 2 a−6 x, &c. Q.E.D.

SCHOLIUM I.

NOR can we doubt but that this will always be so, how far soever the Progression be continued; if you consider that the last Term contains in it self the first, and moreover the difference so many times taken, as is the number of Terms ex­cepting one, but that the first has no difference added to it; and therefore tho the last since one contains one difference less than the last; the second on the contrary has one more than the first, and consequently the Sum of the one will necessarily be equal to the Sum of the other; and in like manner the last except two, contains two Differences less than the last; but, on the contrary, the third exceeds the first by a double Diffe­rence, the double Difference being added to it, &c. as is ob­vious to the Eye in our first universal Example. Hence you have these

CONSECTARYS.

I. YOU may obtain the Sum of any Terms in Arithmetical Proportion, if the Sum of the Extremes be multiplied by half the number of Terms, or (which is the same thing) half the Sum by the number of Terms,

II. To obtain therefore the Sum of 600, or never so many such Terms, you need only have the Extremes and the number of Terms: So that you have a very compendious Way of pro­ceeding in Questions that are solvible by these Progressions, if, having the first Term and Difference of the Progression given, you can obtain the last, neglecting the intermediate ones.

III. But you may obtain the last Term, by Multiplying the given given Difference by the given Number of Terms lessened by Unity, and then adding the first Term to the Product; as i [...] evident from the preceding Scholium.

IV. Hence we may easily deduce this Theorem, that the Sum of any Arithmetical Progression beginning from o, is subduple of the Sum of so many Terms, equal to the greatest, as is the number of Terms of that Progression. For if the first Term is o and the last x, and the given number of Terms a, the Sum of the Progression will be ½ ax, by Consect. 1. but the Sum of so many Terms equal to the greatest, ax. Q.E.D.

SCHOLIUM II.

NOw if any one would be satisfied of the truth of this last Consect. without the literal or specious Notes, let him consider, that if the first Term be supposed to be o, the last (whatever it is) will be the sum of the Extremes. The last therefore mul­tiplied by half the number of Terms, gives the Sum of the Pro­gression, by Consect. 1. and the same last Term multiplied by the whole number of Terms, gives the Sum of so many Terms equal to the greatest. But that this must needs be double of the precedent 'tis evident, because any Multiplicand being multiplied by a double multiplier, must needs give a double Product. Now as this Consectary will be of singular Use to us hereafter for De­monstrating several Propositions, so the three former are the very same Practical Rules of Arithmetick, which are commonly made [Page 69] use of in Arithmetical Progressions; for the Illustration whereof Swenterus gives us several Ingenious Examples in his Delic. part 1. Quest. 70. &c.

CHAP. IV. Of Geometrical Proportion in General.

Proposition XVII.

IF there are three Quantities continually (α) Proportional, the Rectangle of the Extremes, is equal to the Square of the mean Term.

Demonstration.
Such are e. g. a, ea, e (powerof2) a, The mean Term, [...]
The Extremes
Rectangle Square Q.E.D.
SCHOLIUM.

MOreover if three Quantities on each side are in the same Continual Proportion, as

[figure]

[Page 70] the Rectangles of the Extremes made Cross-ways, are equal to the Rectangle of the mean Term; being every way e (powerof2) ab.

Whence by the way may appear that Proposition of Archimedes Eucl. 17 l. 6 & 20 l. 7. lib. I. de Sphaer. & Cyl. Prop. 14. That the Surface of a Right Cone is equal to the Circle, whose Radius is a mean Proportional between the Side of that Cone and the Semidiameter of the Base. For suppose EF to be a mean Pro­portional between the side of the Cone BC ( Fig. 57.) and the Semidiameter of the Base CD, since an equal number of Peripherys answer to an equal number of Radii in the same Propor­tion; half the Product of the first Line BC into the last Periphery, ½ e (powerof2) ab (that is, by Consect. 4. Definit. 18. the Surface of the given Cone) will be equal to half Product of the mean Line into the mean Periphery, ½ e (powerof2) ab (i. e. by Consect. 2. Definit. 15.) to the Area of the Circle of the mean Proportional EF. Q. E. D.

The same Proposition of Archimedes may also be Demonstrat­ed after this Way: If the side of the Cone BC be called b, and the Semidiameter of the Base a) so that the Periphery may, by Consect. 1. Definit. 31. be 2 ea, and so the Surface of the Cone, by Consect. 4. Definit. 18. eab) the √ ab will be a mean propor­tional between b and a, by this 17th Proposition; which being taken for Radius, the whole Diameter will be √2 ab, and the Pe­riphery 2 e√ ab; therefore by Consect. 2. Definit. 15. half the Ra­dius ½√ ab multiplied by the Periphery (since √ ab multiplied, by √ ab necessarily produces ab) will give you the Area of the Cir­cle by that mean lib. I. de Sphaer. & Cyl. prop. 15. Proportional, equal to the Surface of the given Cone, which before was expressed in the same Terms. Q. E. D.

Hence also naturally flows this other Proposition, That the Surface of the Cone (½ e (powerof2) ab) is to its Base (½ ab) as the Side of the Cone ( e (powerof2) b) is to the Radius of the Base b, as may appear from the Terms.

Proposition XVIII.

IF E [...]cl. 16 l. 6 & 19. l. 7. 4. Quantities are Proportional, either continuedly or dircretely, the Product of the Extremes is equal to the Product of the Means.

Demonstration.

Suppose one Continual Proportional, a, ea, e (powerof2) a, e (powerof3) a.

Extremes e (powerof3) a Means e (powerof2) a
a ea
Prod. e (powerof3) aa = Prod. e (powerof3) aa. Q.E.D.
SCHOLIUM.

ON this Theorem is founded the Rule of Three in Arithme­tick; so called because having 3 Numbers, (2. 5. 8.) it finds an unknown fourth Proportional. For altho this fourth be, as we have said, unknown, yets its Product by 2 is known, because the same with the Product of the Means, 5 and 8. Wherefore the Rule directs to multiply the third by the second, that you may thereby obtain the Product of the Extremes: which divided by one of the Extremes, viz. the first, necessa­rily gives the other, i. e. the fourth sought.

Proposition XIX.

IF 2 Products (on the other side) arising from the Multipli­cation of 2 Quantities, are equal, those 4 Quantities will be at least directly Proportional.

Demonstration.

Suppose eba be the equal Product of the Extremes, and eab of the Means; the Extremes will either be eb and a, or e and ba, or b and ea, as also the Means. But what way soever either is taken, there can be no other Disposition or placing of them, than one of the following.

1 eb eb a a
— ee ab  
— ea b; or inversly.
— a eb  
— ab e  
— b ea  
2 e e e ba ba
— eb a  
— ea b; or inversly.
— ba e  
— a eb  
— b ea  
3 b b ea ea
— a eb  
— ba e; or inversly.
— ea b  
— eb a  
— e ba; or inverting the Order of them all.

In all these Dispositions there may be immediately seen a Geometrical Proportion, by what we have in Definition 3 [...] and 33.

CONSECTARSY.

I. AS we have shewn one Sign of Proportionality in the Definition of it, viz. That the same Quotient will a­rise by dividing the Consequents by the Antecedents; so now we have another Sign of it, viz. The Equality of the Products of the Extremes and Means.

II. By a bare Subsumption may hence appear the Truth of Prop. 14. lib. 6. Euclid. at least partly: Which we shall yet more commodiously shew hereafter.

Proposition XX.

IF there are never so many Continual Proportionals, the Pro­duct of the Extremes is equal to the Product of any 2 of the Means that are equally distant from the Extremes, as also to the Square of the mean or middle Term, if the Terms are odd.

Demonstration.

Such are e. g. a, ea, e (powerof2) a, e (powerof3) a, e (powerof4) a, e (powerof5) a, e (powerof6) a, &c. and the Product of the Extremes, and of any two Terms equally remote [Page 73] them, and the Square of the mean or middle Term, every where e (powerof6) aa. Q. E. D.

SCHOLIUM I.

NOR can there be any doubt but this will always be so, how far soever the Progression is continued; if [...] con­sider that the last Term always contains the first, [...] way of Reason, so many times multiplied as is the place of that Term in the rank of Terms, excepting one. Altho therefore the last Term but one is in one degree of its Reason less than [...] last, [...]he second on the contrary, is in one more than the [...] re­ [...]ore the Product of the one will necessarily [...] [...]e Product of the other. Thus also the las [...] [...] Degrees of Proportion lower than the [...] being to be multiplied into that, exc [...]ds [...] fi st [...] of the Proportion, &c. as may be seen thou U [...]iver [...]al Ex [...] Hence you have the following

CONSECTARIES.

[...]. HAving some of the Terms given in a Continual Pro­portion ( e. g. suppose 10) you may easily find any other that shall be required ( e. g. the 17th) as the last; If [...]he 2 Terms given, being equally remote from the first and [...]hat required (as are e. g. the eigth and tenth) be multiplied by one another, and this Product, like that also of the Extremes, be divided by the first.

II. But this may be performed easier, if you moreover take [...]n this Observation, That if, e. g. never so many places of pro­portionals, passing over the the first, be noted or marked by Ordinals or Numbers according to their places (as in this uni­versal Example)

a, ea, e (powerof2) a, e (powerof3) a, e (powerof4) a, e (powerof5) a, e (powerof6) a,
  I. II. III. IV. V. VI.

The place of the 7th Term is (e. g.) VI. (and so the place of any other of them being less by Unity than its number is among the Terms) and also composed of the places of any other equal­ [...]y distant from the Extremes, e. g. V. and I. IV. and II. or [...]wice III. &c.

[Page 74]III. Here you have the Foundation of the Logarithms, i. [...] of a Compendious Way of Arithmetick, never enough to b [...] praised. For if, e. g. a rank of Numbers from Unity, con [...] ­nually Proportional, be signed or noted with their Ordinals, as w [...] have said, as Logarithms,

1. 2. 4. 8. 16. 32. 64. 128. 256, &c.
  I. II. III. IV. V. VI. VII. VIII.  

and any two of them (as 8 and 32) are to be multiplied together; add their Logarithms III and V, and their Sum VII [...] gives you the Logarithm of their Product 256, as the Te [...] equally remote from the 2 given ones and the first, and [...] whose Product with the first (which is Unity) i. e. it self w [...] be equal to the Product of the Numbers to be multiplied: A [...] contrariwise, if, e. g. 128 is to be divided by 4, subtracting t [...] Logarithm of the first II from the Logarithm of the second V [...] the remaining Logarithm V points out the number sought 3 [...] so that after this way the Multiplication of Proportionals [...] by a wonderful Compendium, turned into Addition, and the Division into Subtraction, and Extraction of the Square Ro [...] into Bisecting or Halving, (for the Logarithm of the Squa [...] Number 16 being Bisected, the half II gives the Root sough [...] 4) of the Cube Root into Trisection (for the Logarithm of th [...] Cube 64 being Trisected, the third part gives the Cubi [...] Root sought 4).

SCHOLIUM II.

THat we may exhibit the whole Reason of this admirabl [...] Artifice (which about 35 years ago was found out b [...] the Honourable Lord John Naper Baron of Merchiston in Scotland and published something difficult, but afterwards render'd much easier and brought to perfection by Henry Briggs, the first S [...] vilian Professor of Geometry at Oxford.) I say that we may exhibit the whole Reason of it in a Synopsis, after an easie way when its use appear'd so very Considerable in the great Num­bers in the Tables of Sines and Tangents, nor yet could they be useful without mixing vulgar Numbers with them, especial­ly in the Practical Parts of Geometry, the business was to ac­ [...]ommodate [Page 75] this Logarithmical Artifice to them both. First [...]herefore that Artists might assign Logarithms to all the com­mon Numbers proceeding from 1 to 1000 and 10000, &c. [...]hey first of all pick out those which proceed in continued Geo­metrical Proportion, and particularly, tho arbitrariously, those which increase in a Decuple Proportion, e. g. 1. 10. 100. 1000. 10000, &c.

But now to fit them according to the Foundation of Consect. 8. a Series of Ordinals in Arithmetical Progression, we do'nt only substitute the simple Number 1, 2, 3, &c. but augmented with several Cyphers after them, that so we may also assign [...]heir Logarithms in whole Numbers to the intermediate Num­ [...]ers between 1 and 10, 10 and 100, &c. Wherefore, by [...]his first Supposition, Logarithms in Arithmetical Proportion, [...]nswer to those Numbers in Geometrical Proportion, after the [...]ay we here see,

1 10 100
Log. 0000000 10000000 20000000
  1000 10000
  30000000 40000000, &c.

As that they also exhibit certain Characteristical initial Notes, whereby you may see, that all the Logarithms between 1 and [...]0 begin from 0, the rest between 10 and 100 from 1, the [...]ext from 100 to 1000 from 2, &c.

The Logarithms of the Primary Proportional Numbers being [...]hus found, there remain'd the Logarithms of the intermediate Numbers between these to be found: For the making of which, [...]fter different ways, several Rules might be given drawn from [...]he Nature of Logarithms, and already shewn in Consect. 3. See Briggs's Arithmetica Logarithmica, and Gellibrand's Trigonometria Britannica; the first whereof, chap. 5. and the following, shews [...]t length both ways delivered by Neper in his Appendix. But [...]he business is done more simply by A. Ʋlacq. in his Tables of Sines &c. whose mind we will yet further explain thus: If you are [...]o find, e. g. the Logarithm of the Number 9, between 1 and [...]0, augmented by as many Cyphers as you added to the Lo­garithm of 10, or the rest of the Proportionals ( h. e. between [Page 76] 10000000 and 100000000) you must find a Geometric [...] Mean Proportional, viz. by multiplying these Numbers togeth [...] and extracting the Square Root out of the Product, by Pr [...] 17. Now if this Mean Proportional be less than 9 augmen [...] by as many Cyphers, between it and the former Denary Nu [...] ­ber you must find a second mean Proportional, then betwe [...] this and that same a third; and so a fourth, &c. but if it [...] greater, then you must find a mean Proportional between and the next less, &c. till at length after several Operatio [...] you obtain the number 9999998, approaching near 90000000. Now if between the Logarithm of Unity a [...] Ten ( i. e. between 0 and 10000000) you take an Arithm [...] tical Mean Proportional (05000000) by Bisecting their S [...] by Prop. 14. and then between this and the same Logarithm Ten, you take another mean, and so a third and a fourth, [...] at length you will obtain that which answers to the last abo [...] mentioned, viz. 9. See the following Specimen.

A TABLE of the Geometrical Proportionals betwe [...] 1 and 10, augmented by 7 Cyphers, and of t [...] Arithmetical Proportionals between 0 and 10000000 being the Logarithms corresponding to them.
Geometrical Mean Pro­portionals.   Arithmetical Logar. mean Proportionals
31622777 First, 05000000
56234132 Second, 07500000
74989426 Third, 08750000
86596435 Fourth, 09375000
93057205 Fifth, 09687500
89768698 Sixth, 09531250
91398327 Seventh, 39609375
90579847 Eighth, 09570312
90173360 Ninth, 09550781
89970801 Tenth. 09541015

Which is thus made: In the first Table a Geometrical Mean [...]oportional between 10 000 000 and 100 000 000 the first Number of it; then another Mean between that and [...]e same last 100000000, gives the second; and so to the [...]th, 93057205. Which, since it is already greater than the [...]ovenary, another Mean between it and the precedent fourth, [...]comes in order a sixth, but sensibly less than the Novenary. [...]herefore between it and the fifth you will have a seventh [...]ean yet greater than the Novenary; and between the sixth [...]d seventh, an eighth, somwhat nearer to the Novenary, but [...]t yet sensibly equal, but somewhat bigger; moreover between [...]e sixth and eighth you will have a ninth, between the ninth [...]d sixth a tenth gradually approaching nearer the Novenary, but [...]t somewhat sensibly differing from it. Now if you con­ [...]ue this inquiry of a mean Proportional between this tenth, [...] somewhat too little, and the precedent ninth as somewhat [...]o big, and so onwards, you will at length obtain the Num­ [...]r 8999 9998, only differing two in the last place from the [...]ovenary Number augmented by seven Cyphers, and conse­ [...]ently insensibly from the Novenary it self. But for the Lo­ [...]arithm of this in the second Column, by the same process you [...]e to find Arithmetical Mean Proportionals between every 2 [...]ogarithms answering to every two of the superiour ones, till you [...]nd, e. g. the Logarithm of the tenth Number 09541015, [...]d so at length the Logarithm of the last, not sensibly differ­ [...]g from the Novenary, 09542425.

Thus having found, with a great deal of labour, but also [...]ith a great deal of advantage to those that make use of them, [...]e Logarithms of some of the numbers between 1 and 10, and [...]0 and 100, &c. you may find innumerable ones of the other [...]ntermediate Numbers with much less labour, viz. by the help [...]f some Rules, which may be thus obtain'd from Consect. 3 of [...]e precedent Proposition. The Sum of the Logarithms of the [...]umber Multiplying and the Multiplicand, gives the Logarithm of the [...]roduct. 2. The Logarithm of the Divisor subtracted from the Lo­ [...]arithm of the Dividend, leaves the Logarithm of the Quotient: [...]he Logarithm of any number doubled, is the Logarithm of the Square, [...]ripled of the Cube, &c. 4 The half Logarithm of any number is [...]he Logarithm of the Square Root of that number, the third part of [...] the Cube Root, &c. Thus, e. g. if you have found the Lo­garithm [Page 78] of the number 9, after the way we have shewn, by th [...] same reason you may find the Logarithm of the number 5 ( vi [...] by finding mean Proportionals between the second and the fi [...] number of our Table, and between their Logarithms, &c. and by means of these 2 Logarithms you may obtain several others: First, since 10 divided by 5 gives 2; if the Logarith [...] of 5 be subtracted from the Logarithm of 10, you'l have th [...] Logarithm of 2, by Rule the second. Secondly, since 10 m [...] tiplied by 2 makes 20, and by 9 makes 90, by adding th [...] Logarithms of 10 and 2, and 10 and 9, you'l have the L [...] garithms of the numbers 90 and 20, by Rule 1. Thirdly Since 9 is a Square, and its Root 3, half the Logarithm of [...] gives the Logarithm of 3, by Rule 4. since 90 divided by [...] gives 30, the Logarithm of this number may be had by s [...] tracting the Logarithm of 3 from the Logarithm of 90, b [...] Rule the second. Fifthly, 5 and 9 squared make 25 and 8▪ the Logarithms of 5 and 9 doubled, give the Logarithms [...] these numbers, by Rule 3. In like manner, sixthly, the Su [...] of the Logarithms of 2 and 3, or the Difference of the L [...] garithms of 5 and 30, give the Logarithm of 6, and the Su [...] of the Logarithms of 3 and 6, or 2 and 9, gives the Log [...] rithm of 18; the Logarithm of 6 doubled, gives the Logarithm of 36, &c. And after this way you may find and reduce it to Tables, the Logarithms of Vulgar Numbers from 1 to 100 [...] (as in the Tables of Strauch. p. 182, and the following) or [...] 100000 (as in the Chiliads of Briggs) But as to the manner [...] deducing the Tables of Sines and Tangents from these Logarithms of Vulgar Numbers, we will shew it in Schol. of Pr [...] 55, only hinting this one thing before-hand; that this Artifi [...] of making Logarithms is elegantly set forth by Pardies in hi [...] Elements of Geometry, pt 112. by a certain Curve Line then [...] called the Logarithmical Line; by the help whereof he suppose Logarithms may be easily made; and having found those o [...] the numbers between 1000 and 10000, he shews, that all o­thers may be easily had between 1 and 1000. Wherefore w [...] shall Discourse more largely in Schol. Definit. 15. lib. 2.

Proposition XXI.

IF the first Term of never so many Continual Proportionals, be sub­tracted from the last, and the Remainder divided by the name of the Reason or Proportion lessen'd by Ʋnity, the Quotient will be equal to the Sum of all except the last.

Demonstration.
  ea
  e (powerof2) a
  e (powerof3) a
  e (powerof4) a
  e (powerof5) a
The last Term less the first e (powerof6) a− a
Divided by the name of the Reason lessen'd by unity.
e−1
* Quote. e (powerof5) a+ e (powerof4) a+ e (powerof3) a+ e (powerof2) a+ ea+ a; And it is evident from the Operation, that the same will always happen tho the number of Terms be con­tinued never so far.
e−1
e−1
  e−1
  e−1
  e−1
  e−1
CONSECTARYS.

I. WHerefore in adding never so great a Series of Geo­metrical Proportionals, since it is enough that the first and last Term, and the Name of the Reason be known, by this Prop. and having found at least some of the Terms of the Proportion, any other may be afterwards found, whose place will be compounded of the places of the two Antecedent ones, according to Consect. 2. Prop. 20. viz. by Multiplying the Terms answering to the two above-mentioned places, and dividing the Product by the first Term; thence it will be very easie to add a great Series of Proportionals into one Sum, tho the particular separate Terms remain almost all of them unknown.

SCHOLIUM.

THese are the same Practical Arithmetical Rules concerning Geometrical Progressions; for the illustration of whic [...] Swenterus in Delic. has given us so many pleasant Examples, li [...] 1. Prop. 59. and fol. First of all, that famous Example is of th [...] kind which relates to the Chequer-work'd Table or Board t [...] fling Dice on, with its 64 little Squares, which Dr. Wa [...] has translated out of the Arabick of Ebn Chalecan, into Latin in Oper. Mathem. part. 1. Chap. 31. for the illustration of whic [...] we have heretofore composed an Exercitation, and shall he [...] only note these few things: If there are supposed 64 Terms [...] double Proportion from Unity, and the first of them, note [...] with their local Numbers, are these that follow;

1 2 4 8 16 32 64 128
  I II III IV V VI VII

You may have the Term of the 13th place, 8192, by mu [...] tiplying together the VIth and VIIth place; and the Ter [...] of the XXVIth place, by squaring or multiplying this new Product again by it self, and moreover the Term of the L [...] place, by multiplying that Product again by itself; and furthe [...] more the Term of the LIXth place, by multiplication of the number last found by the number of the VIIth place, and lastly the Term of the LXIIId place ( i. e. the last in the proposed Series) by multiplying this last of all by the number of the IV [...] place.

II. Moreover you may, by this Art, collect infinite Seri [...] of Proportional Terms into one Sum, altho it is impossible [...] run over all the Terms separately, because infinite. e. g. in [...] continued Series of Fractions, decreasing in a double Propo [...] tion ½ ¼, ⅛ [...], [...], &c. ad infinitum, if you take them bac [...] wards, you may justly reckon a Cypher or 0, for the fi [...] Term (for between ½ and 0 there may be an infinite Numbe [...] of such Terms) and the infinite Sum of these Terms will b [...] precisely equal to Unity; for subtracting the first 0, from the last ½, and the remainder ½ being divided by the name of th [...] Reason lessened by 1, i. e. by I. which divides nothing, th [...] [Page 81] Quotient ½ is the Sum of all the Terms excepting the last, by Prop. 21. and so the last ½ being added, the Sum of all in that Series will be I. Now if the last is not ½ but I, the Sum of all will necessarily be 2; if 2 be the last, the Sum of all will be 4; in a word, it will be always double the last Term.

III. And since in this case the Sum of all the precedent Terms is equal to the last Term, the one being subtracted from the other, there will remain nothing, i. e. ½−¼−⅛− [...]− [...], &c. in Infinitum, is = 0, and also 1−½−¼, &c. or 2−1−½−¼. &c. = 0.

IV. In like manner the Sum of infini [...]e Fractions decreasing in triple Reason in an infinite Series (⅓+ [...]+ [...]+ [...], &c.) will be equ [...]l to ½: for if from the last ⅓ (again in an inverted Order) you subtract the first 0, and the Remainder ⅓ be divided by the name of the Reason lessen'd by Unit, i. e. by 2, the Quotient ⅙ will be the Sum of all the antecedent Terms, and adding to this last ⅓ or [...] the Sum of all will be [...] or ½.

V. Thus an infinite Series of Fractions decreasing from ¼ in a Quadruple Proportion (¼+ [...]+ [...] &c.) is equal to ⅓; for subtracting the first 0 from the last ¼, and the remainder ¼ being divided by the name of the Proportion, i. e. by 3, you will have [...] the sum of all except the last, and adding also the last ¼ or [...], you'l have the whole Sum [...] or ⅓.

VI. Thus also an infinite Series decreasing from ⅕ in a Quin­tuple Proportion (⅕+ [...]+ [...], &c.) is equal to ¼: ⅙+ [...]+ [...], &c. is equal to ⅕ &c. and so any Series of this kind is equal to a Fraction, whose Denominator is less by an Unit than the Denominator of the last Fraction in that Series.

VII. Generally also, any infinite Series of Fractions decreasing according to the Proportion of the Denominator of the last Term, and having a common Denominator less by an unit than the Denominator of the last Term (e. g. ⅔+ [...]+ [...], &c. or ¾+ [...]+ [...], &c. or ⅘+ [...]+ [...], &c.) is equal to Unity, after the same way as the Series Consect. 2. which may be compre­hended under this kind, and which may be demonstrated in all its particular cases by the same method we have hitherto made use of, or also barely subsumed from Consect. 4, 5, and 6. For since ⅓+ [...]+ [...], &c. is equal to ½; ⅔+ [...]+ [...] will be equal to [...], or 1, and so in the rest.

[Page 82]VIII. Particularly the sum of [...], &c, decreasing in a Quadruple Proportion, is equal to [...]; and the sum of [...], &c. is equal to [...]; and the sum of [...], &c. decreasing in Octuple Proportion, is equal to ⅛: For subtracting the first Term 0, and dividing the remainder by the name of the Reason lessen'd by 1, i. e. by 3, the Quotient [...] gives the sum of all except the last. This therefore ( viz. [...]) being ad­ded, the sum of all will be [...] or [...]: In like manner [...] being divided by the name of the Reason lessen'd by Unity, the Quo­tient will give [...], and adding the last, the sum of all will be [...] i. e. ⅛. So that hence it is evident, that [...], &c. or − [...], &c. in Infinitum, will be equal to noth­ing; also ⅛− [...]− [...]− [...] &c. = 0.

IX. The Sum of a simple Arithmetical Progression (i. e. ascending by the Cardinal Numbers) continued from 1, ad Infinitum, is S [...] ­duple of the Sum of the same number of Terms, each of which is equ [...] to the greatest; or on the contrary, this latter Sum is double of the fo [...] ­mer. We might have subsumed this in Consect. 4. Prop. 16. for, prefixing a Cypher before Unity, it will be a case of that Con­sectary, the Sum of the Progression remaining still the same. B [...] that this is true, in an infinite Series beginning from Unity (f [...] in a finite or determinate one, the proportion of the Sum is al­ways less than double, tho it always approaches to it, and come so much the nearer by how much greater the Series is) [...] shall now thus Demonstrate: To the Sum of three Terms, [...] 2, 3, i. e. 6, the sum of as many equal in number to the greatest, i. e. 9, has the same Proportion as 3 to 2; but t [...] the sum of six Terms, 1, 2, 3, 4, 5, 6, i. e. 21, the su [...] of as many equal to the greatest, i. e. 36, has the same pro­portion as 3 to 1+¾, that is, as 3 to 2−¼, the decrease be­ing ¼: but to the sum of 12 Terms, which may be found b [...] Consect. 1. Prop. 16.=78, the sum of so many equal to the greatest, viz. 144. has the same proportion (dividing both sid [...] by 48) as 3 to 1 [...], i. e. 3 to 1+½+⅛ (for 24 make ½, a [...] the remainder [...] is the same as ⅛) that is, as 3 to 2−¼− [...] the decrement being now ⅛. Since therefore, by doubling th [...] number of Terms onward, you'l find the decrement to be [...], an [...] so onwards in double Proportion; the sum of an infinite Num­ber of such Terms, in Arithmetical Progression, equal to th [...] greatest [Page 83] will be to the sum of the Progression from 1, ad Infinitum, as 3 to 2−¼−⅛− [...], &c. that is, by Consect. 2 and 3, as 3 to 2−½, that is, as 3 to 1 ½, or as 2 to 1. Q.E.D.

X. The Sum of any Duplicate Arithmetical Progression ( i. e. a Progression of Squares of whole numbers ascending) continued from 1 ad Infinitum, is subtriple of the Sum of as many Terms equal to the greatest as is the number of Terms: For any such finite Progression is greater than the subtriple Proportion, but approaches nearer and nearer to it continually, by how much the farther the Series of the Progression is carried on. Thus the Sum of 3 Terms 1, 4, 9=14 is to thrice 9=27 as 1 [...], or 1 [...], or 1+½+ [...] to 3 (dividing both sides by 9,) the Sum of six Terms, 1, 4, 9, 16, 25, 36, viz. 91. to six times 36, i. e. to 216 (dividing both sides by 72) is as 1+¼+ [...] to 3; and the Sum of 12 Terms 650, to 12 times 144, i. e. to 1728 (dividing both sides by 576) is as 1+⅛+ [...] to 3, &c. the Fractions adhering to them thus constantly decrea­sing, some by their half parts, others by three quarters (for [...] is [...]; therefore the first decrement is [...] and [...], is [...]; there­fore the second decrement is [...], &c.) Wherefore the Sum of the Infinite Progression will be to the Sum of the like number of Terms equal to the greatest, as [...], &c. to 3, that is, by Consect. 3 and 8, as 1 to 3. Q.E.D.

XI. The Sum of a triplicate Arithmetical Progression ( i. e. ascending by the Cubes of the Cardinal Numbers) proceeding from 1 thro' 27, 64, &c. ad Infinitum, is Subquadruple of [...]he Sum of the like number of Terms equal to the greatest. For the Sum of 4 Terms, 1, 8, 27, 64, i. 100, to 4 times 64, i. e. 256 (dividing both sides by 64) will be found to be as 1+½+ [...] to 4; but the Sum of 8 Terms, 1, 8, 27, 64, 125, 216, 343, 582, i. e. 1296 to 8 times 512, that is, 4096 (dividing both Sides by 1024.) will be found to be as 1+¼+ [...] to 4, &c. The adhering Fractions thus [Page 84] constantly decreasing, the one by their ½ part, the others by [...] (for [...] is [...], and [...] is [...], &c. Wherefore the Sum of the In­finite Progression will be to the Sum of a like (Infinite) num­ber of Terms, equal to the greatest, as [...], &c. [...], &c. to 4; that is, by Consect. 3 and 8, as 1 to 4. Q. E. D.

XII. The Sum of an Infinite Progression, whose greatest Term is a Square Number, the others decreasing according to the odd numbers 1, 3, 5, 7, &c. is in Subsesquialteran Pro­portion of the Sum of the like number of equal Terms, i. e. as 2 to 3. For the Sum of three such Terms, e. g. 9, 8, 5, i. e. 22 to thrice 9, i. e. 27. is (dividing both sides by 9) [...] 2 [...], viz. [...] to 3, or 2+½− [...] to 3. But the Sum of s [...] such Terms, 36, 35, 32, 27, 20, 11, i. e. 161, to six time 36, i. e. 216 (dividing both sides by 72) is as 2+¼− [...], &c. the adhering Fractions thus always decreasing, some by ½, o­thers by ¾, as above in Consect. 10. Wherefore the Sum of the Infinite Progression will be to the Sum of the like number of Terms equal to the greatest, as [...], &c. to 3, i. e. by Consect. 3 and 8, as 2 to 3. Q E. D.

SCHOLIUM II.

THus we have, after our method, demonstrated the chie [...] Foundations of the Science or Method, or Arithmetick [...] Infinites, first found out by Dr. John Wallis, Savilian Professo [...] of Geometry at Oxford, and afterwards carried further by Det [...] ­lerus Cluverus, and Ismael Bullialdus. And from these Founda­tions we will in the following Treatise demonstrate, and that directly and à priori, in a few Lines, the chief Propositions o [...] [Page 85] Geometry, which the Antients have spent so much labour, and composed such large Volumes to demonstrate, and that but in­directly neither.

Proposition XXII.

THe Powers of Proportionals whether continuedly or discretely, such as the Squares, Cubes, &c. are also Proportional.

Demonstration.
Continual Proportionals. Discrete Proportionals.
a ea e (powerof2)a e (powerof3)a a ea b eb
Squares aa e (powerof2)a (powerof2) e (powerof4)a (powerof2) e (powerof6)a (powerof2) a (powerof2) e (powerof2)a (powerof2) b (powerof2) e (powerof2)b (powerof2)
Cubes a (powerof3) e (powerof3)a (powerof3) e (powerof6)a (powerof3) e [...]a (powerof3) a (powerof3) e (powerof3)a (powerof3) b (powerof3) e (powerof3)b (powerof3)

Q E. D.

SCHOLIUM.

YOu founded in this Truth, 1. the Reason of the Multipli­cation and Division of Surd Quantities: For since from the Nature and Definition of Multiplication, it is certain, that 1 is to the Multiplier as the Multiplicand to the Product (for the multiplicand being added as many times to it self as there are Units in the Multiplier, makes the Product) if the √5 is to be multiplied by √3, then as 1 to the √3, so the √5 to the Product; and, by the present Proposition, as 1 to 3, so 5 to the □ Product, i. e. to 15. Wherefore the Product is √15; and so the Rule for Multiplying Surd Quantities is this: Multiply the Quantity under the Radical Signs, and prefix a Radical Sign to the Product. Eucl. lib. 6. prop. 22. Likewise since it is certain from the Nature of Division, that the Divisor is to the Dividend as 1 to the Quotient (for the Quotient expresses by its Units how many times the Divisor is contained in the Dividend) if the √15 is to be divided by √5, you'l have √5 to the √15 as 1 to the Quo­tient, and, by the present Scholium, 5 to 15, as 1 to the □ of the Quotient, i. e, to 3. Therefore the Quotient is the Root of 3, and so the Rule of dividing Surd Quantities this; viz. [Page 86] Divide the Quantities themselves under the Radical Signs, and pre­fix the Radical Sign to the Quotient.

II. Hence also flows the usual Reduction in the Arithme­tick of Surds, of Surd Quantities to others partly Rational, and on the contrary, of those to the form of Surds, e. g. If you would reduce this mixt Quantity 2 a√ b, i. e. 2 a multiplied by the √ b, to the form of a Surd Quantity; which shall all be con­tained under a Radical Sign; The Square of a Rational Quan­tity without a Sign 4 aa, if it be put under a Radical Sign, in this form √4 aa, it equivalent to the Rational Quantity 2 a; but the √4 aa being multiplied by √ b makes √4 aab. by N [...] 1. of this Scholium. Therefore √4 aab is also equivalent to the Quantity first proposed 2 a√ b. Reciprocally therefore, if th [...] form of a meer Surd Quantity √4 aab, is to be reduced to on [...] more Simple, which may contain without the Radical Sig [...] whatever is therein Rational, by dividing the Quantity com­prehended under the sign √ by some Square or Cube, &c. as here by 4 aa, (i. e. √4 aab by √4 aa, i. e. 2 a) the Quotient wil [...] be √ b, which multiplied by the Divisor 2 a, will rightly ex­press the proposed Quantity under this more simple Form 2 a√ [...] Which may also serve further to illustrate the Scholia of Prop. 7. and 10.

Proposition XXIII.

IF there are four Quantities Proportional, (a, ea, b, eb) they will be also Proportional,

  • 1. Inversly. ea to a as eb to b.
  • 2. Alternatively,
    Eucl. 15, 16. v. 9 10, 13, vi [...].
    a to b as ea to eb.
  • 3. Compoundedly,
    18, v.
    a+ ea to ea, so b+ eb to eb.
  • 4. Conversly, a+ ea to a as b+ eb to b.
  • 5. Dividedly,
    17, v.
    a− ea to
    • ea as b− eb to
    • or a
      • eb
      • or [...]
  • 6.
    Eucl. 15, 16. v. 9 10, 13, vi [...].
    By a Syllepsis, a to ea as a+ b to ea+ eb.
  • 7. By a Dialepsis, a to ea as a− b to ea− eb.

Which are all manifest, by comparing the Rectangles of the Means and Extremes according to to Prop. 19. and its Consect. 1. [Page 87] or by dividing any of the Consequents by their Antecedents, ac­cording to Def. 31.

Proposition XXIV.

IF in a Eucl. 3, 20, 22. lib. v. 14. vii. double Rank of Quantities you have

  • as a to ea,
  • so b to eb,

and also

  • as ea to oa,
  • so eb to ob,

then you'l have also by proportion of Equality orderly pla­ced,

  • as the first a, to the last oa, in the first Series;
  • so the first b, to the last ob, in the second Series.

Which is manifest from the Terms themselves.

Proposition XXV.

BUt Eucl. 21, 23. lib. v. if they are disorderly plac'd

  • as oa to ea
  • † so eob to ob

* as ea to a † so ob to eb, * you'l have here again by proportion of Equality,

  • as the first oa to the last a, in the first Series;
  • so the first eob to the last eb, in the second Series.

As is evident from the Rectangles of the Extremes and Means, as also from the very Terms.

Proposition XXVI.

IF Eucl. 1. 12. v. 5. 6. 12. vii. as the whole ea to the whole a, so the part eb to the part b; then also will

the Remainder Remainder Whole Whole
ea− eb to the a− b, as the ea to the a.

This is evident from the Rectangle of the Extremes and Means, both which are eaa− eab. Q.E.D.

Proposition XXVII.

REctangles or Products having one common Efficient or Side, are one to another as the other Efficients or Sides.

Demonstration.

Suppose the Products to be ab and ac, having the common Efficient a; I say they are

  • as b to c, so ab to ac.

Which is evident at first sight, by comparing the Products of the Extremes and Means, and also fully shews, that other way of proving Proportionality, whereby by dividing the Conse­quents by their Antecedents, the identity or sameness of the Quotients are wont to be demonstrated.

SCHOLIUM I.

I. THe Reduction of Fractions either to more compounded or more simple ones is founded on this Theorem; on the one hand by multiplying, on the other by dividing, by the same quantity, both the Numerator and the Denominator, as, e. g, [...] and [...] and [...], ⅓, [...], [...], &c. are in reality the same Fractions. And

II. The Reduction of Fractions to the same Denomination, as if [...] and [...] are to be changed into two o­thers that shall have same Denominator; Eucl. 5 & 19. lib. v. 7 & 11. l. 7. this is to be done by multiplying the Denominators together for a new Denominator, Besides se­veral other Prop. see also the 17 & 18 lib. vii. and each Numerator by the Denominator of the other for a new Numerator, and you'l have for the two Fractions above — [...] and [...]

SCHOLIUM II.

WE will here for a conclusion of Proportionals, shew the way of cutting or dividing any Quantity in Mean and Extreme Reason, viz. if for the greater Part you put x, the less [Page 89] will be a− x; and so by Hypoth. these three, a, x and a− x, will be proportional, by Def. 34. Therefore by Prop. 17. the Product of the Extremes aa− ax = to the Square of the Mean xx, and (adding on both sides ax) aa= xx+ ax; and more­over adding on both sides ¼ aa, you'l have [...] aa= xx+ ax+¼ aa. Now this last Quantity, since it is an exact Square, whose Root is x+½ a, you'l have √ [...] aa= x+½ a, and (subtracting from both sides [...] a) √ [...] aa−½ a= x.

Now therefore we have a Rule to determine the greater part of a given Quantity to be divided in Mean and Extreme Rea­son, viz. if the given Quantity be a Line, e. g. AB= a (Fig. 58.) join to it Eucl. 11. lib. II. & 30. lib. vi. at Right Angles AC=½ a: Wherefore by the Theorem of Pythagoras from Schol. Definit. 13. the Hypo­thenuse CB, or, which is equal to it, CD=√¾ aa; and conse­quently AC=½ a being taken out of CD, the Remainder AD, or AE, which is equal to it, will be = x, the greatest part sought; according to Euclid, whose Invention this first Specimen of Analysis, by way of Anticipa­tion, reduces to its original Fountain. As for Numbers (tho none accurately admits of this Section) the sense of the Rule, or which is all one as to the thing it self, is this: Add the Squares of a whole Num­ber and its half, and subtract the said half from the Root of the Sum (which can't be had exactly, since it is √ [...].

CAAP. V. Of the Proportion or Reasons of Magnitudes of the same kind in particular.

Proposition XXVIII.

TRiangles and Parallelograms, also Pyramids and Prisms and Paral­lelepipeds, lastly Cones and Cylinders, each kind compared among themselves, if they have the same Altitude, are in the same Proportio [...] to one another as their Bases.

Demonstration.

This and the following Proposition might have been by [...] bare Subsumption added, as Consectarys to the precedent; fo [...] the Altitudes in the one, and Bases in the other, may be looke [...] on as common Efficients, and the Magnitudes mentioned as their Products: But for the greater distinction sake, we will thus De­monstrate them more particularly.

I. If the equal Altitudes of two Triangles, o [...] Eucl. Prop. 1. lib. vi. two Parallelograms, are called b and [...] Base of the one a, and of the other ea; these Products will be ba and bea, the other ½ ba and ½ bea, by Def. 28. Schol. 2.

II. Likewise the equal Altitudes of two Priso [...] Prop. 5.6. lib. xii. 25, 32. xi. & Cons. 30 & 31 of the same or Pyramids, may be called b, and the Pro­portion of their Bases expressed by a and ea; a [...] the Prisms will be among themselves as ba to be [...] and the Pyramids as ⅓ ba to ⅓ bea, Prop. 11. lib. xii. by the sai [...] Schol. Num. 3. Prop. 35, 36, 37, 38, 39, 40, lib. lib. 1 & 29, 30, 31. lib. xi.

III. There is also the same Proportion of Cylin­ders and Cones as of Pyramids and Prisms, by Consect. 4. Definit. 17. But,

as a to ea so is ba to bea.
−½ ba to ½ bea.
−⅓ ba to ⅓ bea.

Q.E.D.

CONSECTARY.

THerefore Magnitudes of the same kind upon the same or equal Bases (dgr;) and of the same heighth, are equal among themselves, and the contrary.

Proposition XXIX.

TRiangles and Parallelograms, Pyramids and Prisms and Parallele­pipeds, Cones and Cylinders, being on equal Bases, are in the same Proportion as their heighths. Schol prop. 1. l. 6, 12, 13, 14

Demonstration.

Let all their Bases be called a, and the Proportions of their Heighths be as b to eb: Therefore, 1. the Parallelograms, Pa­rallelepipeds and Cylinders, are one to the other of the same kind, as ba to eba; the Triangles as ½ ba to ½ eba; the Pyra­mids and Cones as ⅓ ba to 3 eba, by Def. 28. Schol. 2. But,

as b to eb, so is ba to eba.
  and ½ ba to ½ eba.
  and ⅓ ba to ⅓ eba.

Q.E.D.

Proposition XXX.

EQual lib. 6. prop. 14. 15. l. 11.34. l. 12.11, and its Coroll. also Prop. 15. Triangles, Parallelograms, Prisms, Parallelepipeds, also equal Pyramids, Cones, and Cylinders, have their Bases and Heighths reciprocally Proportional.

Demonstration.

For if for the equal Triangles you put ½ ab, for the Cones and Pyramids ⅓ ab, and for the rest ab;

whether the Bases of the equal Quan­tities are supposed to be a, and so the Altitudes [Page 92] on both sides b; or if the Base of the one be a and b the Al­titude, but the Base of the other b and the Altitude a, you'l certainly have eitherways,

as a to a, so Reciprocally b to b; the Base of the former to the Base of the latter, as the Altitude of the latter to the Altitude of the former, or,

as a to b, so Reciprocally a to b. Q.E.D.

CONSECTARY.

AND those Magnitudes of the same kind, whose Bases and Altitudes are thus Reciprocal, are equal by Prop. 18. for the Product or Rectangle of the Extremes is ab, and that of the Means ba.

Proposition XXXI.

TRiangles, Parallelograms, Prisms, Parallelepipeds, Pyramids, Co [...] and Cylinders, each kind compared among themselves, are in the Proportion compounded of the Proportion of their Altitudes and Bases. Prop. 23. lib. 6.

Demonstration.

Suppose the Base of the one to be a, and the other ea, and the Altitude of the one b, of the other ib; therefore the one will be to the other,

as a b to eiab,

or ½ ab to ½ eiab,

or ⅓ ab to ⅓ eiab; i. e. every where as a to ei [...] i. e. in Proportion compounded of a to ea, and of b to ib, by Consect. 2. Def. 34. Q.E.D.

SCHOLIUM.

FRom what we have hitherto Demonstrated, we may not only make an estimate of Magnitudes of the same kind compared together, which is easie to any one who attentively considers them; but also with F. Morgues, deduce a General Rule of expressing the Proportions of any Rectilinear Planes or Solids, contained under [Page 93] Plane Surfaces, by the proportion of one Right Line to ano­ther. For since the one may be resolved into Triangles, and the other into Pyramids, having first two Rectilinear Planes given and thus resolved, upon a Right Line I make the ▵ abc (Fig. 49.) Equal to one of the Triangles of either of the Planes e. g. to ABC; then having drawn the Parallel cm, if the ▵ BCD has the same Altitude with the former, you need only joyn the Base BC to the Base ab. But if the Altitude DS is greater than the Altitude of the other e. g. by ⅕, then you must make b f equal to the Base bc augmented by a fifth part, and the Trian­gle bcf will = BCD, and the whole acf = to the Rectilinear Figure ABCD. If now therefore I likewise make another Tri­angle ghi equal to another Rectilinear Figure between the same Parallels, then will the ▵ acf be to the ▵ ghi, that is, the Right Lined Figure ABCD to the Right Lined Figure FGHIK, as af to gh, by Prop. 28. 2. Having 2 Right Lined Solids gi­ven, and having resolved them into Triangular Pyramids, they may be transferr'd between 2 parallel Planes, viz. by augment­ing or diminshing their Triangular Bases reciprocally, ac­cording to the excess or defect of their Altitudes, as was done above with the Linear Bases; then those Triangular Bases on both sides may be converted into one Triangular Base, and con­sequently each Solid into a Pyramid equal to it self; which two Pyramids will be one to the other as their Triangular Bases. And because the Proportions of these Bases may be reduced to the Proportion of two Lines each to the other, by N o 1. of this; therefore also the Reason or Proportion of the two Solids may be expressed by the Proportion of two Lines. Q.E.D.

Proposition XXXII.

CIrcles Eucl. Prop. 2. l. 12. are in the same Proportion to one another as the Squares of their Diameters.

Demonstration.

Suppose a to be the Diameter of one Circle, and b of another; then by Definit. 31. Consect. 1. the Area of the one will be ¼ eaa, and that of the other ¼ ebb. But as aa to bb so is ¼ eaa to ¼ ebb by Consect. 1. Prop. 19. Q E. D.

CONSECTARY I.

THe same will in like manner be manifest of like Sectors Circles, while for the parts of the Periphery you put and ib, as for the wholes we put ea and eb: for thus the A [...] of the one will be ¼ iaa, and of the other ¼ ibb.

CONSECTARY II.

CYlinders whose Altitudes are equal to the Diameters of th [...] Bases, are in proportion to one another as the Cubes their Diameters; for the Cylinders will be ¼ ea (powerof3) and ¼ eb (powerof3), Cubes a (powerof3) and b (powerof3).

CONSECTARY III.

HEnce also (whatever the Reason of the Sphere is to the Cylinder of the same Diameter and Heighth; which will hereafter Demonstrate, and which in the mean while will denote by the name of the Reason y) I say, hence Sphe [...] which have the same Proportion to one another as these Cyli [...] ders ( viz. as ¼ ea (powerof3) to ¼ eb (powerof3), so ¼ yea (powerof3) to ¼ yeb (powerof3)) will also (by C [...] sect. 1.) be in the same proportion as the Cubes, a (powerof3) to b (powerof3) is also evident from these Terms themselves.

Proposition XXXIII.

THE Angle Prop. 18 lib. 12. Eucl. Prop. 20. l. 3. at the Center of any Circle ACB (Fig. 60) to an Angle at the Circumference which has the same Arch its Base ADB, as 2 to 1.

Demonstration.

The truth of this has already appear'd fro [...] Schol. Definit. 10. N o 3. but here we will demo [...] strate it otherwise in its three Cases, after E [...] clids way. In the first Case DE being conceive Parallel to CB, by Def. 11. Consect. 1 and 2. th [...] External Angle ACB is = to the Internal A [...] gle [Page]

Pag. 95.

61

62

63

64

65

66

67

68

69

70

71

72

[Page 95] ADE, and the Angle BDE, is equal to the alternate Angle BCD, i. e. to the other at the Base CDB, by Consect. 2. Definit. 13. Therefore BDE is as 1, and CDE, i. e. ACB as 2.

In the second Case the whole ECB is double of the whole EDB, and the subtracted Angle ECA is double of the sub­tracted Angle EDA, by Case 1. Therefore the Remainder ACB is also double of the Remainder ADB, by Prop. 26. In the third Case the part ECA is double of the part EDA, and also the part ECB is double of the part EDB, by Case 1. Therefore the whole ACB is double the whole ADB. Q.E.D.

CONSECTARYS.

I. HEnce all Angles ADB Eucl. prop. 21. 27. 31. lib. 3. in the same Segment are equal, and the Angle ADB ( Fig. 61.) in a Semicircle is a Right one; because the Aperture at the Center answering to it, ACB contains two Right Angles: The Angle in a less Segment than a Semicircle EDF, is greater than a Right one; because the Aperture at the Center EGHFC answering to it, comprehends more than two Right Angles. An Angle, last­ly, in a Segment greater than a Semicircle GDH, is less than a Right one; because its double at the Center GCH is less than two Right ones. All which we have already otherwise demon­strated in Schol. Def. 10. N o 6.

II. All the three Angles prop. 32. l. 1 of any Triangle ABD taken together, are equal to two Right ones; because they are the half of the three at the Center C, which always make 4 Right ones, by Definit. 8. Consect. 2.

III. Therefore any external Angle IAB, is e­qual to the two Internal opposite ones at B and D; because that, as well as they with the other contiguous to them BAD, make two Right ones, by Consect. 1. of the same Definit.

IV. And the greatest Side of a Triangle, because it insists on a Eucl. prop. 21. 27. 31. lib. 3. greater Arch of a Circumscribed Circle, does also ne­cessarily subtend a greater Angle, by vertue of Consect. 1. hereof.

Proposition XXXIV.

IN Equiangular Triangles (ACB and abc, Fig. 62.) the Sides [...] bout the equal Angles are Proportional, viz. as AB to BC, so ab to bc, and as BC to CA so is bc to ca. &c. (β)

Demonstration.

For having described Circles thro' the Vertex of each Trian­gle, according to Consect. 6. Definit. 8. by reason of the supp [...] sed equality of the Angles A and a, B and b, C and c, th [...] Arches also AB and ab, &c. will necessarily agree in the number of Degrees and Minutes, by the foregoing 33 Prop, a [...] so also the Chords AB and ab, BC and bc, &c. will agree in th [...] number of Parts of the Radius or whole Sine ZA and za, [...] Consect. 2. Definit. 10. Wherefore as many such Parts as A [...] has, whereof az has also 10000000, so many such also will a [...] have, whereof az has also 10000000, &c. Therefore AC [...] to CB as ac to cb, &c. Q. E. D.

CONSECTAYS.

I. WHerefore by the same necessity the Bases of such T [...] angles AB and ab, will be proportional to their A [...] titudes CD and cd, as being Right Sines of the like Arches ( [...] and cb, or rather CE and c e; and so for similar or like Tria [...] gles (and consequently also Parallelograms) we may rightly suppose that their Bases are as a to ea, and their Heighths as b [...] eb; tho we must not immediately conclude on the contrary, tha [...] because their Bases and Altitudes are so, therefore they are S [...] milar.

II. As also in Similar Parallelepipeds it will be manifest [...] any attentive Person, that the Bases are in a duplicate Propo [...] tion of the Altitudes. For since the Planes of Similar Solids a [...] equal in number, and Similar each to the other, if for A [...] ( Fig. 63.) we put a, and for BC b, AB will = ea and BC= eb; and so that Basis will be to this as ab to eeab. Moreover having let fall the Perpendiculars EH and EH, the Triangle [Page 97] [...]BH and EBH are similar, and by putting c for BE, BE will be ec, putting also d for EH, EH will consequently be ed. But the Reason of the Base ab to the Base eeab, is du­plicate of the Reason of d to e d, by Def. 34. Wherefore in Similar Parallelepipeds we may rightly suppose, that their Bases are as a b to eeab, or as a to eea, and their Altitudes as d to ed.

SCHOLIUM I.

FRrom this Proposition flows first of all the chiefest part of Trigonometry for the Resolution of Right Angled ▵▵: For since in any Right Angled Triangle, if one side, e. g. AB ( Fig. 64.) be put for the whole Sine, the other BC will be the Tangent of the opposite Angle at A (and in like manner if CB be the whole Sine, BA will be the Tangent of the Angle C;) but if the Hypothenuse AC be made Radius or whole Sine, then the Side BC will be the Right Sine of the Angle A, or the Arch CD described from the Center A, and AB the Right Sine of the Angle C, or the Arch AE, described from the Center C, (we will omit mentioning the Secants, because the business may be done without them) which all follow from Def. 10. Wherefore you may find,

  • I. The Angles.
    • 1. From the Sides by inferring As one leg to the other, so the whole Sine to the Tangent of the Angle opposite to the other Leg.
    • 2. From the Hy­poth. & one side, by inferring As the Hyp. to the W.S. (whole sine) so the given leg to the S. of the opp. angle
  • II. The Sides.
    • 1. From the Hy­poth. and Angles: As the W. S. to the Hypoth. so the Sine of the Angle, opposite to the Leg sought, to the Leg it self.
    • 2. From one Leg and the Angles: As the W. S. to the given Leg, so the Tan­gent of the Angle adjacent to it, to the Leg sought.
    • 3. From the Hypoth. and one of the Sides: Having first found the Angles, it's done by the 2, 1. or by the Pythagorick Theorem.
  • [Page 98]III. The Hypothenuse.
    • 1. From the An­gles and one of the Legs. As the S. of the Angle, opposite to the given Leg, to that Leg, so the W. S. to the Hypoth.
    • 2. From the Legs given; Having first found the Angles its done by the 1. or by the Pythagorick Theorem.

III. Inversly also, if two Triangles ABC and ABC ( [...] the Figure of the present Proposition) have one Angle of o [...] equal to one Angle of the other ( e. g. B and B) and the Sid [...] that contain these equal Angles proportional ( viz. as AB to B [...] so AB to BC) then the other Angles (A and A, C and C will be also equal, and the Triangles similar Eucl prop. 6. lib. 6. for to [...] like Chords AB and AB, BC and BC, there answer by t [...] Hypoth. like or similar Arches, i. e. equal in the number [...] Degrees and Minutes; and to these also there answer equal A [...] gles both at the Periphery and Center.

IV. ( Fig. 65. N o 1.) If Eucl. 2. lib. 6. the Sides of the Angle BA [...] are cut by a Line DE, parallel to the Base BC, the Segments [...] those, Sides will be proportional, viz. AE to EC as AD to BD [...] for by reason of the Parallelism of the Lines BE and BC, th [...] Triangles ADE and ABC are Equiangular: Therefore as th [...] whole BA to the whole AC, so the part AD to the part A [...] and consequently also the remainder EC to the remainder D [...] as the part EA to the part AD, by Prop. 26. and alternative [...] by Prop. 24. EC will be to EA as BD to AD.

SCHOLIUM II.

THere are several useful Geometrical Practices depend [...] this Consectary and its Proposition. 1. That Eucl. l. 9 & 10. l. 6. where [...] we are taught to cut off any part required, e. [...] ⅓ from a given Line AB, and so generally to [...] or divide any given Line AC, in the same pr [...] portion as any other given Line, is supposed [...] be divided in D, (and consequently into as ma [...] equal parts as you please;) viz. if in the fi [...] Case, having drawn any Line AF, you take AD [...] 1, and make DB 2, and having joined CB, dra [...] [Page 99] the Parallel LE: for as AD to DB so is AE to AC; that is, as 1 to 2, by this 4th Consect. therefore AE is one third of the whole AC, &c.

II. A Rule Eucl. 11 & 12, l. 6. to find a third Proportional to the 2 Right Lines given AB and BC (N o 2. Fig. 65.) (or a fourth to three given;) if, viz. having drawn AF at pleasure, you make AD equal to BC, and Joining DB, draw the Parallel EC: For as AB to BC, so AD i. e. BC) to DE. Now if AD be not equal to BC but to another (viz. a) third Proportional, then by the same Reason DE will be a fourth Proportional.

III. Another Rule Eucl. 13. lib. 6 & and Eucl. 8. lib. 6. to find a mean Proportional between two Right Lines given AC and CB; which is done by join­ [...]ng both the Lines together, and from the middle of the whole AB describing a Semicircle, and from C erecting the Perpendi­cular CD: For since the Angle ADB is a Right one, by Con­sect. 1. of the preceding Proposition, and the two Angles at C are Right ones, and those at A and B common to the whole Triangle ADB, and to the two partial ones ACD and BCD, [...]hese two will be Equiangular and Similar to the great one, and consequently to one another: Therefore by the present Proposition, as AC to CD, so CD to CB, Q. E. D. and also as AB to BD so BD to BC, and as AB to AD so AD to AC, &c.

IV. The Analytical Praxis of multiplying and dividing Lines [...]y Lines, so that the Product or Quotient may be a Line; and [...]lso the way of Extracting Roots out of Lines: Which Des Cartes, gives us, p. 2. of his Geom. viz. assuming a certain Line [...]or Unity, e. g. AB (in Fig. 65. N o 2.) if AC is to be multi­ [...]lied by AD, having joined BD, and drawn the Parallel CE, [...]he Product will be AE; for it will be as 1 to [...]he Multiplier AD, so the Multiplicand AC to [...]he Product AE; or if AE is to be divided [...]y AC, having joined EC and drawn the Pa­ [...]allel BD, the Quotient will be AD; (for AC [...]he Divisor, will be to AE the Dividend, as an [...]nit AB to the Quotient AD;) all which are e­ [...]ident from the Nature of Multiplication and Division, and the [Page 100] Precedent Praxes. As also taking CB (in the same Fig. N o 3.) for Unity, if the Square Root is to be extracted out of any o­ther Line AC, this being joined to your Unity in one Line AB▪ and having described thereon a Semicircle, the Perpendicular CD will be the Root sought, as being a Mean Proportional be­tween the two Extremes CB and AC, according to Prop. 17.

V. A Right Line AG which divides Eucl. 3. lib. 6. any given Angle A into two equal Parts ( Fig. 66.) being prolonged, divides the Base BC proportionally to the Legs of the Angle AB and AC For having prolonged CA to E, so that AE shall be = to AB [...] the Angles ABE and AEB will be equal, by Consect. 2. Def. 13. and consequently also equal to each of the halves of the exter­nal Angle CAB, by Consect. 3. of the antecedent Proposition Therefore the lines AG and EB will be parallel, by Cons. 1. Def. 11. Therefore as AC to AE, i. e. to AB, so GC to GB, by Co [...] sect. 3. of this Proposition. Q.E.D.

VI. Hence also there follows further, by conversion of th [...] last inference, as AC+AB to AC, so GC+GB ( i. e. BC) [...] GC; and inversly GC to BC as AC to AC+AB; and lastly alternatively, GC to AC as BC to AC+AB.

N. B. This last Inference follows also immediately from the preceding Consectary. For by reason of the Similitude of the ▵▵ ACG and ECB, as GC to AC so BC to CE, i. e. to AC+AB.

SCHOLIUM III.

FRom these two last Consectarys there an [...] these or two or three Practical Rules, t [...] first whereof shews, how having the two Legs A [...] and AC given, and also the Base BC, to find the Se [...] ­ments GC and GB, made by the Bisection of the Intercrural Angle [...] (viz. by this inference, according to Consect. 6: As the Sum [...] the Sides to one Side (e. g.) AC;) so the Sum of the Segmen [...] of the Base, i. e. the whole Base to one of the Segments, vi [...] that next the said Side GC. 2. It shews on the contrary, how having the Base and one of its Segments given, and moreover the S [...] of the Sides, to find separately the Side AC next the known Segment [Page 101] by inferring as the Sum of the Segments, or the Base BC to the Sum of the Sides, so the given Segment GC to the sought AC: or also, 3dly, Having only the Base and Sum of the Sides given, but not the Segment GC, yet to express its Proportion to the next side AC, — viz. in the Quantities of the given Terms, by putting (by Consect. 6.) for GC the value of the Base BC, and for AC the value of the Sum AB+AC; the great use of which last Rule will appear hereafter in the Cy­clometry (or Quadrature of the Circle) of Archimedes.

VII. In any Triangle ABC ( Fig. of the present Proposition) the Sides are to one another as the Sines of their opposite An­gles: For they are as the Chords of the double Angles at the Center, by Prop. 33. therefore they are also one to another as half those Chords, i. e. by Definit. 10. as the Sines of the half Angles.

SCHOLIUM IV.

HEnce flow two new Rules of Plane Trigonometry, for Oblique-angled Triangles to find, viz.

  • 1. The other Angles:
    • From 2 gi­ven Sides, & an Angle op­posite to one of them: by inferring As the Side opposite to the given Angle to the other Side, so is the Sine of the given Angle to the sine of the angle opposite to the other Side; which being given, the third is easily found.
  • II. The other Sides:
    • From one side and the angles given, As the Sine of the Angle opposite to the given side, to that side; so is the Sine of the Angle op­posite to the side sought to the side sought.

So that this way we have reduced all the Cases excepting one of Plane Trigonometry, and consequently all Euthymetry to their original Foundations (for in that Case of having two Sides, and the included Angle given, we may find the rest by [Page 102] the Resolution of the Obliqueangled Triangle into two Right Angled ones; and so it's done by the Rules we have deduc'd in ( Schol. 1.) I say, excepting one, in which from the three sides of an Obliqueangled Triangle given, you are required to find the Angles: the Rule to resolve which we will hereafter deduce in the 2d Consect. of Prop. 45. from that Theorem which Euclid gives us, lib. 2. Prop. 13.

VIII. Because in in the Right Angled ▵ BAC ( Fig. 67) BC is to CA as CA to CD, by N o 3. of the 2d Schol. of this Prop. the □ of CA will be = ▭ CE, by Prop. 17. In like manner because as CB to BA so is BA to BD; the □ of BA will be = to ▭ BE: Wherefore the two Rectangles BE and CE taken together, that is, the □ of the Hypothenuse BC, will be = to the two □'s BA and CA taken together: Which is the very Theorem of Pythagoras demonstrated two other ways in Schol. of Definit. 13.

SCHOLIUM V.

THis Theorem of Pythagoras as it furnishes us with Rules of adding Squares into one Sum, or subtracting one Square from another; so likewise it helps us to some Foundations where­on, among the rest, the structure of the Tables of Sines relies, &c. Whose use we have already partly shewn in Schol. 1 and 4. 1. If several Squares are to be collected into one Sum, having joined the Sides of two of them so as to form a Right Angle, e. g. AB and BC ( Fig. 68. N o 1.) the Hypothenuse AC being drawn, is the Side of a Square equal to them both; and if this Hypothenuse AC be removed from B to D, and the Side of the third Square from B to E, the new Hypothenuse DE will be the Side of a Square equal to the three former taken together. 2. If the Square of the side MN (N o 2.) is to be subtracted from the Square of the side LM. Having described a Semicircle upon LM, and placed the other MN within that Semicircle, then draw the Line LN and that will be the Side of the remain­ing Square. 3. Having the Right Sine EG of any Arch ED gi­ven (but how to find the Primary Sines we will shew in another place), you may obtain the Sine Complement CG or EF, by the preceding [Page 103] Numb. viz. by subtracting the □ of the given Sine from the □ of the Radius; and moreover the versed Sine GD by subtract­ing the Sine Complement CG from the Radius CD. 4. The Squares of the versed Sine GD, and of the Right sine EG being added together, give the □ of the Chord ED of the same Arch, (which all are evident from the Pythagorick Theorem) and half of that EH gives the Right Sine of half that Arch. 5. From the Right Sine EG you have the Tangent of that Arch, if you make, as the Sine Complement CG to the Right Sine GE, so the whole Sine CD to the Tangent G I. 6. Lastly, From these Data you may also have the Secants (if required) thus, as the Sine Complement CG to the W. S. CE, so the W. S. CD to the Secant CI; or as the Right Sine EG to the W. S. E.C. so the Tangent ID to the Secant IC; both which are e­vident by our 34th Proposition.

Consect. 9. If the Quadrant of a Circle (CBEG, Fig. 70.) be inclined to another Quadrant (CADG) and two other Per­pendicular Quadrants cut both of them, viz. FBAG and FEDG, and the latter do so in the extremities of them both) having let fall Perpendiculars from the common Sections E and B, thro' the Planes of the Perpendicular Quadrants, and the inclined Quadrant, ( viz. on the one side EG and BH, as Right Sines of the Segments EC and BC; on the other EI and BK, as Right Sines of the Segments ED and BA) you'l have 2 Tri­angles EIG and BKH Right Angled at I and K, Equiangular at G and H (by reason of the same inclination of the Plane CBEGC) and consequently similar, by our 34th Proposition; wherefore as the Sine EG to the Sine EI, so the Sine BH to the sine BK, or as EG to BH so EI to BK, and contrariwise.

SCHOLIUM VI.

HEnce you have several Rules of Spherical Trigonometry for resolving Right Angled ▵▵ Lansberg. Geom. Triang. lib. 4. Prop. 12. 1. Having given in the Rightangled ▵ ABC the Hypothenuse BC and the Oblique Angle ACB, for the Leg AB op­posite to this Angle, make: as the sine T (EG) to the sine of the Hypoth. (BH) so the sine of the given Angle (EI) to the sine of the Leg sought (BK). 2. Having given the Hypothe­nuse [Page 104] BC and the Leg AB for the opposite Angle ACB make as the sine Hypoth. (BH) to S. T. (EC) so the sine of the given Leg ( BK) to the sine of the Angle sought (EI) 3. Having given the side AB and the Angle opposite to it ACB, for the Hypothenuse BC (supposing you know whether it be greater or less than a Quadrant) make as the sine of the given Angle EI to the sine T. (EG) so the sine of the given Leg ( BK) to the sine of the Hypoth. BH). 4. Having given in the Right Angled ▵ EBF (which we take instead of ABC that so we may not be obliged to change the Figure) one Leg EB and the Hypothenuse BF for the other Leg EF, you may find its complement, if you make as the sine Complement o [...] the given side (BH) to S. T. (EG) so the sine Compleme [...] of the Hypothenuse (BK) to the sine Compl. of the side sought (EI) 5. Having both Legs EB and EF given, for the Hy­pothenuse BF its Compl. BA may be found thus: as S. T. (EG) is to the sine Cpmpl. (BH) of one side EB, so the sine Compl. (EI) of the other side (EF) to (BK) the sine Compl. of the Hypothenuse.

6. Having given in the same Right Angled Triangle EBF one Leg EF, and the Angle adjacent to it ETB, first prolong into whole Quadrants BA to f, that A f may = BF Hypoth. & BC to e that Ce may = EB, and AC to d that C d may = D [...] the measure of the given Angle EFB: secondly from d thro' e and f let fall a Quadrant thro' the extremities of the Quadrant B f and Be, that so the ▵ C de may be Right Angled, in which there are given the Hypoth. C d = to the given Angle, and the Angles C = to the Compl. of the given Leg ( viz. to the Arch ED) and so, thirdly, there is sought the side de, as the Complement of the Arch ef, or of the Angle sought ABC, or EBF; viz. by the first case of this, by inferring, as S. T. to the sine Hypoth. cd (i. e. of the given Angle EFB;) so the An­gle dce (i. e. DE the Compl. of the given Leg RF) to the sine de (as the Compl. of the Angle fB e or EBF.

7. Having given, in the same Triangle, the side EF and the opposite Angle EBF ( i. e. the Arch ef) for the other Angle EFB (that is the Hypoth. cd in the ▵ cde) make by the third of this:

As the Sine of the Angle dce (i. e. the sine Compl. of the gi­ven Leg DE) to the S. T. so the sine of the Leg de (i. e. the [Page 105] fine Compl. of the Angle EBF) to the Hypoth. cd (i. e. the sine of the Arch DA or Angle EFB.).

8. Having the Oblique Angles given to find either of the sides, viz. EF; which may be done thus by the second of this:

As the sine of the Hypoth. cd (i. e. the sine of the Angle at F) to the W. S. so the sine de (i. e. the sine Compl. of the Angle at B) to the sine of the Angle dce (i. e. the sine Compl. of the side sought EF.)

Consect. 10. The same being given as in Consect. 7. if instead of the Right Sines EI and BK, you erect Perpendicularly DL and AM( Fig. 71) because of the similitude of the Triangles DGL and AHM, you'l have, as DG sine T. to DL the Tangent of the Arch DE, so AH the Right Sine of the Arch AC to AM the Tangent of the Arch AB; or as DG to AH, so DL to AM, and contrariwise.

SCHOLIUM VII.

HEnce flow the other Rules of Spherical Trigonometry for Re­solving Right Angled Triangles, viz. 9. Having given the side AC in the ▵ ABC, and the adjacent Angle ACB, for the other side AB, make as the W. S. (DG) to the sine of the given side (AH) so the Tangent of the given Angle (ACB) to the Tangent of the Angle sought (AB.) 10. Having given the side (AB) and the opposite Angle (at C) for the other side (AC, so you know whether it be greater or less than a Quadrant) make as the Tangent of the given Angle (DL) to the Tangent of the given Leg (AB) so the whole S. (DG) to the sine of the Leg sought ( viz. at AH.) 11. Both sides being given, for the Angles, make, as the sine of one Leg (AH) to the W. S. (DG) so the T. of the other Leg (AM) to the Tangent of the Angle opposite to the same (at C.) 12. Having given moreover in the Right Angled Triangle EBF the Hypothenuse (BF) and the Angle (EFB) for the adjacent side EF, make, as the sine Compl. of the given Angle (AH) to the W. S. so the Tangent Compl. of the Hypoth. (AM) to the Tang. Compl. of the Leg sought (DL) 13. Having given the side (EF) and the adjacent Angle F for the Hypoth. BF make; as the W. S. to the sine Compl. of the given Angle (AH) so the Tangent Complement of the given Leg (DL) [Page 106] to the Tangent Compl. of the Hypoth. (AM.) 14. Having given the Hypoth. (BF) and one side EF for the adjacent An­gle (F) make as the Tang. Compl. of the given Leg (D. L.) to the W. S. so the Tang. Compl. of the Hypoth. (AM) to the Sine Compl. of the Angle sought (AH). 15. Having gi­ven the Hypoth. (BF, i. e. the arch Af, or the angle at d) and either of the oblique angles (at F) for the other angle (EBF) make by help of the new Triangle cde, by the 12th of this.

As the Sine Compl. of the angle cde (i. e. the Sine Compl. of the Hypoth. (AH) to the W. S. so the Tang. Compl. of the Hypoth. cd (i. e. Tang. Compl. of the given angle) to the Tang. Compl. of the Side de (i. e. to the Tang of the angle sought ABC or EBF.)

16. Having given the oblique Angles to find the Hypoth. (BF, or the arch A f, or the angle cde) it is done by the 14 of this Schol.

As the Tang. Compl. of the Leg de (i. e. the Tangent of the angle ABC or EBF) to the W. S. so the Tangent Com­plement of the Hypoth. cd (i. e. the Tangent Complement of the other angle EFB) to the Sine Compl. of the angle cde (i. e. the Sine Compl. of the Hypoth. BC sought.)

So that now we have with Lansbergius (but much more com­pendiously) Scientifically Resolved all the Cases of Rigt-angled Triangles; the Resolution of Oblique-angled ones only now remaining.

Consect. 11. In Oblique-angled Spherical Triangles, as well as Right-angled ones, the Sines of the angles are directly proportio­nal to the Sines of the opposite Sines. 1. Of the Right-angled ones this is evident from N o 3. Schol. 6. and from the 9th Consect. For as the Sine of the angle A ( Fig. 72.) to the Sine of BD, so the W. S. ( i. e. of the angle D) to the Sine of AB. 2. The same is immediately evident of an Oblique-angled Tri­angle ABC, resolved into 2 Right-angled ones. For,

The Sine of the angle C is to the Sine of BD as the sine of the gle D to the Sine of AB; and also,

The Sine of the angle C to the Sine of BD as the Sine of the angle D to the sine of BC, by the 1.

In each Proportionality the means are the Sines of BD and D; therefore the Rectangles of the Extremes of the Sines of AB in­to the Sine of A, and the Sine of BC into the Sine of C, will be equal among themselves, since the Rectangles of the same Means are equal, by Prop. 18. therefore by Prop. 19. as the Sine of A to the Sine of BC, so the Sine of C to the Sine of AB, Q. E. D.

SCHOLIUM VIII.

THe latter may appear of Oblique-angled Triangles after this way also; since the Sine of the angle A is to the Sine of BD as the Sine of the angle D to the sine of AB, call the first a, the second ea, the third b, the fourth eb; and because the Sine of the angle C (which we call c) is likewise to the Sine of BD ( i. e. to ea) as the Sine of D ( i. e. b) to the Sine of BC (which will consequently be [...]) it will be manifest, that the Sine of the angle

A is to the sine of BC as the sine of the angle C to the sine of AB, i. e. as a to .. [...], i. e... c to .. eb. by multiplying the Means and Extremes, whose Rectangles are on both sides eab. Therefore as by the present and precedent Consectary 7, it is universally true, That in any Triangle whether Right Lined or Spherical, Right-Angled or Oblique-angled, the Sides or their Sines, are to one another, as the Sines of their opposite Angles (which therefore is commonly called a Common Theorem:) so also hence flow 2 new Rules of Spherical Trigonometry for Oblique-angled Triangles, like those we found in Schol. 4.

To find

  • I. The other Angles.
    • From 2 sides given of an an­gle opposite to one of them, by inferring As the sine of the side opposite to the gi­ven angle to the sine of the other side, so the sine of the given angle to the sine of the angle sought.
  • [Page 108]II. The other Sides.
    • From one side and the angles given, by inferring As the sine of the angle opposite to the given side to the sine of that side, so the sine of the angle opposite to the side sought to the sine of the side sought.

And thus we have reduced all the Cases and Rules of Sphe­rical Trigonometry to their original Fountains (for from 2 Sides given and the interjacent Angle, or 2 Angles and their adjacent side, we may find the rest in Oblique-angled Triangles, by resolving them into 2 Right-angled ones; and so by the Rules we have deduc'd in Schol. 6 and 7) excepting two Cases, viz. when from 3 sides given, the Angles, or from 3 Angles the Sides are sought; to resolve which, we are supplied with Rules from the following

Cons. 12. In the given Oblique angled Spherical Triangle ABC ( Fig. 73.) whose Sides are unequal and each less than a Quadrant, having produced the sides AB and AC to the Quadrant AD and AE, and effected besides what the Figure directs, then will

The Arch DE be the Mea­sure of the angle A, AF=AC, and so FB the difference of the Sides AB and AC.

BC=BG, and so GF the difference of the third side, and the differences of the rest FB.

But now, 1. As EH or DH to CM or FM so will PH be to NM (by reason of the Equi­angular Triangles EPH and CNM;) therefore by Prop. 26. so will also DP be to FN. Make therefore DH= a FM= ea, DP= b FN= eb.

AI the R. Sine of the side AB. GM the Sine of the Side AC GL the Sine of GB, or of the side BC.

FK the R. Sine of the Arch F B. BI the versed Sine of A B.

BL the vers. Sine of G B or BC. BK the versed Sine of F B. KL or NO the difference of the versed Sines we have now men­tioned.

EP the Right Sine and DP the versed Sine of the arch DE.

CN the R. Sine and FN the ver­sed Sine of the arch FC.

Pag. 108.

73

74

75

76 n. 1.

76 n. 2.

77 n. 1.

77 n. 2.

[Page 109]2. By reason of the Equiangular ▵▵ FNO and HAI (for FNO is Equiangular to the ▵ FKQ, and that to the ▵ HQM, by reason of the Vertical Angles at Q; and that also to the ▵ HAI by reason of the Common Angle at H) and you have also,

As HA,

Or DH to AI, so FN to NO.

[...].

Wherefore now, 3. you'l have evidently, the □ DH to FM into AI as DP to NO.

[...]. DH NO DP.

And Inversly as oea to a so oeb to b.

SCHOLIUM IX.

SInce therefore the Radius DH or a is known, and also NO the Difference of the versed Sines BL and BK, it is evident, that DP the versed Sine of the angle A will be known also; supposing that the first Quantity oea is likewise known. But this may be had by another Antecedent Inference, if you make, as AH to FM so AI to a fourth oea.

a ea oa

Hence therefore arises, 1. the Rule: Having given the 3 Sides of an Oblique-angled Triangle, to find any one of the Angles, viz. by inferring,

1. As the Sine of T to the Sine of R, one of the sides com­prehending AC; so the sine of the other side AB to a fourth, [...]

2. As this fourth to the sine of T, so the difference of the versed sines of the third side, BC, and the differences of the o­thers to the versed sine of the Angle sought, viz.

[...].

But since the sides of a Spherical Triangle may be changed [Page 110] into Angles, and contrariwise the sides being continued [as [...] the side AB of the given Triangle ABC ( Fig. 47.) be con [...] nued a Circle, the rest into Semicircles from the Poles b and [...] and likewise the Semicircle HI from the Pole A, and the Sem [...] circle FG from the Pole B, and the Semicircle EA from th [...] Pole C, you'l have a new Triangle a, b, c, the 3 angles o [...] which will be equal to the 3 sides of the former ABC; as th [...] angle a or its measure IG, is equal to the side AB, by reaso [...] each makes a Quadrant joined with the third arch AG; b [...] the measure of the angle b, is the side AC ( viz. in this cas [...] wherein the side AC is a Quadrant, in the other wherein [...] would be greater or less than a Quadrant, it would be th [...] measure of the angle of the Compl. for then the Semicircle H [...] described from the Pole A, would not pass thro' C but beyon [...] or on one side of C. See Pitisc. lib. 1. Prop. 61. p. m. 25.) — the angle c or its measure KL, is equal to the [...] BC, because with the third KC they make the Quadrants BK and CL] Therefore, 2. Having given the three Angles [...] the Oblique-angled Triangle abc, you may find any side, e. g. ac, if there be sought the Angle ABC, or rather its Comple­ment KBF, or its measure FK= ac, — from th [...] 3 sides given of the ▵ ABC, by the preceding Rule, by infer­ring, viz. 1. As S. T. to the sine R, of one side comprehending the angle of one side AB ( i. e. of one angle a adjacent to the side sought) so the sine of the other side BC ( i. e. of the other an­gle C) to a fourth.

2. As the fourth to the S. T. so the difference of the ver­sed Sines of the third side AC, and the differences of the others ( i. e. of the 3d angle b, and the differences of the rest) to the versed Sine of the comprehended angle, or Complement to a Semicircle ( i. e. of the side sought ac.)

Proposition XXXV.

SImilar Plane Figures (α) are to one another in Duplicate Proportion of their Homologous Sides.

Demonstration.

For, 1. the Bases of 2 similar Triangles or Parallelograms [Page 111] for which any 2 Homologous Sides, e. g. AB and AB ( Fig. 75.) may be taken) and Perpendiculars let fall thereon DE and DE, will be by Consect. 1. Prop. 34, as a to ea, b to eb. Therefore the Parallelograms and Triangles themselves, will be as ba to eeba, by Consect. 7 and 8. Def. 12. i. e. by Def. 34. in duplicate Reason of their Perpendiculars or assumed Sides, which is most conspicuous in Squares, which putting a for the [...]ide of one, and ea for the other, are to one another as aa to [...]eaa.

2. Like Polygons are resolved into like Tri­ [...]ngles, when the Triangles ABC and ABC, Eucl. 19 & 20 lib. 6. and [...]lso AED and AED are Equiangular, by Con­sect. 3. Prop. 34. but CAD and CAD, are also Equiangular, because each of their angles are the remainder of [...]qual ones, after equal ones are taken from them. Wherefore [...]he first Triangles are in duplicate Proportion of the sides BC [...]nd BC; the second likewise of the sides CD and CD the [...]hird are also in the same Proportion of the sides DE and DE, &c. i. e. (since by the Hypoth. BC has the same reason to BC [...]s CD to CD, and DE to DE) each to each is in duplicate Pro­portion of the sides BC to BC, or CD to CD, by the first of [...]his. Therefore by a Syllepsis, the whole Polygons are in du­plicate Proportion of the same Sides: Which is the second thing [...]o be demonstrated.

3. Circles and their like Sectors, are as the Squares of their Diameters, by Prop. 32. therefore in duplicate Proportion of them, by the first of this: Which is the third thing: Therefore simi­ [...]ar Plane Figures, &c. Q. E. D.

CONSECTARYS.

THerefore 2 similar Plane Figures are one to another, as the first Homologous Side, to a third Proportional, by vertue of Definition 34.

II. Any two Figures described on 4 Proportional Lines Eucl. prop. 22. lib. 6. and similar to 2 others, are likewise Proportional, and contrariwise; for if the simple Reasons or Proportions of Lines be the same, their duplicate Proportions will be the same also, and reciprocally.

SCHOLIUM.

BUT as this second Consectary confirms Prop. 22 and i [...] Scholium, so the first teaches us a twofold Geometric [...] Praxis. 1. A Way to express the Proportion of similar F [...] gures by two Right Lines, viz. by finding a third Proportional to their Homologous Sides. For as the side of the first [...] this third, so will be the first Figure to the second. 2. A wa [...] to augment or diminish any given Figure in a given Reaso [...] or Proportion, viz. by finding a mean Proportional between a [...] side of the given Figure, and another Line which shall be [...] that in a given Proportion, and then by describing thereon [...] similar or like Figure.

Proposition XXXVI.

SImilar or like Solid Figures, are to one another in triplicate Pr [...] portion of their Homologous Sides.

Demonstration.

For, 1. The similar Bases of two similar Parallelepipedo [...] (and consequently also of Prisms and Cylinders, by Consect. [...] and 5. Definit. 16. and also of Pyramids and Cones, by Consect 3 and 4, of Definit. 17.) are, as ab to eeab, by (α) the pr [...] ceding Proposition, and their Altitudes as c to ec, by Consect. [...] Prop. 34.

Therefore Parallelepipeds, Cylinders and Prisms (and so the thi [...] part of these, Cones and Pyramids) will be as abc to e (powerof3) abc, by Co [...] sect. 3, 4, 5. Definit. 16. i. e. they will be, by Definit. 34. [...] Consect. 1 and 2. Prop. 34. in Triplicate Proportion of the [...] Perpendiculars or Homologous Sides. Which is especially Co [...] spicuous in Cubes; which, putting a for the Side of one, an [...] ea for the side, are to one another as a (powerof3) to e (powerof3) a (powerof3).

2. Polyedrous or many sided Figures, may be resolved in [...] Pyramids of similar Bases and Altitudes; which is evident [...] Regular ones, from the Consect. of Definit. 21. and cannot b [...] difficult to understand also of Irregular ones; because the lik [...] [Page 113] [...]nclination of their Planes every where similar and equal in number, necessarily require that the whole Altitudes of similar Polyedrous Solids, as well as similar Parallelepipedons, by Con­sect. 2. Prop. 34. should be in subduplicate Proportion of their Bases, and so these being likewise divided in C and C ( Fig. 76. N o 1.) the parts of their heigths GC and GC, will be in the same Proportion: Whence, e. g. 2 Pyramids standing on simi­ [...]ar Bases ABDEF and ABDEF, and having like Altitudes GC and GC, will necessarily be like or similar; and the same [...]hing may be likewise judged of others.

Or yet, to shew it more evidently, the Polyedrous Solids may be resolved into like Triangular Prisms; for, e. g. each of the Trian­gles of their similar Bases ABDEF and ABDEF (N o 2.) are similar, viz. abf and abf, ABF and ABF, by the pre­ceding Prop. N o 2. The Planes ABba and A Bba, also AafF and A af F, are similar by the Hypoth. and consequently also the Planes BbfF and B bfF ( fb is to ba as fb to ba, and also in the one ba to bB, as ba to bB in the other; therefore ex equo as fb to bB so f b to bB, &c.) and so the whole Triangular Prisms will be similar, by Definit. 35. and so of others. There­fore similar Polyedrous Solids will be in the same Proportion as similar Pyramids, or Triangular Prisms, i. e. by the first of this in Triplicate Proportion of their Sides. Eucl. prop. 12. lib. 12. of Cones and Cylinders.

3. Spheres are as the Cubes of their Diame­ters Eucl. 18. lib. 12. by Consect. 3. Prop. 32. Therefore they are by the first of this, as a (powerof3) to e (powerof3) a (powerof3). Therefore similar or like Solids are in Triplicate Proportion of their Homologous Sides. Q. E. D.

CHAP. VI. Of the Proportions of Magnitudes of divers sorts com­pared together.

Proposition XXXVII.

THE Parallelogram ABCD (Fig. 77. N. 1.) is to the Triangle BCD upon the same base DC, and of the same heighth as 2 [...] 1. This has been already Demonstrated in Consect. 3. De [...] ­nit. 12. Here we shall give you another

Demonstration.

Suppose, 1. the whole Base CD divided into four equ [...] parts by the transverse Parallel Lines EG, HK, LN, then w [...] (by reason of the similitude of the ▵▵ DGF, DKI, DN [...] DCB) GF be 1, KI 2, NM 3, CB 4; and having further­more continually Bisected the Parts of the Base, the Indivisible or the Portions of the Lines drawn transversly thro' the Trian­gle will be 1, 2, 3, 4, 5, 6, 7, 8, &c. ad infinitum, all a­long in an Arithmetical Progression, beginning from the Poi [...] D, as 0; to which the like number of Indivisibles always an­swer in the Parallelogram equal to the greatest, viz. the Li [...] BC. Wherefore by the 4th Consect. of Prop. 16. all the In [...] visibles of the Triangle, to all those of the Parallelogram take [...] together, i. e. the Triangle it self to the Parallelogram, is as [...] to 2. Q. E. D.

SCHOLIUM.

NOW if any one should doubt whether the Triangle [...] Parallelogram may be rightly said to consist of an in [...] nite number of Indivisible Lines, he may, with Dr. Wa [...] [Page] [Page]

Pag. 115.

78

79

80

81

82

83

84

85

86

87

88

[Page 115] instead of Lines, conceive infinitely little Parallelograms of the same infinitely little Heighth, and it will do as well. For ha­ving cut the Base (N o 2) into 4 equal parts by transverse Pa­rallels, there will be circumscribed about the Triangle so many Parallelograms of equal heighth, being in the same Proportion as their Bases, by Prop. 28. i. e. increasing in Arithmetical Progression. In the following Bisection, there will arise 8 such Parallelograms approaching nearer to the Triangle, in the next 16, &c. so that at length infinite such Parallelograms of in­finitely less heighth, and ending in the Triangle itself, will constitute or make an infinite Series of Arithmetical Propor­tionals, beginning not from 0 but 1; to which there will an­swer in the Parallelogram infinite little Parallelograms of the same heighth, equal to the greatest. Whence it again follows, by Consect. 9. Prop. 21. that the one Series is to the other, i. e. the Triangle to the Parallelogram as 1 to 2; which being here thus once explained, may be the more easily applied to Cases of the like nature hereafter.

CONSECTARYS.

I. SInce in like manner in the Circle ( Fig. 79.) the Periphe­rys at equal intervals from one another, as so many Ele­ments of the Circle, increase in Arithmetical Progression; the Sum of these Elements, i. e. the Circle it self will be to the Sum of as many Terms equal to the greatest Periphery, i. e. to a Cylindrical Surface, whose Base is the greatest Periphery, and its Altitude the Semidiameter, as 1 to 2.

II. Hence the Curve Surface of a Cylinder circumscribed a­bout a Sphere, i. e. whose Altitude is equal to the Diameter, is Quadruple to its Base.

III. Also the Sector of the Circle bac, to a Cylindrical Sur­face, whose Base is the Arch bc, but its Altitude the Semidia­meter ac, is as 1 to 2.

IV. And because the Surface of the Cone BCD is to its cir­cular Base, as BC to CA, i. e. as the √2 to 1. by Schol. Prop. 17. the Cylindrical Surface, the Conical Surface and the Cir­cular [Page 116] one we have hitherto made use of will be as 2, √2 and 1, and consequently continually Proportional.

SCHOLIUM

ALL which may also abundantly appear this way, viz. by putting for the Diameter of the Circle a, for the Semidiameter ½ a and for the Circumference ea, you'l have the Area of the Circle ¼ eaa by Consect. 1 Definit. 31. and Multiplying [...]he heighth of the Cylinder AB i. e. ½ a by the Periph [...] ea yo [...]'l have the Cylindrical Surface ½ eaa by Consect. 6. Definit 1 [...] as now is evident also by Consect. 1, 2 and 3. Now if y [...] would also have the Surface of the Cone, since it's side by the Pythagorick Theorem is [...] and the half of that [...] i. e. (b [...] N o 2 of Schol. Prop. 22.) [...] and this half being multiplied by the Periphery of the Base ea, you'l have (by virtue o [...] Consect. 4. Definit. 1.8.) the Su [...]f [...]ce of the Cone [...] i. e. (b [...] the Schol. just now cited) [...]: So that now appears also th [...] 4th Consect. of this; because the Rectangle of those Extreme [...] eaa and ¼ eaa is ⅛ eaa (powerof4) as well as the square of the mean.

Proposition XXXVIII.

A Parallelepiped Euclid. Prop 7 Coroll. lib. 12. BF (Fig. 77. N o 3.) is to a Pyra [...] ABCDE upon the same Base BD and of the same heighth, [...] 3 to 1. This was D [...]monstrated in Consect. 3. Definit. 17. bu [...] here we shall give you another.

Demonstration.

Suppose 1 the whole Al [...]itude BE divided into 3 equal Pa [...] by transverse P [...]ains Parallel to the Base, then will (by reason [...] the Similitude of the Pyramids abcd ACBD [...] and AECDE) the Bases abcd ABCD and ABC [...] be by Consect. 2. Prop. 34 and Consect. 3. Defi [...] 17 in duplicate Proportion of the Altitudes i. e. in duplicate Arithmetical Progression 1, [...] 9, mo [...]eover 2, bisecting the parts of the Altitude, the qu [...] d [...]angular Sections now double in Number (as the Indivisibles o [...] Elements of the proposed Pyramid) will be as 1, 4, 9, 16, 2 [...] [Page 117] 36, &c. ad Infinitum, all along in a duplicate Arithmetical Pro­portion; while in the mean time there answer to them as many Elements in the Parallelepiped equal to the greatest ABCD, wherefore by Consect. 10. Prop. 21. all the Indivisibles of the Pyramid taken together will be to all the Indivisibles of the Parallelepiped also taken together, i. e. the Pyramid it self to the Parallelepiped, as 1 to 3. Q. E. D.

CONSECTARY.

THis Demonstration may be easily accommodated to all other Pyramids and Prisms, and also Cones and Cylinders, Euclid. Prop. 10. lib. 12. since here also ( Fig. 78.) the circular Planes ba, BA, and BA are as the squares of the Diameters, and so as 1, 4, 9. and so likewise all the other Elements of the Cone by continual bisection are in duplicate Arithmetical Progression; when in the mean time there answer to them in the Cylinder as many Ele­ments equal to the greatest BA &c.

Proposition XXXIX.

A Cylinder is to a Sphere inscribed in it i. e. of the same Base and Altitude as 3 to 2.

Demonstration.

Suppose 1 ( Fig. 80.) the half Altitude GH (for the same pro­portion which will hold when demonstrated of the half Cylinder AK and Hemisphere AGB, will also hold the same of the whole Cylinder to the whole Sphere) to be divided into 3 equal parts, then will AH, C1, E2, be mean proportionals between the Segments of the Diameter by Prop. 34 Schol. 2 N o 3, and so by Prop. 17. the Rectangles LHG, L1 G, L2 G equal to the Squares AH, C1, E2, being in order as 9, 8 and 5. and also 2dly, having bisected the former parts of the heighth, the six Squares cutting the Sphere Cross ways will be found to be as 36, 35, 32, 27, 20, 11. &c. in the progression we have shewn at large in Consect. 12. Prop 21. Wherefore since all the Indivisibles of the Hemisphere, viz. the circular Planes answering to the Squares of the said [...]aniverse [Page 118] Diameters have the same proportion of Progression, by Pr [...] 32. and there answer to them the like number of Elements i [...] the Cylinder equal to the greatest AH: All thes [...] these taken together will be to all the other take [...] together i. e, the whole Cylinder AK to the whol [...] Hemisphere AGB by vertue of the aforesaid Co [...] sect. 12. as 3 to 2 A [...]chim. 32 (al. 31.) lib. I de Sphaer. Q. E. D.

SCHOLIUM. I.

HOnora [...]us Fabri elegantly deduces this Prop. a priori, in a g [...] netick Method in his Synopsis Geom. p. 318. (which a [...] Carotus Renaldinus performs from the same common Foundationi [...] lib. 1. de Compos. and Resol. p. 301, and the following, but aft [...] a more obscure way and from a demonstration further fetch'd Fabri's is after this Method: The whole Figure (81) AL being turned round about BZ, the Quadrant ADLBA will describe a Hemisphere, the Square AZ a Cylinder and the triangle BM [...] a Cone all of the same Base and Altitude. Since therefore Circl [...] are as the squares of their Diameters by Prop. 32. and the Squa [...] of GE=to the Squares of GD and GF taken together (f [...] the Square of GF i. e. GB×□ GD is = □ BD or BA or G [...] by the Pythag. Theor.) and so the Circle described by GE [...] be = to 2 Circles described by GD and GF taken together then taking away the Common Circle described by GF there w [...] remain the circle described by GF within the Cone equal t [...] the Annulus or Ring described by DE about the Sphere. A [...] since this may be demonstrated after the same way in any othe [...] case, viz. that a circle described by g f, will be equal to an A [...] nulus described by de; it will follow, that all Rings or An [...] described by the Lines DE or de (i. e. all that Solid that conceived to be described by the trilinear Figure ADLM turne round) will be equal to all the Circles described by GF or gf (i [...] to the Cone generated by the Triangle BLM;) and so as th [...] Cone i [...] ⅓ part of the Cylinder generated by AL, by the C [...]sect. of Prop. 38. so also the Solid made by the Triline [...] ADLM ( viz. the Excess of the Cylinder above the Sphere will be ⅓ of the Cylinder, and consequently the Hemisphere Q. E. D.

CONSECTAYS.

HEnce you have a further Confirmation of Consect. 2. Prop. 32. and Prop. 36. N. 3.

II. Hence also naturally flows a Confirmation of Consect. 2. Definit. 20. and consequently the Dimension of the Sphere both as to its solidity and Surface. For putting a for the Dia­meter of the Sphere and circumscribed Cylinder, and Ea for the Circumference, the Basis of the greatest Circle will be ¼ eaa, and [...]hat multiplied by the Altitude, gives ¼ ea (powerof3) for the Cylinder. Therefore by the present Proposition, ⅙ ea (powerof3) gives the Solidity of the Sphere (by making as 3 to 2 so ¼ to ⅙) This divided by ⅙ a, will give, by vertue of Consect. 1. of the aforesaid Def. 20. and Consect. 3 Definit. 17. the Surface of the Sphere eaa.

III. Therefore the Archim. lib. I. de Sph. & Cylind. Prop. 31. (al. 30). Surface of the Sphere eaa, is manifest­ly Quadruple of the greatest Circle ¼ eaa.

IV. The Surface of the Cylinder, without the Bases, made by multiplying the Altitude a by the Circular Periphery of the Base ea, will be eea, equal to the Surface of the Sphere.

V. Adding therefore the 2 Bases, each whereof is ¼ eaa, the whole Surface of the Cylinder 1 ½ eaa, will be to the Sur­face of the Sphere eaa as 3 to 2.

VI. The Square of the Diameter aa to the Area of the Cir­cle ¼ eaa, is as a to ¼ ea, i. e. as the Diameter to the 4th part of the circumference.

VII. A Cone of the same Base and Altitude with the Sphere and Cylinder, will be by Consect. 2. of this, Prop. and the Con­sect. of Prop. 38. [...] ea (powerof3), and of the Cylinders ¼ or [...] ea (powerof3). There­fore a Cone, Sphere, and Cylinder, of the same heighth and dia­meter, are as 1, 2, 3. The Cone therefore is equal to the Excess of the Cylinder above the Sphere; as is otherwise evident in Scholium 1. of this.

SCHOLIUM II.

AND thus we have briefly and directly demonstrated the chief Propositions of Archimedes, in his 1. Book de Sphaer [...] & Cylind. which he has deduced by a tedious Apparatus, and [...] [...]directly. And now if you have a mind to Survey the [...] and more perplext way of Archimedes, and compare it with this shorter cut we have given you; take it thus: Archimedes thought it necessary first of all to premise this Lemma; Th [...] all the Conical Surfaces of the Conical Body made by Circum­volution of the Polygon, or many angled Figure A, B, C, D, E, &c. (Fig. 82.) inscribed in a Circle, according to Definit. 19. I say, those Conical Surfaces taken all together, will be equal to a Circle, whose Radius is a mean Proportional between the Diameter AE and a transverse Line BE, drawn from one ex­tremity of the Diameter E to the end of the side AB next to the other extremity. This we will thus demonstrate by the help of specious Arithmetick: Since BN, HN are the Right Sines of equal Arches, CK and CK whole Sines, &c. and the Lines BH, GC, &c. parallel; having drawn obliquely the transverse Lines HC, GD, all the angles at H, C, G, D, &c. will be equal by Consect. 1. Definit. 11. and consequently all the Tri­angles BNA, HNI, ICK, &c. equiangular, both among them­selves, and to the ▵ ABE; since the angle at B is a Right one, by Consect. 1 Prop. 33. and the angle at A common with the ▵ BNA. Wherefore as

  • BN to NA
  • or HN to NI

so

  • CK to CI.
  • & GK to KL.

and so

  • DM to ML
  • FM to ME

so EB to BA; and so by making BN, HN, DM, FM= a CK and GK= b, EB= c, for NA, NI, ML and ME, you may rightly put ea for KI and KL eb for AB, ec. Which being done you may easily obtain the Conical Surfaces of the inscrib'd Solid, and the Area of a Circle whose Radius shall be a mean Proportional between AE and FB, and it will be evidently manifest, that these two are equal. For, 1. (for Conical Surfaces) the Diameter of the Base BH=2 a, and the side of the Cone A B= ec: Therefore (making here o the name of the Reason between the Diameter and Circumference) the circumference will be 2 oa [Page 121] which multiplied by half the side ½ ec, gives the Conical Surface oaec, by Consect. 4. Definit. 18. And since the Circumference BH is as before 2 oa, and the circumference CG = 2 ob, half of the sum 2 oa+2 ob, viz. oa+ ob is the equated Circum­ference: which multiplied by the Side BC= ec gives the Sur­face of the truncated Cone BHGC= oaec− obec, by Con­sect 5. of the aforesaid Def. since, lastly, the Surface of the truncated Cone DFGC is equal to one, and likewise the conical Surface EDF to the other, by adding you'l have the Sum of all 40 oaec+2 obec.

2. (for the Area of the Circle) the Diameter AE is =4 ea+2 eb, and BE= c: the Rectangle of these is =4 eac+2 ebc = (which also is evidently equal to the Rectangle of all the transverse Lines BH, CG, DF into the side A B, as Ar­chimedes proposes in the matter) = to the Square of the Radius in the Circle sought, because the Radius is a mean Proportio­nal between AE and BE, and so equal to [...], so that the whole Diameter is [...]. There­fore, 2. the circumference of this Circle will be [...], i. e. [...]: which multiplyed by half the Semidiameter, i. e. by [...] gives the Area of the Circle sought [...]. But this Root extracted is [...], equal to the superiour Sum of the conical Surfaces. Q. E. D.

Having thus demonstrated the Lemma, we will easily demon­strate with Archimedes (tho not after his way) That the Sur­face of any Sphere, is Quadruple of the greatest Circle in it, which is already evident from the 3d Consect. For since all the conical Surfaces of the inscribed Solid taken together, by the preceding Lemma, are equal to the Area of a Circle whose Ra­dius is a mean Proportional between the Diameter AE and the Transverse EB; and this mean Proportional approaches always so much nearer to the Diameter AE, and those Surfaces so much nearer to the Surface of the Sphere, by how many the more sides the inscribed Figure is conceived to have, by Con­sect. 1 and 2. Def. 18. if you conceive in your mind the Bi­section of the Arches AB, BC, &c. to be continued in Infinitum, it will necessarily follow, that all those conical Surfaces will at length end in the Surface of the Sphere it self, and that mean [Page 122] Pro­portional in the diameter AE, and so the Surface of the Sphere will be equal to a Circle, whose Radius is the Diameter AE, But that Circle would be Quadruple of the greatest Circle in the present Sphere, by Prop. 35. Therefore the Surface of the Sphere is Quadruple of that Circle also. Q. E. D.

Hence also it would be very easie to deduce with Archime­des (tho again after another way) that celebrated Proposition, which we have already demonstrated from another Principle in the Prop. of this Schol. viz. That a Cylinder is to a Sphere of the same Diameter and Altitude, as 3 to 2. For by putting a for the Diameter and Altitude, and ea for the Circumference, the Area of the Circle, will be ¼ eaa; and this Area being multiplied by the Altitude a, gives ¼ ea (powerof3) for the Cylinder, by Consect. 5. Definit. 16. and the same Quadruple, i. e. eaa mul­tiplied by ⅙ a gives ⅙ ea (powerof3) for the Sphere, by Consect. 1. Definit. 20. and Consect. 3. Definit. 17. Wherefore the Cylinder will be to the Sphere as ¼ to ⅙, i. e. in the same Denominator as [...] to [...], i. e. as 6 to 4, or 3 to 2. Q.E.D.

Whence it is evident, that the Dimension of the Sphere would be every ways absolute if the Proportion of the Diame­ter to the Circumference were known; which now with Ar­chimedes, we will endeavour to Investigate.

Proposition. XL.

THE Proportion of the Periphery of a Circle Archim. Cyclom. prop. 2 to the Diame­ter, is less than 3 [...] or [...] to 1. and greater than 3 [...] to 1.

Demonstration.

The whole force of this Proposition consists in these, that, 1. Any Figure circumscribed about a Circle, has a greater Peri­phery than the Circle, but any inscrib'd one a less. 2. The Peri­phery of a circumscribed Figure of 96 sides, has a less Pro­portion to the Diameter, than 3 [...] to 1.

To demonstrate this second, we will enquire 3. the Proportion of one side of such a Figure whether circumscribed or inscribed after the fol­lowing way.

For the first part of the Proposition.

Suppose the Arch BC ( Fig. 84. N. 1.) of 30 degrees, and its Tangent BC making with the Radius AC a Right Angle, to make the Triangle ABC half of an Equilateral one, so that AB shall be to BC in double Reason, viz. as 1000 to 500; which being supposed, AC will be the Root of the Difference of the Squares BC and AB, i. e. a little greater than 866, but not quite [...].

Then continually Bisecting the Angles BAC by AG, GAC by AH, HAC by AK, KAC by AL, BC is half the side of a circumscribed Hexagon, GC the half side of a Dodecagon (or 12 sided Figure) HC of a Polygon of 24 sides, KC of one of 48; lastly, LC of one of 96 sides; and by N. 3. Schol. 3. Prop. 34. GC will be to AC as BC to BA+AC, and also HC to AC as GC to GA+AC, &c. Wherefore

In the first Bisection, of what parts GC is 500, of the same will AC be 1866 and a little more, and AG (which is the Root of the Sum of the □ □ GC and AC) 1931 [...]+.

In the second Bisection, of what parts HC is 500, of the same will AC be found to be 3797 [...]+and AH 3830 [...]+.

In the third Bisection, of what parts KL is 500 of the same will AC be 7628 [...]+ and AK 7644 [...]+.

In the 4th Bisection, of what parts LC is 500 of the same will AC be 15272 [...]+.

Now therefore LC taken 96 times, will give 48000 the Semi-periphery of the Polygon, which has the same Proportion to the Semi-diameter AC 15272 [...] as the whole Periphery to the whole Diameter. But 48000 contains 15272 [...], 3 times, and moreover 2181 [...] remaining parts, which are less than [...] part of the division, for multiplied by 7 they give only 1526 [...]+.

Therefore it is evident, that the Periphery of this Polygon (and much more the Periphery of a less Circle than that) will have a less Proportion to the Diameter, than 3 [...] to 1. Which is one thing we were to demonstrate.

For the 2d Part of the Prop.

SUppose the Arch BC to be (N o 2.) of 60 Degr, that is the Angle BAC at the Periphery of 30 Since the Angle at B is a right one by Consect. 1. Prop. 33. the Triangle ABC will be again half an Equilateral one, and BC the whole side of an Hexagon, and GC of a Dodecagon, &c. So that putting for BC 1000 (as before we put 500 for the side of the Hexagon) let AC be 2000 and AB the Root of the difference of the Squares BC and AC i. e. less than 1732 [...] viz. 1732 and not quite [...].

Then bisecting continually the Angles BAC, GAC, &c. since the Angles at the Periphery BAG, GAC, GCB, standing on equal Arches BG and GC, are equal by Prop. 33. and the Angle at C, (common to the Triangles GCF and GCA) and the o­thers at H, K, L are all right ones by Consect. 1. of the afore­said Prop. these 2 Triangles CGF and CGA are equiangular and consequently by Prop. 34. the Perpendicular GC in the one will be to the Perpendicular GA in the other as the Hypothenuse CF in the one to the Hypoth. AC in the other i. e. (by the foun­dation we have laid in the former part of the Demonstration of N o 3, Schol. 3. Prop. 34.) as BC to AB+AC; and in like man­ner in the following HC will be to HA as GC to AG+AC, &c. Wherefore.

In the first Bisection, of what parts GC is 1000 of the same AG will be a little less then 3732 [...]; and AC (which is the Root of the Sum of the □ □ AG and GC) will be a little less then 3863 [...].

In the second Bifection, of what parts GC is 1000 of the same will AH be a little less then 7595 [...] and AC a little less then 766 [...].

In the third Bisection, of what parts KC is 1000 of the same will AK be a little less then 15257 [...] and AC a little less 15290 [...].

In the fourth Bisection, of what parts LC is 1000 of the same will AL be a little less then 30547 [...]; and AC a little less then 30564, and consequently if LC be put 500, AC will be less then 15282.

Now therefore LC taken 96 times will give 48000 for the Periphery of the inscrib'd Polygon, and 15282 and a little less for the Diameter AC. But 48000 contains 15282 thrice, and moreover a remainder of 2154 parts, which are more then [...] of the Divisor; for [...] of this number makes 215 [...] and so [...] makes 2150 [...] i. e. 2152 [...]. Therefore it is evident that the Peri­phery of this Inscribed Polygon (and much more the Periphery of a Circle greater then that) will be to it's Diameter in a greater proportion then 3 [...] to 1, which is the 2d thing.

SCHOLIUM

IF any one had rather make use of the small numbers of Archimedes, which he chose for this purpose, by putting in the first part of the Demonstration, for AB 306 and for BC 153, in the second for AC 1560 and for BC 780, by the like process of Demonstration, he may infer the same with Archimedes. We like our Numbers best, tho' somewhat large, because they may be remember'd, and are more proportionate to things, and make also the latter part of our Demonstration like the former. The Proportion in the mean while of the Dia­meter to the Periphery of the Circle by the Archimedean way is included within such narrow Limits, that they only differ from one another [...] or [...] parts; for [...] Subtracted from [...] leave [...], as 3 [...] or [...] and 3 [...] or [...] if they are reduced to the same Deno­mination make on the one hand [...] on the other [...]. Hence it would be easy, having divided the difference [...] into 2 Parts, to express a middle proportion of the Periphery to the Diameter be­tween the 2 Archimedean and Extreme ones as in these Numbers, 1561 to 4970, (or by dividing both sides by 5) as 3123 to 994, or (dividing both sides by 7) as 446 [...] to 142, or (by dividing again by 2) as 223 [...] to 71, &c.

While these Numbers become as fit for use as those of Archi­medes, which we therefore use before any other, particularly in Dimensions that dont require an exact Niceness; where they do those may be made use of exhibited by Ptolomey Vieta, Ludolphus a Ceulea, Metius, Snellius, Lansbergius, Hugeus, &c. as if

The Diameter be
The Circumference will be
10,000,000
31, 416, 666. Ptolomey.
10000,000,000
31, 415, 926, 535. Vieta.
100,000,000,000,000,000000
314, 159, 265, 358, 979, 323, 846 ½, &c. Ludolph. a Ceulen.
Proposition XLI.

THe Area of a Circle has the same proportion to the Square of it Diameter, as the 4th part of the Circumference to the Dia­meter.

Demonstration.

Though we have before Demonstrated this Truth in Consect. 6. Prop. 39. yet here we will give it you again after another way. Since therefore the Circumference is a little less then 3 [...], and [...] little more than 3 [...] Diameters, for this excess putting z if the Diameter be 1 we will call the Circumference 3+z therefore the 4th Part of it will be [...]: And the Area of the Circle (by Multiplying the half Semidiameter) i. e. ¼ by the Circumfe­rence, you'l have both [...] and the square of the Diameter = [...] Q. E. D.

CONSECTARY.

THerefore if the Archim. Prop. 3. Cy­clom. Proportion of Archimedes be near e­nough truth to be made use of, viz. 22 to 7; the Ar [...] of the Circle will be to the Square of the Diameter as 11 to 14 because the quarter part of 22, i. e. 5 ½ or [...] to the Diam. 7 [...] is in the same Proportion.

Proposition XLII.

THE Diameter Eucl. last Prop. lib. 10. of a Square AC ( Fig. 83.) is incom­mensurable to the side AB (and consequently also to th [...] whole Periphery) i. e. it bears a Proportion to [...] that cannot be exactly expressed by Numbers.

Demonstration.

For, if for AB you put 1, BC will be als [...] [Page 127] 1, and by the Pythagorick Theorem AC will be = √2. There­fore by Consect. 4. Definit. 30. AC is incommensurable to the side AB, &c. Q.E.D.

CONSECTARY I.

IT is notwithstanding Commensurable in Power; for its Square is to the Square of the Side as 2 to 1.

CONSECTARY II.

NOW if the Proportion of the side or whole Periphery ABCDA, to the Diam. AC is to be expressed by Num­bers somewhat near, as we have done in the Diameter and Cir­cumference of a Circle; then making the side AB=100, the Diameter is greater than 141 [...], less than 141 [...].

Proposition XLIII.

THE Area of a Circle is Incommensurable to the Square of the Dia­meter.

Demonstration.

For dividing the Semidiameter CD ( Fig. 85.) into two equal Parts (and consequently the Diameter DF into 4) AC will be 2. viz. √4 and AC √3. by Schol. 2. Prop. 34. N. 3. the sum √4+√3, and the sum of as many equal to the greatest AC 4. Having moreover Bisected the Parts of the Semidiameter, AC will =4 or the √16, ac=√15, AC=√12, AC=√7; the sum [...]; and the sum of as many e­qual to the greatest AC is =16, &c. And thus the last sums will be the Square Numbers increasing in Quadruple Pro­portion; but the former Sums will be always composed of the Ra­tional Root of every such Square, and of several other irrational Roots of Numbers unevenly decreasing; so that it will be im­possible to express those former Sums by any Rational Number, by what we have said in Schol. 2. Definit. 30. Wherefore all the Indivisibles of the Quadrant ADC are to as many of the Square ACDE equal to the greatest, i. e. the Quadrant it self [Page 128] ADC to the Square ACDE (and consequently the whole Area of the Circle to this circumscrib'd Square) will be as a Surd Quantity to a true and truly Square Number, i. e. the Area of the Circle will be Incommensurable to the Square of the Diameter, by Consect. 4. of the said Definit. Q. E. D.

CONSECTARY.

AND because the fourth part of the Circumference ha [...] the same Proportion to the Diameter, as the Area of the Circle to the Square of the Diameter, by Prop. 41. There­fore also that will be Incommensurable to this, and consequently the whole Circumference will be so to the Diameter.

SCHOLIUM.

WHerefore it is somewhat Wonderful, which G. G. Leib­nitius Acta E­rudit. Ann. 82. p. 44 & the following. tells us, that the Square of the Diameter be­ing 1, the Area of the Circle will be [...], &c. ad Infinitum. i. e. by adding 1 [...] and [...], &c.) to [...], &c. i. e. to the Sum of infinite Fractions whose common Numerator is 2. But their Denominators Squares les­sen'd by Unity, and taken out of the Series of the Squares of Natural Numbers by every fourth, omitting the Intermediate ones: Which Sum might seem expressible in Numbers, since all its parts are Fractions redu­cible to a common Denominaton; while notwith­standing Leibnitius himself confesses, that the Cir­cle is not Commensurable to the Square, nor ex­pressible by any Number.

CHAP. VII. Of the Powers of the Sides of Triangles, and other Regular Figures, &c.

Proposition XLIV.

IN Right-angled Triangles Euc. 47. Lib. 1. (ABC, Fig. 86.) the Square of the Side (BC) that subtends the Right-angle, is equal to the Squares of the other Sides (AB and AC) taken together.

Demonstration.

Though we have demonstrated this Truth more than once in the foregoing Proposition; yet here we will confirm it again as follows. Having described on each side of the Square BE a Se­micircle, which will all necessarily touch one another in one point, and be equal to the Semicircle, BAC, if you conceive as many Triangles inscribed also equal to BAC; it will be evident that the Square BE will contain the said 4 Triangles; and besides the little Square FGHI, whose side FI, v. g. is the difference be­tween the greater side of the Triangle CI, and the less CF, (for because the less side CF=BA, lying in the first Semicircle, if it be continued to I in the se­cond Semicircle makes CI=CA the greater side of the other Triangle, and so in the others. From thence it is evident, That as the Angles ABC, and ACB together make one right one; so likewise BCF (= CBA) and ECF make also one right one; and consequently ECF is =ACB, and the Arch and the Line EI= to the Arch and the Line AB, &c.) Where­fore, if the greatest side of the given Triangle BC or BD, &c. be called a, and AC, b and the least AC, or CF, &c. be called c; the [...] of the side BC, will be = aa, and the Area of each Triangle ½ bc: and so the 4 Triangles together 2 bc: but the side of the middle little Square will be b− c, and its Square bb+ cc=2 bc: Wherefore if you add to this the 4 Trian­gles▪

The Sum of all, i. e. the whole Square BE will be bb+ cc= aa. Q. E. D.

CONSECTARYS.

I. HEnce having the sides that comprehend the Right-ang [...] given, AC= b and AB= c, the Hypothenuse or Ba [...] that subtends the Right-angle BC will be = √ bb+ cc.

II. But if BC be given = a and AC= b, and you are [...] find AB= x; because xx+ bb= aa; you'l have (taking awa [...] from both sides bb) xx= aa− bb: therefore x, i. e. AB=√ aa− bb.

III. If 2 Right-angled Triangles have their Hypothenus [...] and one Leg equal, the other will also be equal.

Proposition XLV.

IN Obtuse-angled Triangles (Fig. 87. N. 1. the Square of the Ba [...] or greatest Side BC that subtends the Obtuse-angle BAC, is eq [...] to the Squares of Eucl. prop. 12. lib. 2. the other 2 Sides (AB and AC) taken togethe [...] and also to 2 Rectangles (CAD) made by one of [...] Sides which contain the Obtuse-angle (AC) and [...] continuation AD to the Perpendicular BD let fall fr [...] the other side.

Demonstration.

If BC be called a, AB= c, AC= b, AD= x, CD will b [...] = b+ x. Therefore □ BD= cc− xx, by Consect. 2. of the pr [...] ceding Prop. In like manner if □ CD= [...] be subtracted from the □ BC= aa, you'l have [...] to the same □ BD. Therefore [...], i. e. (adding on both sides xx) [...]. i. e. (adding on both sides bb and 2 bx) [...]. Q. E. D.

CONSECTARY.

IF in this last Equation you subtract from both Sides cc+ bb then will [...] and (if you moreover divide both Sides by 2 b) you'l have [...]: Which is the Rule, when you have the Sides of an Obtuse-angled Triangle given, to find the Segment AD, and consequently the Perpen­dicular BD.

Proposition XLVI.

IN Acute-angled Triangles Eucl. Prop. 13. l. 2. the Square of any side ( e. g. B. C, Fig. 87. N. 2.) subtending any of the Angles, as A is equal to the Squares of the other 2 sides (AB and AC) taken together, less 2 Rect­angles (CAD) made by one side, containing the Acute-angle (CA) and its Segment AD reaching from the Acute-angle (A) to the Perpen­dicular (BE) let fall from the other side.

Demonstration.

Make again BC= a, AC= b, AB= c, AD= x; then will CD= b− x. Therefore cc− xx= □ BD, and [...] ( i. e. □ BC−□CD) will also be = □ BD.

Therefore [...].

i. e. (adding to both sides xx)

[...],

i. e. (adding on both sides bb, and subtracting 2 bx)

[...]. Q. E. D.

CONSECTARYS.

I. IF in the last Equation, except one, you add on both sides bb, and subtract aa, you'l have [...], and, if moreover you divide hoth sides by 2 b, you'l have [...]: Which is the Rule, having 3 sides given in an Acute-angled Triangle, to find the Segment AD, and consequently the Perpendicular BD.

Knowing therefore the Segments AD and CD, and also the Perpendicular BD in Oblique-angled Triangles, whether Ob­tuse-angled or Acute-angled, when moreover the sides BC an [...] AB are likewise given, the Angles of either Right-angled Tri­angles or Oblique-angled ones, will be known; so that the la [...] Case of Plane Trigonometry, which we deferr'd from Prop. 34 to this place, may hence receive its solution.

Proposition XLVII.

THE Square of the Tangent of a Eucl. 36. Lib. 3. Circle, is equal to a Rectangle contain'd under the whole Secant DA, and that part of it whi [...] is without the Circle DE, whether the Secant pass thro' the Centre o [...] not.

Demonstration.

For in the first Case, if CB and CE are = b, DE= x, then will CD= b+ x, & AD=2 b+ x: therefore

□ ADE=2 [...]. there­fore, if from the □ CD you substract □ CB= bb the remainde [...] will be [...] =□ BD=□ ADE. Q.E.D.

In the second Case, the lines remaining as before, make D [...]= y, FE or FA=Z: therefore the □ ADE will be = [...], but the □ FC equal to the □ EC−□FE= [...] i. e. [...]. But the same Square FD is = [...]. Wherefore taking away from these equal Square [...] the common one ZZ you'l have 2 [...] i. e. pr. 1st Case =□ BD. Q. E. D.

CONSECTARYS.

I. THerefore the Rectangles of diverse secants (as of AD [...] & ade in Fig. 88, n. 1.) which are equal to the sam [...] Square. BD, are equal also to one another; which the la [...] Equation in our Demonstration ( [...] is an ocular proof of.

Pag. 133.

89

90

91

92

93

94

95

96

[Page 133]II. Therefore by Prop. 19. as a D to AD so is reciprocally DE to D e in the Fig. of the 2 Case.

III. Tangents to the same Circle from the same Point, as DB & D b (n. 3.) are equal; because the Square of each is equal to the same Rectangle.

IV. Nor can there be more Tangents drawn from the same point then two: For if besides DB & D b, D c could also touch the Circle, then it would be equal to them. By Cons. 2. but that is absurd by Cons. 3. Def. 7.

SCHOLIUM I.

HEnce is evident, the Original of the Geometrical Constru­ctions, which Cartes makes use of, p. 6, and 7. in resol­ving these 3 Equations, [...], & [...]. For in the First case, since he makes NO or NL ( Fig. 89. 11. 1.) or NP ½ a, and the Tangent ZMb, make MNO through the Center xx will =Z the quantity sought; which thus appears: Making MO=Z, NM will =Z½ a, and its □ [...]. But the □ OMP (which is by the present Prop. = □ LM, i. e. bb) together with the □ NL or NP i. e. ¼ aa is = □ NM by the Pathagor. Theor. Therefore [...]. i e. (taking away from both sides ¼ aa) ZZ− a Z= bb i. e. (by adding on both sides aZ) ZZ= aZ+ bb: Which is the Equation proposed. In the Second Case, if you make PM=Z (as Cartes makes it) you'll have [...], and to this again as before you'l have = bb+¼ aa. Therefore ZZ+ aZ= bb: Therefore ZZ=− aZ+ bb: Which is the very Equation of the second Case. In the third Case, whether you make the w ole Secant RM (N. 2) or that part of it without the Circle QM=Z, the Root sought, there will come out on both sides the same Equa­tion of the third Case; and so it is manifest, that this Equation has those 2 Roots. For if RM be =Z (adding + to the Fig. of Cartes the Line NO which shall bisect QR, and makes OM=LN) OR or OQ will be =Z−½ a, and so the □ OQ=ZZ− aZ+ ¼ aa, and this together with the □ RMQ (which is by vertue of the pres. Prop. = □ LM) =□ NQ. i. e. OM, i. e. ZZ− [Page 134] [...], i. e. (by adding aZ and taking away ½ aa) [...]; i. e. (taking away bb) ZZ=aZ− bb: Which is the the very Equation of the third Case. But if QM be made =Z, OQ or OR will = ½−Z, and its □ = [...], as well as the former, and so all the rest. Q. E. D.

SCHOLIUM. II.

NOW if you would immediately deduce these Rules by the present Proposition, without the Pythagorick Theorem. It may ( e. g. in the first Case) be done much shorter thus: If MO=Z and NO or NP=½ a, then will PM=Za: There­fore □ OMP= [...], or the □ LM, by the pres. Pro­position; by adding therefore to both sides aZ, you'l have ZZ=aZ+ bb, which is the very Equation of the first Case. In the second Case, if MR be Z, QM or PR will = a−Z: Therefore □ RMP [...] = bb, i. e. aZ= bb+ZZ, i. e. aZ− bb=ZZ; but if QM=Z, RM will = a−Z. There­fore □ RMP aZ−ZZ= bb, as before, &c.

SCHOLIUM III.

FRom the 2 Consect. of the present Proposition, flows ano­ther Rule for solving the last Case of Plain Trigonometry, which we solved in the Consect. of the foregoing Prop. viz. If you have all the three sides of the Oblique-angled Triangle BCD ( Fig. 90) given, if from the Center C, at the distance of the lesser side CB you describe a Circle, then will, by Consect. 2. of the present Proposition, BD the Base of the Triangle (here we call the greatest side of the Triangle the Base, or in an Equicrural Triangle, one of the greatest) will be to AD, (the sum of the Sides DC+CB) as DE the difference of the Sides, to DF the Segment of the Base without the Ciecle; which being found, if the remainder of the Base within the Circle be divided into two equal parts, you'l have both FG and GB, as also DG; which being given, by help of the Right-angled ▵▵ GBC and GDC all the Angles required may be found.

Proposition XLVIII.

IN any Quadrilateral Figure Ptol. lib. I. Almagesh. ABCD (Fig. 91. N. 1.) in­scribed in a Circle, the ▭ of the Diagonals AC and BD is equal to the two Rectangles of the opposite sides AB into CD, and AD into BC.

Demonstration.

Having drawn AE so that the Angle BAE shall be equal to the Angle CAD, the Triangles thereby formed (for they have the other Angles EBA and ACD in the same Segment equal, by ver­tue of Consect. 1. Prop. 33.) will be Equiangular one to another and consequently (by Prop. 34) as AC to AD so AB to BE. Wherefore by making AC= a and CD= ea, and AB= b, BE will be = eb. In like manner when in the ▵ ▵ BAC and EAD, the respective Angles are equal ( viz. adding the common part EAF to BAE and CAD, equal by Constr.) and besides the an­gles BCA and EDA in the same Segment are also equal; those Triangles will also be Equiangular, and AD will be to DE as AC to CB; wherefore by putting, as before, a for AC, and oa for CB, and c for AD, DE will = oc. Therefore the whole BD= eb+ oc. The Rectangle therefore of AC into BD will = eba+ oca=; the Rectangle of AB into CD= eba+□ of AD into BC= oac. Q. E. D.

SCHOLIUM

IN Squares and Rectangles ( N. 2.) the thing is self-evident. For in Squares if the side be a, the Diagonals AC and BD will be √2 aa, and so their Rectangle =2 aa will be manifestly equal to the two Rectangles of the opposite sides. In Oblongs, if the two opposite sides are a and the others b. the Diago­nals will be [...], and their Rectangle aa+ bb manifestly equal to the two Rectangles of the opposite Sides.

Proposition XLIX.

THe side (AB) of an Equilateral Triangle (ABC, Fig. 92. N. 1.) inscribed in a Ptol. lib. I. Almagesh. Circle, is in Power triple of the Radius (AD) i. e. of the □ of AD.

Demonstration.

MAke AD or FD= a, and so its Square aa. Since There­fore, having drawn DF thro' the middle of AB, or the middle of the Arch AFB let DE be = ½ a; for the angles at E are right ones, by Consect. 5. Definit. 8. and the Hypothe­nuses AD, AF, are equal, by Schol. of Definition 15. but the side AE is common. Therrefore the other Sides FE and ED are equal by, by Consect. 3. Prop. 43.) and the □ of the latter is ¼ aa, which subtracted from aa leaves ¾ aa for the □ of AE. Therefore the line AE is [...], and conse­quently AB [...], i. e. [...], i. e. [...]: therefore □ AB=3 aa. Q E. D.

CONSECTARYS.

I. IF the Radius of a Circle be = a, the side of an Inscribed Regular Triangle will be [...], e. g. if AD be 10, A B will be [...]; and if AD be 10,000,000, A B will be [...], i. e. 17320508, and the Perpen­dicular DE 5000,000.

II. Hence it is evident, that in the genesis of a Tetraedrum pro­posed in Def. 22, that the elevation CE ( Fig. 44, N. 1.) is to the remaining part of the Diameter of the Sphere CF as 2 to 1; for making the Radius C B= a and its □ aa, the □ of A B or BE will =3 aa, by the present Proposition. Therefore the □ of CB being subtracted from the □ of BD or BE, there remains the □ of CE=2 aa. But since CE, C B, CF, are continual Proportionals, by N. 3. Schol. 2. Prop. 34. CE will be to CF as the □ of CE to the Square of C B, by vertue of Prop. 35. i. e. as 2 to 1.

SCHOLIUM.

HEnce you have the Euclidean way of generating Eucl. 12 lib. 12. a Te­traedrum, and inscribing it in a given Sphere, when he bids you divide the Diameter EF of a given Sphere so that EC shall be 2 and CF 1, and then at EF to erect the Perpendicular CA, and by means thereof to describe the Circle A BD, and to inscribe therein an Equilateral Triangle, &c.

Proposition L.

THE Side (AB) of a Regular Tetragon (or Square) (ABCD, Fig. 92. N. 2.) is double in Power of the Radius (AD).

Demonstration.

For having drawn the Diameter AC and BD, the Triangle AO B is Right-angled, and consequently, by the Pythagorick Theo­rem, if the □ of AO and BO be made equal to aa, then will the □ of A B=2 aa, Q. E. D.

CONSECTARY.

THerefore when the Radius of the Circle AO is made = a, the side of the □ A B will =√2 aa, e. g. if AO be 10, A B will be √200; and if AO be 10,000,000, A B will be √200,000,000,000,000, i. e. 14142136.

Proposition LI.

THE side AB of a Regular Pentagon Eucl. Prop. 13. lib. 13. (ABCDE) (Fig. 93. N. 1.) is equal in Power to the side of an Hexagon and Deca­gon inscribed in the same Circle, i. e. the □ of AB is equal to the Squares AF and AO taken together.

Demonstration.

Make AO= a and AF= b, A B= x: We are to demonstrate that xx= aa+ bb: which may be done by finding the side A B by the parts BH and HA, Eucl. Prop. 10. lib. 13. after the following way: First of all the angle AO B is 72°, and the o­thers in that Triangle at A and B 54°. But BGG is also 54°, as subtending the Decagonal Arch BF of 36°, and also one half of it FG of 18°. Therefore the ▵▵ A BO and H BO are Equiangular, and you'l have [Page 140] As A B to BO so BO to BH [...]

Secondly, in the Triangle BFA the Angles at B and A are e­qual by N. 3. Consect. 5. Def. 8. and by vertue of the same also the Angles at F and A in the ▵ FHA are so too. Wherefore the ▵▵ BFA and FFA are Equiangular, and you'l have

As BA to AF so AF to AH [...]

Therefore the whole line A B (because the part AH is found = [...] and BH= [...]) will be [...], which was first made = x; so that now [...] is = x, and multiplying both sides by x, aa+ bb= xx. Q. E. D.

CONSECTARY I.

THerefore if the Radius of a Circle be (a) the side of a Pen­tagon A B will be [...].

CONSECTARY II.

THerefore the □ AI= [...] and □ OI= [...] i. e. [...]. Therefore OI= [...]. Which yet may be expressed otherwise, viz. OI= [...].

Demonstration.

Make Eucl. Prop. 1. lib. 14. OA or OF ( N. 2.) as before = a, AF= b, and FI now = x; then will OI= a− x and having drawn the Arch FK at the Interval AF, so that AK may be equal to this, and FI= [Page 141] IK= x; then will the angle IKA=F72°. Therefore the an­gle AKO=108; and since KOA is 36°, KAO will be also 36°, and so KA=KO=AF= b. Wherefore OI is = b+ x, which was above a− x. Therefore 2OI= [...], i. e. a+ b. Therefore OI= [...].

CONSECTARY III.

THerefore the difference between the Perpendicular of the Triangle DE ( Fig. 92 and 93. N. 1.) and the Perpen­dicular of the Pentagon OI is =½ b, by vertue of Consect. 2. of this and of the Demonstrat. of Prop. 49.

CONSECTARY IV.

HEnce is also evident, by the present Proposition, that which in Fabri's Genesis of an Icosaedr. Def. 22. we said, viz. that (See Fig. 46.) B a is equal to the side of a Pentagon BA, because, viz. Fa is =to the Semidiameter OB, and BF is the side of a Decagon.

Proposition LII.

THE side of an Hexagon is in Power equal to the Radius, as being it self equal to it by N. 1. Schol. Def. 15.

Proposition LIII.

THE side of a Regular Octagon (ABCD, &c. Fig. 94.) is equal in Power to half the side of the Square, and the difference (PB) of that half side from the Radius, taken together.

Demonstration.

For that the □ of AB is = to the □ of AP+□ BP, is evident from the Pythag. Theor. But that PO is =PA half the side of the Square, is evident from the equality of the Angles PAO and POA, since each is a half right one or 45°. Where­fore [Page 142] the side of the Octagon is equal in Power to half the side of the Square, &c. Q. E. D.

CONSECTARY.

THerefore, if the Radius be = a, AP will be, by vertue of the Pythag. Theor. = [...] and P B= [...] of that □ ½ aa; and of this □ [...]. Therefore the Sum of □ AB= [...]. Therefore the side of the Octagon = [...]. e. g. If AO be 10, AB will be [...], and if the Radius AO 10000000, AB will be [...], by vertue of the present Cons. or from the Prop. it self, if AO be 10000000, AP will be vertue of the Cons. of Prop. 50.=7071068, and consequently BP=2928932 The Squares of these added together give the □ AB, and the Root thence extracted A B=7653 668.

Proposition LIV.

THE side of a Regular Decagon Eucl. 9. lib. 13. Corol. is equal in Power to the greatest part of the side of an Hexagon cut in mean and extreme Reason.

Demonstration.

Suppose BD ( Fig. 95.) divided in mean and extreme Reason in E, and BA to be joined to it long ways = to the side of a Decagon inscribed in the same Circle, whose Radius is BC or AC= BD. Now we are to demonstrate that DE the greatest part of the de of the Hexagon BD divided in mean and extreme Rea­son, is equal to the side of a Decagon BA, and the Power of the one equal to the Power of the other. Because the Angle AC B is 36°, A BC and A72°, and consequently C BD 108°, BCD and D will be each 36°, and so the whole ACD 72°. (that so CD may pass precisely thro' the other end of the side of the Pen­tagon [Page 143] AF.) Wherefore the ▵ ▵ ABC and ADC are Equi­angular, and

  • AD to AC i. e. BD
    • as AC i. e. BD

to AB. Therefore the whole line AD is divided in mean and extreme Reason. But BD is al­so divided in the same Reason by Hyp. Wherefore

  • As AD to D B and D B to BA,
  • So DB to DE and DE to E B.

Therefore D B is in the same Proportion to DE as D B to BA. Therefore DE is = BA, and the Power of the one to the Power of the other. Q. E. D.

CONSECTARYS.

THerefore, if the Radius of the side of the Hexagon is a the side of the Decagon will be [...], by Schol. 2. Prop. 27. e. g. if the Radius be 10, the side of the Decagon will be [...], and if the Radius be put 10000000, the side of the Decagon will be = [...], viz. by adding the Square of the Radius and the Square of half the Radius into one Sum; whence you'l have the side of the Decagon =6180340; the half whereof 3090170 gives the difference between the Perpendiculars of the Triangle and the Pentagon, by Cons. 3. Prop. 51.

II. The side therefore of the Pentagon is by Prop. [...]; for the Square of the Hexagon is aa or [...] aa, the □ of the Decagon [...]: the Root extracted out of the Sum of these is the side of the Pentagon, viz. [...] e. g. if the Radius be 10, the side of the Pentagon will be [...], and if the Radius be put 10000000, since the side of the Hexagon is equal to it, and the side of the Decagon 6180340, their Squares being added into one Sum, the Root extracted out of that Sum will give the side of the Pentagon, 1755704 nearly; and the sides being collected into [Page 144] one sum, the half of it 8090170 will give the Perpendicular in the Pentagon OI, by Consect. 2. Prop. 51.

SCHOLIUM I.

TO illustrate what we have deduced in the Consectarys of Prop. 51, you may take the following Notes. If a be put =10 or [...], the side of the Decagon will be = [...], i. e. [...] nearly = b; therefore the □ aa= [...] and the □ bb= [...]: therefore aa+ bb or the □ AB= [...], the Perpen­dicular OI= [...] divided by 2, that is, [...]. Now the □ of AI is ¼ of the □ A B= [...] the □ OI= [...] = [...]. Now if you add the □ AI and the □ OI, the sum will be = □ AO= [...], i. e. [...], or near 100. Thus likewise, since the Perpendicular above found OI, in Consect. 1. may be also determined by [...], since aa is = [...], and 3 aa [...], subtracting from it bb= [...] the Remainder will be [...], and this be­ing divided by 4, you'l have the □ OI= [...]. and the Root of it extracted [...] nearly; so that those two different quantities in Consect. 1. will rightly express the same Perpendicular OI.

SCHOLIƲM II.

NOW therefore as we have Practical Rules to determine A­rithmetically the sides of the Pentagon and Decagon, so also they may be found Geometrically by what we have demon­strated. For if the Semidiameter C B ( Fig. 96. N. 1.) be divided into 2 parts, EC will =½ a; and erecting perpendicularly the Radius CD= aDE will = [...]. Moreover if you cut of EF equal to it, FC will be = [...]= to the side of the Decagon, by Consect. 1. Having therefore drawn DF, which is equal in Power to the Radius or Side of the Hexagon DC, and the side of the Decagon FC together, by the Pythag. Theorem [Page 145] it will be the side of the Pentagon sought. Much to the same purpose is also this other new Construction of the same Problem, wherein BG ( Numb. 2.) is the side of the Hexagon B D the side of the Square, to which GF is made equal, so that FC is that side of the Decagon, and DF of the Pentagon; which we thus demonstrate after our way: Having bisected GH the side of an Equil. Triangle, the Square of GE will be [...], by Prop. 48. which being subtracted from the Square of GF=2 aa, viz. [...] aa, by Prop. 49. there will remain for the Square of EF [...] aa, and for the line EF [...], and for FC [...], which is the side of the Decagon, as DF of the Pentagon, after the same way as before.

Proposition LV.

THE side of a Quindecagon (or 15 sided Figure) is equal in Power to the half Difference between the side of the Equil. Triangle and the side of the Pentagon, & moreover to the Difference of the Perpendi­culars let fall on both sides taken together.

Demonstration.

For if AB ( Fig. 97.) be the side of an Inscribed Triangle, and De the side of a Pentagon parallel to it; AD will be the side of the Quindecagon to be inscribed, by Consect. 4. Def. 15. But this side AD in the little Right-angled Triangle, is equal in Power to the side AH (which is the half Difference between AB and DE) and the side HD (which is the difference be­tween the Perpendiculars CF and CG) taken together by the Pythag. Theor. Q. E. D.

CONSECTARY.

HEnce if we call the Side A B of the Equil. ▵ c, and mak [...] the side of the Pentagon DE= d, AH will = [...] HD is =½ b, by Consect. 3. of Prop. 51. Since therefore the ▵ AH is = [...] and the ▵ HD= [...] the ▵ AD w [...] [...]

Therefore the side of the Quindecagon = [...]that is, Collecting the Square of the half difference of the side of the Triangle and Pentagon, and the Square of the difference of the Perpendiculars into one Sum, and then Extracting the Square Root of that Sum, you'l have the side of the Quindeca­gon sought. E. 9. if rhe Radius CI be made 10000000 the difference of the sides of the Triangle 17320508, and of the side of the Pentagon 11757704 will be 5564804, and the half of this 2782402; but the difference of the Perpendicular CF from the Perpendicular CG, is 3090170.

The Squares therefore of these Two last Numbers being Collected into one Sum, nnd the Root Extracted will give the side of the Quindecagon 4158234 nearly.

SCHOLIUM.

HEre we will shew the Excellent use of these last Proposition [...] in making the Tables of Signs. For having found above, supposing the Radius of 10000000 parts, the sides of the chief Regular Figures, if they are Bisected, you will have so many Primary Sines; viz. from the side of the Triangle the side of 60 Degrees 8660754, from the Side of the Square, the Sine of 45° 7071068; from the Side of the Pentagon, the Sine of 36° 5877853; from the Side of the Hexa­gon, the Sine of 30° 5000000; from the Side of the Octagon, the Sine of 22° 30 3826843; from the Side [Page 147] of the Decagon the sine of 18°=3090170; from the side of the quindecagon lastly the sine of 12°=2079117. From these seven primary sines you may find afterwards the rest, and consequently all the Tangents and Secants according to the Rule we have deduc'd, n. 3. Schol. 5. Prop. 34. and which Ph. Lansbergius illustrates in a prolix Example in his Geom. of Triangles Lib. 2. p. 7. and the following. But af­ter what way, having found these greater numbers of sines, Tangents, &c. Logarithms have been of late accommodated to them, remains now to be shewn, which in brief is thus; viz. the Logarithms of sines, &c. might immediately be had from the Logarithms of vulgar numbers, if the tables of vul­gar numbers were extended so far, as to contain such large numbers; and thus the sine e. g. of o gr. 34. which is 98900 the Logarithm in the Chiliads of Vlacquus is 49951962916. But because the other sines which are greater than this are not to be found among vulgar numbers (for they ascend not be­yond 100000, others only reaching to 10000 or 20000) there is a way found of finding the Logarithms of greater num­bers, than what are contained in the Tables. E g. If the Logarithm of the sine of 45° which is 7071068 is to be found, now this whole number is not to be found in any vulgar Ta­bles, yet its first four notes 7071 are to be found in the vul­gar Tables of Strauchius with the correspondent Logarithm 3. 8494808, and the five first 70710 in the Tables of Vlac­quus with the Log. 4. 8494808372. One of these Loga­rithms, e g. the latter, is taken out, only by augmenting the Characteristick with so many units, as there remain notes out of the number proposed, which are not found in the Tables, so that the Log. taken thus out will be 6. 8494808372. Then multiply the remaining notes of the proposed number by the difference of this Logarithm from the next following, (which for that purpose is every where added in the Vlacquian Chiliads, and is in this case 61419) and from the Product 4176492 cast away as many notes as adhere to the propo­sed number beyond the tabular ones, in this case 2; for of the remainder 41764, if they are added to the Logarithm before taken out, there will come the Logarithm requi­red 6. 8494850136, viz. according to the Tables of Vlac­q [...]us, wherein for the Log of 10 you have 10,000,000, [Page 148] 000; but according to those of Strauchius which have for the Logarithm of 10 only 10,000,000, you must cut off the three last notes, that the Logarithm of the given sine may be 6.8494850; as is found in the Strauchian and other tables of sines, except that instead of the Characteristick 6 there precedes the Characteristick 9, whereof we will add this reason: If the Characteristicks had been kept, as they were found by the rule just now given, the Logarithm of the whole sine (which is in the Strauchian Tables 10,000,000) would have come out 70,000,000, incongruous enough in Trigo­nometrical Operations. Wherefore that Log. of the whole sine might begin from 1, for the easiness of Multiplication and Division they have assumed 100,000,000; the Cha­racteristick being augmented by three, wherewith it was con­sequently necessary to augment also all the antecedent ones; and hence e. g. the Logarithm of the least sine 2909 begins from the Characteristick 6, which otherwise according to the Tables of vulgar numbers would have been 3.

Having found after this way the Logarithms of all the sines (altho' here also if you have found the Logarithms of the signs of 45° and moreover the Logarithm of 30, the Loga­rithms of all the rest may be compendiously found by addition and substraction from a new principle which now we shal omit) the Logarithms of the Tangents and Secants may easi­ly be found also, only by working, but now Logarithmically, according to the Rules of Schol. 5. Prop. 34. n. 5. and 6.

Proposition LVI.

THE side of a Tetraëdrum Eucl. 13. lib. 13. or equ [...] Pyramid is in power to the Diameter of a circumscribed Sphere, as 2 to 3.

Demonstration.

For because by the genesis of the Tetraëdrum Def. 22 (see its Fig. 44 n. 1.) and Schol. Prop. 49. OC is ⅓ of the se­midiameter OB, which we will call a the □ of CB will be = [...] aa by the Pythag. Theor. and so the power (or Sq.) of the side of the Tetraëdrum = [...] aa by Prop. 49. but the pow­er [Page 149] of the Diam. 2 a or [...] a is [...] aa. Therefore the power of the side of the Tetraëdrum is to the power of the Diam. as 24 to 36. i. e. (dividing each side by 12) as 2 to 3. Q. E. D.

Or more short.

The □ of CB is =2 by Schol of Prop. 49. and the □ of EC=4. Therefore the □ EB=6. But the □ EF is =9. Therefore the □ of EB is to the □ of EF as 6 to 9, i. e. as 2 to 3. Q. E. D.

CONSECTARY.

THerefore if the Diam. EF be made = a, the side EB will be = [...].

Proposition LVII.

THE side of the Octaëdrum Eucl. 14. lib. 13. is in power one half of the Diameter of the circum­scribed Sphere.

Demonstration.

For since by the genesis of the Octaëdrum Def 22. (see Fig. 44. n. 2.) CA, CB, CF, &c. are so many radii of great circles, if for Radius you put a, the square of AF will be =2 aa by the Pythag. Theor. But the square of the Diam. FG=2 a is 4 aa. Therefore the power of the side is to the power of the Diam as 2 to 4, i. e. as 1 to 2. Q. E. D.

More short.

Because AF is also the side of a square inscribed in the grea­test circle by the gen. of the Octaëdrum; the □ of AF will be by Prop. 50. to the □ of FC as 2 to 1: Therefore to the square of FG as 2 to 4, by Prop. 35. Q. E D.

CONSECTARY.

THerefore if the Diam. of the sphere be made (a) the side of the Octaëdrum AF will be [...].

Proposition LVIII.

THE side of the Hexaëdrum or Cube Eucl. 15. lib. 13. [...] in Power subtriple of the diameter of the circumscribed sphere.

Demonstration.

Making a the side of the inscribed cube GF or FE ( Fig. 98.) the square of the diagonal GE of the base of the cube will be 2 aa by the Pythag. Theor. and by the same Reason the square of the diam. of the cube and the circumscribed sphere GD will be = to the square of GE+□ DE=3 aa. Q. E. D.

CONSECTARYS.

I. THerefore if the diameter of the sphere be made = a, the side of the cube AB will be [...].

II. The diameter of the sphere is equal in power to the side of the Tetraëdrum and cube taken together. For if the power of the diam. of the sphere be made aa the power of the side of the Tetraëdrum will be ⅔ aa by Consect. Prop. 56. and the power of the side of the cube ⅓ aa by the pres. Consect. 1. Wherefore these two powers jointly make aa. Q. E. D.

Proposition LIX.

THE side of the Dodecaëdrum Eucl. Consect. 1. Prop. 17. lib. 13. is equal in power to the greater part of the side of the cube divided in mean and extreme reason.

Pag. 150.

97

98

99

100

101

102

103

Demonstration.

For if to the side of the cube AB ( vid. Fig. 45. n. 1.) and to its upper base ABCD you conceive to be accommodated or fitted a regular Pentagon according to the genesis of a Dodeca­ëdrum laid down in Def. 22, and at the interval B e you make the arch ef, the ▵ ▵ AB e and A ef will be equiangular; (for the angles at A and B being 36°, and A eB 108, having drawn ef, the angles B ef and B fe are each 72°; therefore A ef the remaining angle will be 36°) wherefore as AB to B e ( i. e. B f) so A e ( i. e. B e or B f) to A f. Therefore the side of the cube AB is divided in mean and extream reason in f, and B e the side of the Dodecaëdrum is = to the greater part B f. Q. E. D.

SCHOLIƲM.

HEnce would arise a new method of dividing a given line in mean and extream reason, viz. if you apply to the given line a part of the equilateral Pentagon by means of the angles A and B 36°, and at the interval B e you cut off B f. This angle may be had geometrically, if another regular Pen­tagon be inscribed in a circle, and having drawn also a like subtense, if the angles at the subtense are made at A and B equal, by Eucl. 23. lib. 1.

Proposition LX.

THE side of an Icosaëdrum Eucl. Prop. 16. Co­roll. lib. 13. is equal in power to the side of a Pentagon in a circle containing only five sides of the Icosaëdrum; and the semi diameter of this circle is in power sub­quintuple of the Diam. of the sphere of the circumscribed Icosa­ëdrum.

Demonstration.

Both these are evident from the genesis of the Icosaëdrum in Def. 22. The first immediately hence, because all the other [Page 152] sides of the triangles ( Fig. 99.) A a, B a, &c. are made equal to the side of the Pentagon AB by Consect. 4. Prop. 51. The latter from this inference; if for OA the radius of the circle you put a (since the side of the Pentagon, which here is also the side of the Icosaëdrum, it will be equal in power to the radius and side of the Decagon taken together by the aforesaid Prop.) the altitude OG will be the side of the Decagon = [...] by Consect. 1. Prop. 54. to which the equal in­ferior part oH being added, and the intermediate altitude O o = a, you'l have the whole diameter of the circumscribed sphere GH= [...] i. e. [...] i e. [...] i. e. [...] and so the square of the diameter of the sphere will be 5 aa: Therefore the square of the diameter of the sphere is to the square of the semi-diam. of the circle containing the five sides of the Icosaëdrum as 5 to 1. Q. E. D.

SCHOLIƲM.

IT is also evident that a sphere described on the diameter GH will pass thro' the other angles of this Icosaëdrum; for assuming the center between O and o the radius FG will be = [...]. But FA is also = [...]; for the □ of FO is =¼ aa, and the □ AO= aa: Therefore the sum is = [...] aa = □ FA. Q. E. D.

CONSECTARY I.

THerefore, if the radius of the circle ABCDE remain a, you'l have the altitude OG [...], and the side of the Icosaëdrum [...], by Cons. 1. and 2. Prop. 54. and the diam of the circumscribed Sphere 2 [...], as is evident from the Demonstration.

CONSECTARY II. Being a general one of the five last Propositions.

IF AB ( Fig. 100) be the diameter of a sphere Eucl. Prop. 18. and los [...], lib. 13. divided in D so that AD shall be ⅓ AB, then (having erected the perpendicular DF) BF will be the side of the Tetraëdrum by Prop. 56. and AF the side of the Hexaëdrum by Prop. 58. Cons. 2. and BE or AE (erecting from the center the perpendicular CE) will be the side of the Octaëdrum by Prop. 57. Now if AF be cut in mean and extreme reason in O, you'l have AO the side of the Dodecaëdrum by Prop. 59. Lastly, if you erect BG double of CB, HI will be double of CI, and the □ of HI=4 □ of CI; consequently the □ CH or CB=5 □ CI. Therefore the □ of AB (double of CH) is also=to 5 □ of HI (which is double of CI) therefore HI is the radius of the circle circumscribing the Pentagon of the Icosaëdrum, and IB the side of the Decagon inscribed in the same circle, and HB the side of the Pentagon, and also the side of the Ico­saëdrum, by Prop. 60.

The End of the first Book.

THE SECOND BOOK.

SECTION I. Containing DEFINITIONS.

Definition I.

IF a Cone ABC ( Fig. 101.) be conceived to be cut by a plane at right angles to the side of the cone, e. g. BA; the Plane EFGHE arising by this section, and terminated on the external surface of the cone by the curve line HEG, &c. was anciently by Euclid, Archimedes, &c. called the Conick Section; and if the sides of the cone AB and BC made a right angle at B, as n. 1. the section was particularly called the Se­ction of a right-angled Cone; but if it made an obtuse angle, as n. 2. it was called the Section of an obtuse-angled Cone; if, lastly, it made an acute one, as num. 3. it was called (3) the Section of an acute-angled Cone.

Definition II.

BUT afterwards their Successors, particularly Apollonius Pergaeus, found from the properties of these Curves, which their Predecessors had happily discovered, that the same (all of them) might be generated in one and the same cone whether right-angled, obtuse-angled, or acute-angled, if the section EF ( Fig. 102.) is made in the first case parallel to [Page 155] the opposite side BC; in the second case, if it meet that side produced upwards; for the third, when it meets downwards with the diameter of the base AC produced to D. And to give new names (for the old ones would do no more now) to these Sections, to distinguish them one from another, nomi­nating them from their Properties hereafter demonstrated, they called the first a Parabola, the second an Hyperbola, the third an Ellipsis.

Definition III.

BUT it is evident, that only the plane making the secti­on of the second case, being according to the line FED produced or carried on, ( Fig. 103.) will fall upon the verti­cal Cone aBc comprehended under the sides AB, CB, &c. continued onwards, and there produce another Section oppo­site to the former; whence these, viz. GEHG and gehg are called opposite Sections.

Definition IV.

BEsides these names of the sections, there are several others made use of to denote various lines drawn and considered both within and without those sections, the chief whereof we shall here explain. And first of all, in general the line EF so let fall thro' the middle of the section from its top E (which is called the Vertex of the section) to the diameter of the base of the Cone AC (produced if occasion be) that it shall bisect the line GH and all others parallel to it, is called the Di­ameter of that Section; and particularly it is called the Axis of the Section if it cuts them at right angles or perpendicularly; as also they name those lines GH, KN, &c. which are cut in­differently by the diameters, but at right angles by the Axis, those, I say, they call Ordinates, or Ordinate Applicates, and their halves, FG, IK, &c. Semiordinates, (or some also call the latter Ordinates, and the former double Ordinates) and the parts of the Ax or Diameter EF, EI, &c. are called Abscis­sa's (by some intercepted axes and diameters)

Definition V.

PArticularly in the hyperbola they call the continuation of the ax or diameter ED 'till it meet the opposite side cB, i. e. to the vertex of the opposite section, the Latus transver­sum of the Hyperbola, to which there answers in the Ellipsis the axis or longest diameter, and so by latter Authors is called by the same name, but by Apollonius the transverse Ax or the transverse Diameter, as also the shortest ax or diameter is cal­led the Conjugate, and its middle point O is called the Center of the Section or of the opposite Sections.

Definition VI.

THey called also a certain line EL ( Fig. 101.) by the name of Latus Rectum, which is particularly to be found in all the sections, as we shall hereafter shew: And be­cause this Latus Rectum is a sort of a Rule or Measure, ac­cording to which the squares or powers of the ordinates used to be estimated or valued (as we will shew in its proper place) therefore the Ancients used to call it by a Periphrasis Linea se­cundum quam possunt Ordinatim applicatae, or the measure of the powers of the Ordinates; by some latter Authors it is cal­led the Parameter. Now a mean proportional PQ found be­tween this Latus Rectum and the Latus Transversum (Fig. 104 n. 1. and 2) (see also hereafter Consect. 2. Prop. 8.) and drawn thro' the centre O parallel to the Ordinates is cal­led the second Axis or Diameter, or the Conjugate of the Hy­perbola.

Definition. VII.

NOW if we conceive the diameter or conjugate ax PQ brought down to the Hyperbola so that its middle point O shall touch its vertex in E, and from the center O you draw the right lines OR, OS, thro' the ends of this tangent line p and q, these are the lines which Apollonius, Prop. 1. lib. 2. de­monstrates, that tho' by being continued, they always ap­proach nearer and nearer to the curve GEH, and come so [Page]

Pag. 156.

104

105

106

107

[Page] [Page 157] much the nearer by how much the farther they are continued, yet they will never concur with it or touch it, for which rea­son they are called Asymptotes or non-coincident lines; and by some the Intactae. Which non-coincidency appears most manifestly where the hyperbolical section of the cone is made parallel to the triangular section thro' the axis of the cone ABC ( n. 3.) along the line e. f. parallel to the ax BF. For if we conceive the hyperbola geh to move forwards always pa­rallel to it self, according to the direction of the equal and pa­rallel lines gG, fF, and hH, 'till it stands in the position GEH; it is manifest that the curve line GEH is distant on both sides from the right lines BC, BA, the length of the versed sine of the equal arches hC and gA in the circumference of the circu­lar base, while in the mean time it is evident that they ap­proach nearer and nearer to them. So that hence there flow the following

CONSECTARYS.

I. IN this case the sides of the cone are the Asymptotes of the hyperbola, while it is manifest, that the point B is its centre, and EB half the transverse diameter; which appears from n 1. and 2. of the pres. Fig. for the section ef being made parallel to the the ax of the cone DF by def. 5. de (which in the case n. 3. would coincide with dq) is the trans­verse diameter, but the triangles dpq and O pE are equiangular, and consequently as pE is to ½ pq, so is OE to ½ dq.

II. The lines AG and HG ( num. 3.) are equal, as being the versed sines of equal arches; and in like manner ( n. 1. and 2.) RG and HS are equal, since FR and FS as well as E p and E q are so also (for

  • as OE to E p, −E q
  • so OF to FR −FS

and the semiordinates FG and FH are also equal.

III. Consequently □ □ of RG into GS and of HS into HR are equal, &c. all which hereafter we will more univer­sally demonstrate.

Definition VIII.

IF a Parabolick plane Archim. de Conoid. & Spher. Def. 1. HEGFH ( Fig. 105. n. 1.) together with a triangle HEG inscribed in it, and a rectangle HL circumscri­bed about it, be conceived to be moved round about the ax EF on the point F; it will be evident that by the triangle there will be generated a cone, by the rectangle a cylinder, and by the parabola a parabolick solid, which with the comprehended cone, and the comprehending cylinder, will have the common base HIGK and the same altitude EF, and was by Archimedes named a Parabolick Conoid.

Definition IX.

IF moreover an Archim. de Conoid. & Spher. Def. 3. hyperbolick plane HEGFH ( n. 2.) with the inscribed trian­gle HEG, and another circumscribed one ROS made by the Asymptotes OR, OS, be concei­ved to be turned about the common ax OEF on the point F; it will be evident that there will be described by the inscribed triangle a cone comprehended within side, and by the hyper­bola an Hyperbolical Conoid upon the same base HIGK and of the same heighth EF; and by the ▵ ROS another cone which Archimedes calls the comprehending cone, whose base is RTSV, and its altitude composed of the axis of the hyperbo­la EF and half the transverse ax OE (which Archimedes cal­led the additament of the ax of the hyperbola) and which we may commodiously divide into two parts, viz into the cone OPMQL, whose base has for its diameter the conjugate ax PQ, and its altitude equal to half the transverse ax; and into a Curti-cone or detruncated cone terminated by the two bases PMQL and RTSV, but answering in altitude to the conoid and inscribed cone: From which, as comprehending it, if you take away the included conoid, there will remain the hollow curticone terminated below by the Annulus or ring RGIVHSKT and above by the circular base PMQL, and generated in the circumvolution by the intermediate lines EP, GR, &c. or the mixtilinear plane EGRP. Now if we suppose instead of [Page 159] this comprehending cone a circumscribed cylinder on the same base and of the same altitude with the conoid and included cone, you'l have every thing like as in Def. 8.

CONSECTARYS.

NOW if we suppose the 1. case of Def. 9. to be expres­sed by the Fig. of Def. 7. n. 3. and conceive the pre­sent figure brought thence to be turned round about the ax BEF ( Fig. 106.) we may deduce these following things in the room of Consectarys for the foundation of our future de­monstrations.

I. The lines EQ, RS, HC, &c. of the mixtilinear space comprehended between the hyperbola and the Asymptotes ( viz. the excess of the ordinates) altho' they are unequal, and by descending always grow less, yet in this circumvolution they will describe equal circular spaces, viz. EQ a whole circle, (or circular plane) but RS and HC, &c. circular Annuli or rings all of the same bigness; which will thus appear to any one who compares this figure with the former: Since the spa­ces generated by the lines EQ, FC, &c. are as the squares of those lines, and the □ of F h or FC exceeds the square of fh or FH by the excess of the square F f or E e or EQ, conse­quently the quantity generated by FC will exceed that gene­rated by FH the excess of that generated by EQ; and also that generated by FH by the excess of that generated by HC; it is manifest that the circle generated by EQ will be equal to the annular space generated by HC; and the same will in like manner be evident of any spaces produced by RS.

II. Therefore the hollow detruncated cone generated by the space EHCQ according to Def. 9. will be equal to a cylinder generated by the rectangle FIQE; for all the indivisibles of the one, are equal to all the indivisibles of the other by Con­sect. 1.

Definition X.

IF, lastly, Archim. de Conoid. & Spher. Def. 6. an elliptical plane be turned about either of the axes, viz. either the lon­gest DE ( Fig. 107. n. 1) or the shorter AB ( n. 2.) there will be thence formed an elliptical solid, called by Archimedes a spheroid; which in the first case will be an oblong or erect one, in the other a flat or depres­sed one: And it is self evident, that if before this circumvolu­tion of the ellipsis, there be inscribed in one of its halves a triangle, and also a rectangle circumscribed about it, having the same altitudes and bases with the semi ellipse, there will af­terwards in the circumvolution be described by the triangle a cone, by the rectangle a cylinder, to which afterwards we will also compare the half spheroid; as also both the conoids with their inscribed and circumscribed cones and cylinders.

Definition XI.

IF upon the right line AE ( Fig. 18.) you conceive a wheel or circle to rowl, until its point A, with which it touches the said line, come to touch it again in E; the circle wil mea­sure out the line AE equal to its periphery; but the point A by its motion will describe the curve line AFE, which is cal­led a Trochoid or Cycloid; and the area which this curve with the subtense AE comprehends, is named the Cycloidal Space; and the circle by whose motion they are determined is called the generating Circle.

CONSECTARY.

IT is evident from the genesis of this curve that the descri­bing point a will always be as much distant in the circle from the point of contact d or c, as the point A in the right line passed over AE, is from the same point of contact, i. e. if the point d is distant from A the fourth part of the line AE, the arch da will also be the fourth part of the circle considered in this second station; and the point c being distant from A half of the interval AE, the arch ca will be also half of the [Page]

Pag. 160.

108

109

110

111

[Page] [Page 161] circle, and so the point a coincide in the curve with F: And when the point e is distant from E only an eighth part of the whole line AE, the arch ea will also be the eighth part of the whole circle.

Definition XII.

IF the right line BA ( num. 1. 109.) one end at B remain­ing fixed, be moved round at the other end with an equal motion from A thro' C, D, E to A back again, and in the mean while, there be conceived another moveable point in it to move with an equal motion along the line BA from B to A, so that in the same moment wherein the moveable point A absolves one revolution, the other moveable point shall also have passed thro' its right lined way, coinciding with the point A retur­ned to its first situation; this extremity A by its revolution will describe the circle ACDEA, and that other moveable point another curve B, 1, 2, 3, 4, &c. which with Archi­medes we will call a Helix or spiral Line, and the plane space comprehended under this spiral line and the right line BA in the first station is called a spiral space. Now if we suppose, e. g. the right lined motion of the point moving along BA to be twice slower than in the former case, so that (see num. 2.) in the same time that the point A makes one whole Re­volution, the other moveable point shall come to F, making half the way BA, and then at length shall concur or meet with the extremity, when that shall have performed the other revolution; and so there will be described a double spiral line, whose parts with Archimedes we will so distinguish, that as he calls the part of the right line BF, passed over in the first revolution, simply the first line, and the circle made by the right line BF the first Circle; so we will call that part of the curve which is described in that time or revolution B 136912 the first Helix or the first Spiral, and the area comprehended un­der it the first spiral space: And, as the other part of the right line FA passed over in the other revolution is called the second line, and the circle marked out by the whole line BA the se­cond Circle; so the curve described in the mean while 12, 15, 18, 24, may be called the second spiral line, and the space [Page 162] comprehended under it the second spiral space, and so onwards From these Definitions there flow the following

CONSECTARYS.

I. THE lines B 12, B 11, B 10, &c. drawn out, and making equal angles to the first or second spiral (and after the same manner Archim. Prop. 2. of Spirals. B 12, B 10, B 8, &c. or B 12, B 9, B 6, &c.) are arithmeti­cally proportional, as is evident.

II. The lines drawn out to the first spiral as B 7, B 10, &c. are one among another as the arches of the circles inter­cepted between BA and the said lines Archim. Prop. 4. B 7, B 10, &c. which is also evident to any one who considers what we did suppose; for in the same time as the end A passes over seven parts of the circle, the other moveable point will also run over seven parts of the right line BA, &c.

III. Lastly, The right lines drawn from the initial point Archim. Prop. 15. B to the second spiral e. g. B 19 and B 22 ( num. 2.) will be one to ano­ther as the aforesaid arches together with the whole periphery added to both sides: for at the same time the extremity A runs thro' the whole circle or 12 parts and moreover 7 parts ( i. e. in all 19 parts) in the same time the other moveable point passes through 12 parts of the right line BA (in this case di­vided into 24 parts) and moreover 7 parts, that is, in all 19; and so in the others.

Definition XIII.

IF a right line BAF be conceived so to move within the right angles ADC, CDE, that on the one hand a certain point C fixed in one leg of the Norma or ruler may always glide along, and on the other hand a certain moveable point A may always run along the other side of the Norma (which complicated motion of the describing rule BAF, after what way it may be organically procured, may be seen by [Page] [Page]

Pag. 163.

112

113

114

115

116

117

118

[Page 163] the 110th. Figure;) by the extreme point of the moveable line B there will be described a curve called by its Inventor Nicomedes the first Conchoid, whereof this is a property, that the right lines CB, C b, drawn from its centre C to its Ambi­tus or curvity are not themselves, as in the circle, equal, but yet have all the parts, AB, ab, intercepted between the curve, and the directrix AE are equal; as is evident from its genesis.

Definition XIV.

IF, the diameters of a circle being AB and CD ( Fig. 111. n. 1.) cutting one another at right angles, you take BE or B e and BF or B f equal arches, and from E or e you draw the perpendiculars EG or Eg, and through these from D to F or f you draw the transverse lines DF or D f; the several points of intersection H, h, &c. decently connected together will exhi­bit the curve line D h, H hB (which may be continued also without the circle if you please, and) which is commonly at­tributed to Diocles, and called a Cissoid.

Definition XV.

IF, having divided the right line AE ( Fig. 112.) into the equal parts AB, BC, &c. from the points of the division A, B, C, &c. you draw the parallel lines A a, B b, C c, &c. in geometrical progression (as e. g. let A a be 1, B b 4, C c 16, D d, 64, &c. or B b 10, C c 100, D d 1000, E e 10000, &c.) and further bisecting AB, BC, &c, F, G, H, I, you let fall mean proportionals between the next Collaterals Ff, Gg, Hh, Ii, &c. and continue to do so till the parallels are brought near to one another; the curve line drawn thro' the extremities of these lines a f, b g, c h, d ie, will be the logarithmical line of the moderns, whose properties and uses are very excellent.

SCHOLIƲM.

AMong those uses, that is none of the least, from which this curve borrows its name, viz. in shewing the na­ture and invention of Logarithms. For, e g. 1. If this line was accurately delineated in a large space, the parts AB, [Page 164] BC, &c. being taken so big, that they might be subdivided not only into 100 or 1000 but into 10000 or 100000 parts; making AB 100000 (and so A 00000) AC would be 200000, AD 300000, &c. while in the mean while there answer to these as primary Logarithms in arithmetical pro­gression the geometrical proportional numbers, A a 1, B b 10, C c 100, D d 1000, E e 10000, &c. Whence, 2. Its Loga­rithm may be assign'd to any given intermediate number, e. g. to the number 982, for having cut off this number from Dd by a geometrical scale on the line DM, if you draw Mn pa­rallel to AD, and nN parallel to DM, it will give AN on the same scale, viz. the Logarithm sought, and reciprocally. But if, 3. it seem difficult to delineate a Figure so large, yet at least the clear conception of such a delineation evidently shews the arithmetical method, which those ingenious Men have made use of, who have made the tables of Logarithms with a great expence of Labour and pains, viz. by finding continual mean proportionals, arithmetical ones between any two Logarithms already known, and geometrical ones between two vulgar numbers answering to them, &c. by comparing what we have noted in Schol. 2. Prop. 20. Lib. 1. And we will note, 4 out of Pardies, that, since the Logarithms of numbers di­stant from one another by a decuple proportion, differ by the number 100000, having found the Logarithms of all the numbers from 1000 to 10000 you will at the same time have all the Logarithms of all the other numbers that are between 100 and 1000, between 10 and 100, and between 1 and 10, only changing the characteri­stick, and lessening it in the first case by unity, in the second by 2, in the third by 3; as e. g. if of the number 9, 900 you had found the Logarithm 399, 563, the Logarithm of the subdecuple number 990 would be ( viz. substracting from the former 100000) 299, 563. and the Logarithm again of this subdecuple of this 99 would be 199, 563, &c. Thus in the Chiliads of Briggs to the number

99000 Answ. 4,99563,51946
9900 — 3,99563,51946
990 — 2,99563,51946
99 lastly — 1,99563,51946.

But there will not arise such advantage for making Loga­rithms by this observation as it may at first sight seem to pro­mise, because there are 9000 numbers between 1000 and 10000 whose Logarithms must be found also, and but 900 between 100 and 1000, and but 90 between 10 and 100, and but 9 between 1 and 10, and so in all 999, which is not the ninth part of the former.

Definition XVI.

IF the radius AD ( Fig. 113.) be conceived to move equal­ly about the point A through the periphery of the quadrant DB, while in the mean time the side of the square DC re­maining always parallel to it self, descends also with an equal motion thro' DA, so that in the same moment the radius AD and the aforesaid side DC shall fall upon the base AB; or (if any one should think that this way the proportion of a right line to a circular one is supposed by a sort of Petitio Principii or begging the question) the right line DA as well as the quadrant DB being divided into as many equal parts as you please ( e. g. here both of them into 8) and drawing thro' these from the center A so many Radii and thro' them paral­lel lines; the points of intersection being orderly connected to­gether will exhibit a curve line, whose invention is attributed to Dinostratus and Nicomedes in the fourth Book of Pappus A­lexandrinus, and which from its use is called a Quadratrix, it having among the rest this property, that from AB it cuts off a part AE, which is a third proportional to the quadrant DB and its radius DA; which hereafter we will demonstrate. In the mean while from this description of it, you have these

CONSECTARYS.

I. IF thro' any point H assumed in the Quadratrix you draw the radius AHI, and from the same point the perpendi­culars H h and H e, the whole arch DB will always be to the part IB cut off, as the whole line DA to the part hA cut off, or H e equal to it.

[Page 166]II. Consequently therefore any given arch or angle of the quadrant e. g. IB or IAB may by help of the quadratrix be divided into three equal parts or as many as you please, or in what proportion soever you will; while having drawn the radius AI, the perpendicular H a let fall from the point of the quadratrix H, may be divided into three or as many equal parts as you please, or in any proportion whatsoever, and thro' these sections radius's drawn to divide the arch.

BOOK II.

SECTION II.

CHAP. I. Of the chief Properties of the Conick Sections.

Proposition I.

IN the Parabola (GKEH Fig. 114) the I. Pro­perty of the Pa­rab. Apoll. Prop. 11. Lib. 1. square of the semi-ordinate (IK) is equal to the rectangle IL made by the Latus Rectum EL and the abscissa EI.

Demonstration.

MAke the sides of the cone that is supposed to be cut, AB = a, BC= b, and moreover EB= oa, and EI= eb, and AC= c; therefore NI will be = ec, by reason of the si­militude of the ▵ ▵ BCA and EIN; and EP or IO= oc, by reason of the similitude of the ▵ ▵ ABC and EBP. There­fore ▭ NIO= oecc=□ IK by the Schol. of Prop. 34 ( n. 3) and Prop 17. Lib. 1. Now if a line be sought which with the abscissa EI shall make the ▭ IL=□ IK you will [Page 167] have it by dividing the said square by the Abscissa EI. viz. [...] i. e. [...] = EL. And this is called the Latus Rectum, viz. in relation to the Abscissa EI with which it makes that rectangle, which, it's evident, is = □ IK, and from this equality the section has the name of Parabola, in Apollo­nius.

CONSECTARYS.

I. THis Latus Rectum, expressed by the quantity [...], may be found out after a shorter way, if you make as b to c (the side of the cone parallel to the section BC at the Diame­ter of the base AC) so oc (the side EP called by some the Latus Primarium) to a fourth.

II. But if any one, with Apollonius, had rather express this by meer data in the cone it self as cut (because oc or that La­tus Primarium EP is not a line belonging to the cone it self) he may easily perceive, if the quantity of the Latus Rectum found above, be multiplyed by the other side of the cone a, there will be produc'd the equivalent [...] which instead of the proportion above will furnish us with this other,

  • as ab− to cc− so oa
  • □ of AB into BC−□ AC−EB

to a fourth; which is the very proportion of Apollonius in Prop. 9. Lib. 1. and confirms our former.

SCHOLIƲM I.

HEnce you have an easie and plain way of describing a Parabola, having the top of the ax and the Latus Re­ctum given, viz. by drawing several semiordinates whose extreme points connected together will exhibit the curvity of the Parabola. But you may find as many semiordinates as you please, if having cut off as many parts of the Ax as you please, you find as many mean proportionals between the [Page 168] Latus Rectum and each of those parts or Abscissa's. See n. 2. and 3. Fig. 47. Introduct. to Specious Analysis.

SCHOLIƲM II.

HEnce also we have a new genesis of the parabola in Plan [...] from the spculations of De Witte, viz. if the rectiline­ar angle HBG ( Fig. 115.) conceived to be moveable about the fixed point B be conceived so to move out of its first situ­ation with its other leg BH along the immoveable rule EF, that it may at the same time move also the ruler HG, from its first situation DK, all along parallel to it self, and with the other leg BG let it all along cut the said ruler HG, and with this point of its intersection continually moving from B to­wards G it will describe a curve. That this curve will be the parabola of the antients is hence manifest, because it will have this same first property of the parabola. For, 1. if the an­gle HBG ( n. 1.) be supposed to be a right one, and BD or HI= a, BI or KG= b (viz. in that station of the angle and rule HG by which they denote the point G in the inter­section) you'l have by reason of the right angle at B, BI, i. e. b a mean proportional Between HI i. e. a, and IG or BK, and so this as an abscissa = [...]. Wherefore if BK i. e. [...] be multiplyed by BD= a, the rectangle DBK will be = bb =□KG; which is the first property of the parabola: So that it follows, since the same inference may be made of any other point in this curve, that this curve will be the parabo­la, BD or HI its Latus Rectum, KG a semiordinate, and BK its axis, &c. 2. If the angle HBG be an oblique one ( num. 2) it may be easily shewn from what we have supposed that the ▵ ▵ DBH and BKG will be equiangular: Therefore as BD (i. e. a) to DH sc. BI (i. e. b) so KG sc BI (i. e. b) to BK (i. e. [...]) Therefore again the □ DBK= bb□KG. QED.

Consect. 3. It is also evident in this second case, that BK drawn parallel to the ax, but not thro' the middle of the pa­rabola, [Page 169] will be a diameter which will have for its vertex B, its Latus Rectum BD, and semiordinate GK, &c.

Consect. 4. Therefore you may find the Latus Rectum in a given parabola geometrically, if you draw any semiordinate whatsoever IK ( Fig. 116.) and make the abscissa EF equal to it, and from F draw a parallel to the semiordinate IK, and from E draw the right line EK thro' K cutting off FH the Latus Rectum sought; since as EI to IK so is EF ( i. e. IK) to FH by Prop 34. lib. 1. wherefore having the ab­scissa and semiordinate given arithmetically, the Latus Re­ctum will be a third proportional.

Consect. 5. Since therefore the Latus Rectum found above is [...], if you conceive it to be applyed to the parabola in LM, so that N shall be that point which is called the Focus, LN will be [...] and its square [...] and this divided by the Latus Rectum [...] will give occ [...] for the abscissa EN: So that the distance of the Focus from the Vertex will be ¼ of the Latus Re­ctum.

Consect. 6. Since therefore EN is = [...] if for EF you put ib, NF will be = [...], whose square will be found to be [...], to which if there be added □ GF= oicc, by Prop. 1. the square of NG will be = [...] whose root (as the extraction of it and, without that, the ana­logy of the square NF with the square NG manifestly shews) will be [...]; so that a right line drawn from the Focus to the end of the ordinate, will always be equal to the abscissa EF [Page 170] +EN i. e. (if EO be made equal to EN) to the compounded line FO.

SCHOLIƲM III.

HEnce you have an easier way of describing the parabola in Plano from the given Focus and Vertex, viz. (Fig: 117.) the axis being prolonged thro' the vertex E to O, so that EO shall = EN, if a ruler HI be so moved by the hand G, according to PQ, from OF to HI, that putting in a style or pin, it shall always keep the part of the Thred NGI (which must be of the same length with the rule HI) as fast as if it were glued to it (which perhaps might also be done with the Compasses by an artifice which we will hereafter al­so accommodate to the hyperbola) and at the same time it will describe in Plano the part of the line EGR. That this will be a parabola is evident from the foregoing Consect. because as the whole thred is always = to the ruler IH; so the part GN is always necessarily equal to the part GH, i. e. to the line FO. Moreover from the same sixth Consect. and Fig. 116. may be drawn another easie way of describing the para­bola in Plano from the Focus and Vertex given thro' innume­rable points G to be found after the same way: viz. If from any assumed point in the ax F you draw to the ax a perpendi­cular, and at the interval FO from the Focus N you make an intersection in G. Which innumerable points G will be de­termined with the same facility, having given only, or assu­med the axis and Latus Rectum, by vertue of the present Proposition. For if, having assumed at pleasure the point F in the axis, you find a mean proportional between the Latus Rectum and the abscissa EF, a semiordinate FG made e­qual to it, will denote or mark the point G in the parabola sought.

Proposition II.

IN the hyperbola (GKEH Fig 118) I. Pro­perty of the Hy­perb. Apoll. Prop. 12. Lib. 1. the square of the semiordinate (IK) is equal to the rectangle (IL) made of the Latus Rectum (EL) and the abscissa (EI) together with ano­ther rectangle LS of the said abscissa (EI or LR) and RS a fourth proportional to DE the Latus Transver­sum, (EL) the Latus Rectum, and EI the Abscissa.

Demonstration.

Suppose the side of the cone AB here also = a, and BM parallel to the section = b, and the intercepted line AM= c, and EI= eb; all according to the analogy we have observed in the parabola; and NI will be as there = ec. Making moreover MC= d and the Latus Transversum DE= ob, so that DI shall be = ob+ eb; then will (by reason of the simi­litude of the ▵ ▵ BMC, DEP, and DIO) EP be = od, and IO= od+ ed, and so QO= ed. Therefore □ NIO will be = oecd+ eecd=□IK But this square divided by the Abscissa EI= eb gives [...] or [...] for the line IS which with the abscissa would make the rectangle ES = to the said square [...]K. Now therefore, if here also we call a Line the Latus Rectum found after the same way as in the parabola, viz by making as b− to c− so od to a fourth [...] (as a line parallel to the section — to the intercepted diam. so the Latus Primarium, but that the other part [...] will be a fourth proportional to bc and ed or to eb, ec▪ and ed, or (to speak with Apollonius as we have done in the Prop.) to ob, [...] and eb (for in these three cases you I have the same fourth [...]) Wherefore now it is evident that the square of the se­miordinate [Page 172] oecd+ eecd is equal to the rectangle IL (made by the Latus Rectum [...] into the abscissa eb= oecd) together with ▭ LS of this fourth proportional [...] into the same ab­scissa Eb, which is = eecd. Which was to be found and de­monstrated.

CONSECTARYS.

I. HEnce you have in the first place the reason why Apol­lonius called this Section an Hyperbola; viz. because the square of the ordinate IK exceeds or is greater than the rectangle of the Latus Rectum and the Abscissa.

II. Since therefore the Latus Rectum here also as well as in the parabola is found by making as b to c so od to [...] ( i. e. as the parallel to the section BM is to the intercepted Diam. AM so is the Latus Primarium EP to a fourth EL.) If any one had rather express this Latus Rectum after Apollonius's way, he will easily perceive, this quantity being found and multi­plyed both Numerator and Denominator by b the parallel to the section, there will come out the equivalent quantity [...] which gives us instead of the former proportion this other,

  • as bb− to cd− so ob to a fourth;
  • □BM−▭AMC— Latus Transversum to a fourth;

which is that of Apollonius in Prop. 12. Lib. 1. and consequent­ly herein confirms our former.

III. You may also have this Latus Rectum geometrically, by finding a third proportional (as we have done in the pa­rabola Consect. 4. Prop. 1.) to the abscissa EI ( Fig. 119.) and the semiordinate IK (= EF;) and then find a fourth pro­portional EL to DI (the sum of the Latus Transversum and abscissa) and FH already found, or IS equal to it, and DE (the Latus Transversum) and that will be the Latus Rectum sought.

Pag. 172.

119

120

121

122

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124

125

126

126

127

128

SHCOLIƲM.

FRom this third Consectary, we may reciprocally from the Latus Rectum and transverse given, find out and apply as many semiordinates to the ax as you please, and so describe the hyperbola thro' their (ends or) infinite points: viz. if assuming any part of the abscissa EI, you make as DE to EL so DI to IS; and then find a mean proportional IK between IS and the abscissa EI, and that will be the semiordinate sought: And both this praxis and the Consect. may be abun­dantly proved by setting it down in, and making use of, the literal Calculus.

Proposition III.

IN the Ellipsis (KDEK, Fig. 120.) the I. Pro­pert. of the El­lipse of Apoll. Prop. 13. Lib. 1. square of the semiordinate (IK) is equal to the rectangle (IL) of the Latus Rectum (EL) and the abscissa (EI) (less or) taking first out another rectangle (LS) of the same abscissa (EI or LR) and RS a fourth proportional to (DE) the Latus Transversum (EL) the Latus Rectum and (EI) the abscissa.

Demonstration.

Suppose the side of the cone to be AB here also = a and BM parallel to the section = b and the intercepted AM= c, and EI= eb; and NI will be again = ec, all as in the hyperbola. And mak [...]ng also here as in the hyper­bola MC= d, and the Latus Transversum DE= ob, so that DI will be ob− eb; then will (by reason of the simili­tude of the ▵ ▵ BMC, DEP and DIO) EP be= od, and IO= od− ed. Therefore ▭ of NIO will be = oecd− eecd=□IK. But this square divided by the abscissa EI= eb gives [...] or [...] for that line IS which with the abscissa would make the rectangle ES= to the said square IK. Now therefore if we call the Latus Rectum a right line found after the same way as in the parabola, by making ac­cording to Cons. 1. Prop. 1. [Page 174] as b to c− so od− to a fourth [...] i. e. as the line pa­rallel to the section — to the intercepted diameter — so the Latus Primarium, &c. It is manifest that the Latus Re­ctum is one part of the line just now found; and the other part [...] is a fourth proportional to b, c and ed, or (to speak with Apollonius as we have done in the Prop.) to [...] and eb (for there will come out the same quantity [...]) where­fore now it is evident that the □ of the semiordinate IK is e­qual to the ▭ IL (of the Latus Rectum [...] and the ab­scissa eb= oecd) having first taken out thence the ▭ LS, or eecd out of that fourth proportional [...] by the same abscissa eb; which was to be found and demonstrated.

CONSECTARYS.

I. HEnce yov have first of all the reason of the name of the Ellipse, which Apollonius gave to this section; viz. because the square of the semiordinate IK is defective of, or less than the rectangle of the Latus Rectum and the abscis­sa.

II. Since therefore the Latus Rectum here also as well as in the parabola and hyperbola, is found by making as b to c so od to [...] ( i. e. as BM parallel to the section is to the inter­cept. diam. AM so the Latus Primarium EP to a fourth EL) now if any one had rather express this Latus Rectum after A­pollonius's way, he will easily see that the quantity above found being multiplyed both Numerator and Denominator by b, that there will come out an equivalent one [...], which instead of the former proportion will give this other, [Page 175] as

  • bb− to cd− so ob to a fourth;
  • BM−▭AMC— Latus Transvers. to a fourth;

which is the same with that we have also found in the hyper­bola, and which also Apollonius has Prop. 13. Lib. 1.

III. This Latus Rectum may also be had geometrically, if you find, 1. in the hyperbola a third proportional FH to the abscissa EI ( Fig. 121.) and semiordinate IK (= EF.) 2. But EL a fourth proportional to DI (the difference of the Latus Transversum and the abscissa) and the found FH, or IS equal to it, and the Latus Transversum DE, is the Latus Rectum sought.

SCHOLIƲM.

FRom this third Consect. we may reciprocally, having the Latus Rectum and Transversum given, apply as many semiordinates to the ax as you please, and so draw the ellipsis thro' as many points given as you please, viz. if, taking any ab­scissa EI, you make as DE to EL so DI to a fourth IS; then between this IS and the abscissa EI find a mean proportional IK, and that will be the semiordinate sought: And this Prax­is also and the third Consect. may be abundantly proved by making use of a literal Calculus. For e. g. here a fourth proportional to ob, [...] and ob− eb will be [...] and a mean proportional between this fourth and eb will be [...], &c.

Proposition IV.

IN a Parabola 2. Property of the [...]arab. 20. Prop. Apoll. Lib. 1. Con. the squares of the ordinates are to one another as the ab­scissa's.

Demonstration.

For if EF ( Fig. 122.) be called ib, as above EI was cal­led eb, since the Latus Rectum is [...] the square of FG wil be = oicc. Therefore □ IK will be to □ FG as oecc— oicc

  • e to i or
  • eb to ib.

Q. E. D.

CONSECTARY.

HEence having drawn LO parallel to the ax or diameter EF, if it be cut by the transverse line EG in M and by the curve of the parabola in K; then will OL, ML, and KL be continual proportionals. For EF is to EN as FG to NM or IK, by reason of the similitude of the ▵ ▵ EFG and ENM. But the squares FG and IK are in duplicate proportion of EF to EN by Prop. 35. Lib. 1. and are also in the same propor­tion as the abscissa's EF and EI by the pres. Therefore EF to EI is also in duplicate proportion of EF to EN i. e.

  • EF is to EN as EN to EI Q. E. D.
  • OL is to ML as ML to KL Q. E. D.
Proposition V.

IN the hyperbola and Ellipsis Proper­ty of the Hy­perbola and Ellips. Apol. 21. Lib. I. the squares of the Ordinates are as the rectangles contained under the lines which are intercepted between them, and the Vertex's of the Latus Transversum's.

Demonstration.

For, if EF ( Fig. 118. and 120.) be called ib, as EI was above called eb, then will according to Prop. 2. and the 3d. dedu­ction.

GF= oicd+ eicd in the Hyperb.

oicd− eicd in the Ellips.

and the ▭ DFE= oibb+ iibb in the Hyperb.

oibb− iibb in the Ellips.

Therefore the □ KI is to the square GF as oecd± eecd to oicd ± iicd i. e. as oe± ee to oi± ii.

and ▭ DIE is to the ▭ DFE, as oebb± eebb to oibb± iibb i. e. in like manner as

oe± ee to oi± ii. Q. E. D.

CONSECTARY I.

IN the Ellipsis this may be more commodiously expressed apart thus; the squares of the ordinates (KI and GF) are as the rectangles contained under the segments of the Diame­ter ( viz. DIE and DFE) in which sense this property is also common to the circle, as in which the squares of the ordinates are always equal to the rectangles of the segments.

CONSECTARY II.

THerefore, if the Latus Rectum be conceived to be apply­ed in the hyperbola, so that N shall be the Focus; (see Fig. 123.) then will LN= [...], and its square be [...]. But as the □ KI to the square LN, so is the ▭ DIE to the ▭ DNE i. e. oecd+ eecd to [...] so is oebb+ eebb to [...]. But now the ▭ of the whole DE and the part added EN into the part added EN, i. e. ▭ DNE (= [...] together with the square of half CE (= [...]) is = □ compounded of half and the part added CN= [...] by Prop. 9. lib. 1. Wherefore CN the distance of the Focus from the centre is = [...]. But [...] is the fourth part of the ▭ of the Latus Transversum ob and the Latus Rectum [...] (or the [Page 178] fourth part of the figure, as Apollonius calls it) and [...] is the □ of [...] i. e. of half the Latus Transversum. Wherefore we have found the following Rule of determining the Focus in an hyperbola: If a fourth part of the figure (or the rectangle of the Latus Rectum into the Transversum) be added to the square of half the Latus Transversum, and from the sum you extract the square root; that will be the distance of the Focus from the center CN: And hence substracting half the Latus Transversum CE, you will have distance of the Focus from the Vertex EN.

CONSECTARY III.

IN like manner in the Ellipsis having drawn the ordinate LM ( Fig. 124.) that the Focus may be in N, the □ LN would be [...] as above, and by a like inference □ DNE= [...]. But now □ DNE together with the square of the dif­ference CN is equal to the □ of half CE by Propos. 8. lib. 1. and consequently the □ CN is = □ CE−□DNE, that is, [...]. Wherefore CN the distance of the Focus from the centre is = [...]. Wherefore we have found the following Rule to determine the Focus in the El­lipse. If the fourth part of the figure (or the rectangle of the Latus Rectum into the Latus Transversum) be substracted from the square of half the Latus Transversum, and from the remain­der you substract the square root; that will be the distance of the Focus from the Centre CN: And taking hence half the La­tus Transversum CE, you'l have the distance of the Focus from the Vertex EN.

SCHOLIƲM I.

BOth the Rules are easie in the practice, for since [...] is nothing but the square of CE, and [...] nothing but the re­ctangle of ¼ DE into LM; if between LM and ¼ DE or MO ( Fig. 125.) you find a mean proportional MN, (and so whose □ is equal to that □) and in the hyperbola join to it at right angles MC=CE, the hypothenusa CN will be the distance sought of the Focus from the centre: And the same may be had in the Ellipsis, if ( n. 2.) having described a semi-circle upon CM=CE you draw or apply the mean found MN, and draw CN.

SCHOLIƲM II.

HEnce also we have De Witt Elem. Curv. Lib. 1. Cap. 3. Prop. 13. a new genesis of the Ellipse in Plano about the diameters given, from the speculations of Monsieur de Witt; viz. If about the rectilinear angle DCB ( Fig. 126. n. 1. and 2.) consider'd as immove­able, the rule NLK (which all of it will equal the greater semidiameter CB, and with the prominent part LK the lesser CD) be so moved that N going from C to D, and L from B to C may perpetually glide along the sides of the angle, the extreme point in the K in the mean while describing the curve BKE (and in a like application the other quadrants) and that this curve thus described will be the ellipsis of the ancients is hence manifest, because it has the second property of the ellipse just now described. For, 1. if the angle DCB or NCB be supposed to be a right one (as in Fig. 126. num. 1.) and the rule KN in the same station, it marked out the point K, and having apply'd the semiordinate KI, and drawn the perpendicular LM, from the square KL and the square CE (as being equal) substract mentally the equal squares LM and CI, and there will remain by virtue of the Pythagorick Theor. on the one hand □ KM and on the other by [Page 180] Prop. 8. lib. 1. □ DIE equal among themselves. But now the square of KI is to the square of KM ( i e. to the □ DIE) as the square of KN to the square of KL ( i. e. as the square of CB to the square of CE) by reason of the similitude of the ▵ ▵ KLM and KNI; and since the same may be demonstra­ted after the same manner of any other semiordinate K i viz. that its □ K i is to the □ D iE as the square CB to the square CE. It also follows, that the □ KI is to the □ DIE as □ K i to the □ D iE, and alternatively, the square KI will be to the square K i as the □ DIE to the □ D iE; which is the se­cond property of the ellipse. 2. If the angle NCE be not a right one (as in Fig. 126. n. 2. and the like cases) having drawn NO and KP parallel to the rule nlB in the first station, [in which station the angle NCE, to which the flexible ruler is to be made, is determined, viz. by letting fall the perpendicular B l from the extremity of one diameter upon the other, and moreover by adding or substracting the difference of the semi-diameters ln] having also drawn the Ordinate KIM, and PI parallel to CN; which being done the ▵ ▵ CB l and IKF, and also CB n and IKP will be similar. Where­fore having joined NP, from the parallelism of the lines IP and NC and the similitude of the aforesoid ▵ ▵, as also of NCO and nC l, it will be easie to conclude that NCIP is a parallelogram. Wherefore since KN is = CE and □ KN = □ CE, having substracted the squares of the equal lines NP and CI, there will remain on the one hand □ KP on the other the □ DIE equal among themselves as above. Therefore the square of KI will be to the square of KP ( i. e. to the □ DIE) as the square of BC to the square of B n (i. e. to the square of CE) as in the former case: And since here also the same may be demonstrated after the same manner of any o­ther semiordinate K i; we may infer as above, that the □ □ KI and K i are to one another as the rectangles DIE and D iE, &c.

But after what way the same ellipses may be described by these right lined angles without any of these rulers thro' infi­nite points given, will be be manifest from the same figures to any attentive Person. For having once determined the angle NCE or nCD ( num. 2. e. g.) if NL or nl be applyed where you please by help of a pair of compasses, and continued to [Page 181] K, so that LK or lk shall be equal to lB, you will have eve­ry where the point K, &c.

CONSECTARY IV.

SInce in the hyperbola ( Fig. 123.) the □CN−□CE=□DNE, and in the ellipsis ( Fig. 124.) □CE−□CN=□DNE, by vertue of Prop. 9 and 8. lib. 1 if for CN you put on both sides for brevity's sake m, then will the □ DNE in the hyperbola be rightly expressed in these terms [...], and in the ellipsis in these [...].

Proposition VI.

IN the parabola 3. Pro­perty of the Pa­rab. the Latus Rectum is to the sum of two semiordinates (e. g. IK+FG i. e. HO in Fig. 122.) as their difference (OG) to the difference of the abscissa's (IF or (KO.)

Demonstration.

For if the greater abscissa EF be made = ib, and the less EI= eb, the semiordinates answering to them FG and IK will be √ ooic and √ oecc as is deduc'd from Prop. 1. Where­fore if you set in the same series

1 2 3
The Latus R. −sum of the semiord. −their diff.
[...] −√ oicc+√ oecc −√ oicc−√ oecc
  4  
  −diff. of the absciss.  
  − ib− eb  

and multiply the extremes and means, you'l have on both sides the same product oicc− oecc, which will prove by vertue of Prop. 19. lib. 1. the proportionality of the said quantities. Q. E. D.

SCHOLIƲM.

THis is that property of the parabola, whereon the Clavis Geometrica Catholica of Mr. Thomas Baker is founded, which as unknown to the ancients, nor yet taken notice of by Des Cartes, he thinks was the reason why that incomparable Wit could not hit upon those universal rules for solving all Equations howsoever affected. Concerning which we shall speak further in its place. We will only further here note, that Baker was not the first Inventor of this property, but had it, as he himself ingeniously confesses, out of a Manuscript communicated to him by Tho. Strode of Maperton, Esquire.

Proposition VII.

IN the hyperbola and ellipsis The 3. Property of the Hyperb. and El­lips. Apollon. lib. I. Prop. 21. pars prior. the Latus Rectum is to the Latus Transversum, [...] the square of any semiordinate (e. g. IK in Fig. 118▪ and 120.) to the rectangle (DIE) con­tained under the lines intercepted between it and the Vertex's of the Latus Transversum.

Demonstration.

For the Latus Rectum is on both sides [...], the Latus Trans­versum ob, &c. Wherefore if you make in the same series as the Latus R. to the Lat. Transv. so the □ IK to ▭ DIE

  • [...] in hyperb. [...]
  • in ellips. [...]

The rectangles of the extremes and means will both be ooebcd± oeebcd, and so will prove the proportionality of the said quantities, by Prop. 19. lib. 1. Q. E. D.

CONSECTARY I.

HEnce having given in the ellipsis (see Fig. 124) the La­tus Rectum and the transverse ax, you may easily obtain the second ax or diameter, if you make as the Lat. Transv. to the Lat. Rect. so the □ DCE to □AC [...].

CONSECTARY II.

THerefore the □ of the whole AB will be = oocd=□ of the Latus Rectum into the Lat. Transv. (which Apllo­nius calls the Figure) so that the second Ax (and any second Diameter) will be a mean proportional between the Latus Re­ctum and the Latus Transversum. Hence in the hyperbola also the second or conjugate diameter may be called a mean proportional between the Latus Rectum and Transversum, i. e. √ oocd or a line which is equal in power to the Figure, as A­pollonius speaks.

SCHOLIƲM I.

HEnce may be derived another and more simple way of delineating organically the ellipsis in Plano about the given axes AB, DE ( Fig 127.) which Schooten has given us▪ viz. by the help of two equal rulers CG and GK move­able about the points G and G: If, viz. the portions CF and HK are equal to half the lesser ax AC, but taken with both the augments ( viz. CF+FG+GH) may = ½ the greater ax CD or CI; and the point K moving along the produced line DE the point H may describe the curve EHAD. That this will be an ellipsis will be evident by vertue of this seventh Prop. from a property that agrees to this curve in all its points H. For having drawn circles about each diameter, and the lines IHN, FO perpendicular to CE; and having made the Latus Rectum EL, which is a third proportional to DE and AB by the second Consect. of this Prop. &c. by reason of the similitude of the triangles CFO, CIN, FO will be to FC as [Page 184] IN to IC, and alternatively FO to IN as FC to IC i. e. as AC to CE or AB to DE. Therefore also the square of FO (or HN) will be to the square of IN, as the square of AB to the square of DE, by Prop. 22. lib. 1. i. e. as EL the Latus Rectum to ED the Latus Transversum, by Prop. 35. lib. 1. But the □ IN is = DNE from the proporty of the circle. Therefore □ FO (or of the semiordinate HN) is to the ▭ DNE as EL the Latus Rectum to ED the Latus Transv. therefore by ver­tue of the pres. Prop. the point H is in the Ellipsis, and so any other, &c. Q. E. D.

CONSECTARY III.

NOW if in the ellipsis the □ of AC the second Ax (= [...] by Consect. 1.) and □ CN the distance of the Fo­cus from the centre (= [...] by Censect 3. Prop. 5. the figure whereof you may see n. 124.) be joined in one sum; the □ AN will be = [...], and so the line AN= [...] i. e. to half the Latus Transversum: So that hence having the axes given you may find the Foci, if from A at the interval CD you cut the transverse ax in N and N.

CONSECTARY IV.

NOW if, on the contrary, in an hyperbola ( Fig. 123.) the □ AC or EF= [...] be substracted from the □ CF or CN= [...] by vertue of Consect. 2. Prop. 5. there will remain [...] and its root [...], i. e. half the Latus Trans­versum CD: So that here also, the axes being given, you may find the Focus's, if from the vertex E you make EF a per­pendicular to the ax = to the second Ax AC, and at the in­terval CF from the centre C you cut the Latus Transversum continued in N and N.

SCHOLIƲM II.

BUT now, that the right lines KN and KN drawn from any other point ( e. g. K) to the Foci, when taken toge­ther in the ellipsis, but when substracted the one from the o­ther in the hyperbola, are equal to the Latus Transversum DE, we will a little after demonstrate more universally, and also shew an easie and plain Praxis of delineating the ellipsis and hyperbola in Plano, having the axes and consequently the Fo­ci given.

CONSECTARY V.

SInce we have before demonstrated Consect. 2. and 3. Prop. 5. that the □ DNE in the hyperbola and also in the el­lipsis is = [...]; and here in Consect. 1. the □ of the second semi-diameter AC is also = [...]; it is evident that this □ AC is equal to the □ DNE.

CONSECTARY VI.

IT is hence moreover evident, if the square of half the trans­verse diameter GE= [...] be compared with the square of half the second diameter AC or EF= [...], multiplying both sides by 4 and dividing by o; they will be to one another as obb to ocd i. e. further dividing both sides by b, as ob to [...] the Latus Transversum to the Latus Rectum.

CONSECTARY VII.

BUT since also the □ DIE is to the □ IK as the Latus Transversum to the Latus Rectum, by vertue of the pre­sent 7. Prop. the square of CE the transverse semidiam. will be [Page 186] to the square of AC the second semidiam. (or by the 5th. Con­sect. of this, to the □ DNE) as the □ DIE to the square of IK.

CONSECTARY VIII.

YOU may also now have the □ IK (which otherwise in the hyperb. is eocd+ eecd, in the ellipse oecd− eecd, by vertue of Prop. 2. and 3.) in other terms, if you make as □ CE to the □ DNE so the □ DIE to a fourth; i. e.

as [...] to [...] by vertue of Consect. 4. Prop. 5.

(so oebb+ eebb in the hyperb.

as [...] to [...] so oebb− eebb in the ellipsis.

For hence by the Golden Rule the square IK may be infer'd as a fourth proportional.

In the hyperbola [...];

In the ellipsis [...]:

The use of which quantities will presently appear.

Proposition VIII.

THE

  • Aggregate in the ellipse
  • Difference in the hyperb.

of the right lines Apollon. Lib. 3. Prop. 51. and 52. KN and Kn (Fig. 128.) drawn from the same point K to both the Focus's is equal to the transverse ax DE.

An Ocular Demonstration.

WHich consists wholly in this to find the lines KN and K n by help of the right-angled triangles IKN and IK n (sc. the hypothenuses having the sides given) and afterwards see if the sum of both in the ellipse, and difference in the hyperbola be = ob, i. e. to the transverse ax DE.

I. In the Ellipsis.

Putting for CN (which above Prop. 5. Cons. 3. was found to be [...]) I say putting for it m, you'l have

IN=CI+CN=½ ob− eb+ m

I n=C n−CI= m−½ ob+ eb

Therefore □IN= [...] □I n= [...]

Add to each □ IK which was found in Prop. 7. Consect. 8. in the ellipsis = [...] and you'l have □KN= [...] and by extracting the roots (which is easie) you'l have

[...] and
[...];
Sum ob. Q. E. D.
II. In the Hyperbola.

Putting again m for CN (which above Cons. 2. Prop. 5. was found to be [...]) and you'l have

IN =CI+CN=½ ob+ eb+ m

I n=CI−C n=½ ob+ eb− m. Therefore □IN= [...] and □I n= [...]

Add to both the □ IK which was found in Prop. 7. Consect. 8. in the hyperbola [...] and you'l have □KN= [...] [Page 188] □K n= [...] and extracting the roots out of these (which is easie) you'l have KN= [...].

K n= [...] (which is a false or impossible root, for it would be CE−CN and moreover — another quantity.

Or K n= [...]; which is a true and possible root.

The difference therefore of the true roots is = ob. Q. E. D.

SCHOLIƲM I.

WE first of all tried to make a literal Demonstration by using the quantity of the square IK as you have it expressed Prop. 2. and 3. and the quantity IN as it was com­pounded of CI= ob+ eb+CN= [...], &c. but we found it very tedious in making only the squares of IN and I n. Then for the surd quantity CN we substituted another, viz. m, and we produced the squares of IN and I n as above, but we added the square of IK in its first value; and thus we obtain'd the squares KN and K n, but in such terms, that the exact roots could not be extracted, but must be exhibited as surd quantities, and consequently we must make use of the rules belonging to them to find their sum or difference, which we laid down Cons. 3. Prop. 7. and Consect. Prop. 10. Lib. 1. which tho' it would succeed, yet would be full of trouble and tediousness. Therefore at length when we came to use those other terms which express the square IK, the business succeed­ed as easie as we could wish, and that in a plain and easie way and no less pleasant, which I doubt not but will also be the opinion of the Reader, who shall compare this with other demonstrations of the same thing, which only lead indirectly to this truth, or with them, which de Witte has given us in E­lem▪ Curvar. lin. p. m. 293. and 302. and which he thinks [Page 189] easie and short enough in respect of others both of the ancients and moderns, and which we have reduced into this yet more distinct form, and accommodated to our schemes.

Preparation for the Hyperbola.

Make as

  • CD to CN
  • EE−CN

so CI to CM so that the □ of

  • DCM
  • MCE

will be = □

  • NCI
  • nCI.

Because therefore it will be by Consect. 7. Prop. 7. as □ CD to □ DNE, so the □ DIE to the □ IK.

And also by composition, as □ CD to

  • □CD+□DNE
  • i. e. □ CN per 9. lib I.

so DIE to DIE + □ IK.

Therefore by a Syllepsis, as □ CD to the □ CN so

  • □ CD + DIE
  • i. e. □ CI

to □ CN + DIE + □ IK.

But also by the Hypothesis. as the □ CD to the □ CN so □ CI to □ CM. Therefore □ CM is = □ CN + DIE + □ IK.

Demonstration.

Since therefore it is certain that the difference between DM and EM is the transverse ax DE; if it be demonstrated that DM is = KN and EM = K n, the business will be done, be­cause the difference between KN and K n is also the transverse ax DE.

Resolve the □ KN.

It is certain that NI q + IK q=KN q.

Substitute for NI q, by the 7. Lib. 1. CI q+CN q+2NCI.

Preparation for the Ellipsis.

Make as CD to CN so CI to CM.

So that the □ DCM is = □ NCI.

Because therefore by Consect. 7. Prop. 7. as □ CD to □ DNE so □ DIE to the □ IK;

Then also by dividing, as □ CD to

  • □ CD−DNE
  • i. e. □ CN, by 8. l. 1.

so DIE to DIE−□IK.

Therefore by a Dialepsis, as □ CD to □ CN, so

  • □CD−□DIE
  • i. e. CI □ by 8. cit.

to □ CN−DIE+□IK.

But also by the Hypothesis, as □ CD to □ CN, so □ CI to □ CM: Therefore □ CM is = □ CN−DIE + □ IK.

Demonstration.

Since therefore it is certain that the sum of DM and EM is the transverse ax DE; if it be demonstrated that DM is = KN and EM = K n, the business will be done, because the sum of KN and K n is also equal to the transverse ax DE.

Resolve the □ KN.

It is certain that NI q+ [...]K q=KN q.

Substitute for NI q, by the 7. lib. 1. CI q+CN q+2NC [...]

Then will CI q+CN q+2NC [...]+IK q=KN q.

Substitute for C [...] q, by the 9. lib. 1. CD q+DIE; then will CD q+DIE+CN q+2NC [...]+K q=KN q.

Resolve also □ DM.

It is certain that CM q+CD q+

  • 2DCM
  • 2NCI

= DM, by the 7. lib. 1.

Substitute for CM q its value by the Preparation, and you'l have CN q−D [...]E+ [...]K q+CD q+2NCI=DM q:

Which were before = KN q.

Therefore KN=DM; which is one.

In like manner resolve □ Kn.

It is certain that nl q+ [...]K q=K nq.

Substitute for nI q, by Consect. 1. Prop. 10. Lib. 1. C [...] q+CN q−2 nCI, and you'l have

CI q+CN q−2 nCI+IK q=K nq.

Substitute for CI q, by the 9. lib. 1. CD q+D [...]E, and you'l have [...]=K nq.

Resolve also the □ EM.

It is certain that 2CD q+2CM q−DM q=EM q per 13. lib. 1.

Then will CI q+CN q+2NCI+IK q=KN q.

Substitute for CI q by the 8. lib. 1. CD q−DIE; then will [...].

Resolve also the □ DM.

It is certain that CM q

  • 2DCM
  • 2NCI

= DM q per 7. lib. 1.

Substitute for CM q its value from the preparation, and you'l have CN q−DIE+IK q+CD q+2NCI=DM q:

Which before were = KN q.

Therefore KN = DM; which is one.

In like manner resolve the □ Kn.

It is certain that nI q+IK q=K nq.

Substitute for nI q by Consect. 1. Prop. 10. lib. 1.

CI q+CN q−2NCI, and you'l have CI q+CN q−2NCI+IK q=K nq.

Substitute for CI q per 8. lib. 1. CD q−DIE, and you'l have CD q−DIE + CN q−2NCI+IK q=K nq.

Resolve also □ EM.

It is certain that 2CD q+2CM q−DM q=EM q per 13.1.

Substitute the value of DM q first found above, and you'l have CD q+CM q−2 nCI=EM q.

Substiute for CM q the value as in the preparation, and you'l have CD q+CN q+DIE−2 nCI+IK q=EM q: Which were before = K nq.

Therefore K n=EM; which is the other.

SHCOLIƲM II.

HEnce you have the common mechanical ways of describ­ing the ellipsis and hyperbola about their given axes; and the ellipsis, if the Foci N, N, ( Fig. 129. n. 1.) are gi­ven, or found according to Consect. 3. Prop. 7. and having therein stuck or fixed two pins, put over them a thread NF n tyed both ends together precisely of the length you design the greater ax DE to be of, and having put your pencil or pen in that ▵-string draw it round, always keeping it equally extended or tight. Now because the parts or portions of the thread re­main always equal to the whole ax DE, what we proposed is evident by the present Prop. which may also be very elegantly described by a certain sort of Compasses, a description where­of Swenterus gives us in his Delic. Physico-Math. Part. 2. Prop. 20. which may be also done by a sort of organical Mechanism, by the help of two rulers moveable in the Foci GN and H n (n 2.) and equal to the transverse ax DE, and fastned a­bove by a transverse ruler GH equal to the distance of the Fo­ci, as may appear from the Figure. For if the style F be moved round within the fissures of the cross rulers H n and GN the curve thereby described will be an ellipsis from the pro­perty we have just now demonstrated of it, which it hath in every point F. For the triangles HGN and NH n, which have one common side HN, and the others equal by construction, are equal one to another, and consequently the angles FHN and FNH equal, so also the legs HF and FN, and so likewise FN and F n together are equal to H n = DE; which is the very property of the ellipse we are now treating of. But Van Schoo­ten, who taught us this delineation, hints, that, if thro' the middle point [...] of the line HN you draw the line IFL, it will touch the ellipsis in the point F; for since the angles IFH and IFN are equal, by what we have just now said, the vertical angle LF n of the one IFH, will be necessarily equal to the o­ther IFN: But this equality of the angles, made by the line KL drawn thro' F, with both those drawn from the centres, is here a sign of contact, as is in the circle the equality of the angles with a line drawn from its one centre. So that after this way you may draw a tangent thro' any given point F of [Page]

Pag. 192.

129

130

131

132

133

134

[Page] [Page 193] the ellipsis without this organical apparatus of Rulers; viz. if, having drawn from both the Focus's thro' the given point F the right lines nH, NG equal to the Latus Transversum DE, you bisect HN in I and draw IFL: Or if the line that connects the extremes GH be produced to K, and you draw thence KFL, viz. in that case where GH and N n are not parallel; other­wise a line drawn thro' the point F parallel to them would be the tangent sought.

As to the hyperbola, there is a mechanick method of draw­ing that also, not unlike the others, from a like property in that, communicated by the same Van Schooten, viz. If ha­ving found the Focus's N and n (Fig. 129. n. 3.) you tye a thread NFO in the Focus N and at the end of the ruler nO of the length of the transverse ax DE; then putting in a pen or the moveable leg of a pair of compasses (nor would it be dif­ficult to accommodate the practice we before made use of to this also) draw or move it within the thread NFO from O to E, so that the part of the thread NO may always keep close to the ruler as if it were glued to it. For if we call the length of the thread X, and the transverse ax ob as above, the ruler nO will be, by the Hypoth. = X + ob. Make now the part of the thread OF = ½ X, the remainder or other part will be NF = ½ X and nF = ½ X + ob, and the difference between FN and F n, = ob. Make OF = ¾ X, then will FN be ½ X and F n ¼ X + ob, the difference still remaining ob and so ad infinitum. In short, since the difference of the whole thread and of the whole ruler is ob, and in drawing them, the same OF is taken from both, there will always be the same difference of the re­mainders. Hence also assuming at pleasure the points N and n you may describe hyperbola's so, the thread NFO be short­er than the ruler nFO: For if it were equal there would be described a right line perpendicular to N n, thro' the middle point C.

There yet remains one method of describing hyperbola's and ellipses in Plano, by finding the several points without the help or Apparatus of any threads or instruments, viz. in the ellipsis, having given or assumed the transverse axis DE and the Foci N and n (Fig. 130. n. 1.) if from N at any arbi­trary distance, but not greater than half the transverse ax NF, you make an arch, and keeping the same opening of the com­passes [Page 194] you cut off, from the transverse ax, EG, and then, taking the remaining interval GD, from n you make another arch [...] cutting the former in F, and so you will have one point of the ellipse, and after the same way you may have innume­rable others, f, f, f, &c.

In like manner to delineate the hyperbola, having given o [...] assumed the transverse ax DE and the Focus's N and n (n. 2.) if from N at any arbitrary distance NF you strike an arch, and keeping the same aperture of the compasses from the dia­meter continued, you cut off EG, and then at the interval GD from n make another arch cutting the former in F, you will have one point of the hyperbola, and after the same way innumerable others, f, f, &c.

Proposition IX.

IF the secondary ax, or conjugate diameter of the hyperbola AB (Fig. 131.) be applied parallel to the vertex E, so that it may touch the hyperbola, and OE, EP are equal like BC and AC, and from the centre C you draw thro' O and P right lines running on ad insinitum, and lastly QR parallel to the Tangent OP; you'l have the following

CONSECTARYS.

I. THE parts QG and HR intercepted between the curve and those right lines CQ, CR will be equal; for by reason of the similitude of the ▵ ▵ CEP and CFR as also CEO, CFQ as CE is to EO (and EP) so will CF be to FQ and FR, and consequently these will be equal; and so taking away the semiordinates FG and FH which are also equal, the remainders GQ and HR will be also equal, and consequently the □ □ QGR, GRH, &c. all equal among themselves: Which we had already deduced before in Consect. 2. and 3. Def. 7.

II. The rectangle QGR will be = □ EO or EP = [...] i. e. (as Apollonius speaks) to the fourth part of the figure: For by [Page 195] reason of the similitude of the ▵ ▵ CEO, CFQ, CE will be to EO as CF to FQ: i. e. as the □ CE to the □ EO i. e. as (by Cons 2. 7.) the Lat. Transv. to the Lat. Rect.

  • so the □ CF
  • to th □ FQ

i. e. as (by the 7. Prop.) as the □ DFE to the □ FG

But now if from the □ CF you take the □ DFE, there will remain the □ CE, by Prop. 9. lib. 1. and if from the □ FQ you take the □ FG there will remain the □ QGR, by Prop. 8. lib. 1. wherefore that remaining □ CE to this remain­ing □ QGR will be, as was the whole square CF to the whole square FQ, by Prop. 26. lib. 1. i. e. as was the □ CE to the □ EO; consequently the □ QGR and the square EO (to which the same square CE bears the same proportion) will be equal among themselves.

III. Since this is also after the same manner certain of any other rectangle ggr or grh, &c. it follows that all such rect­angles are equal among themselves.

IV. Wherefore it is most evident, since the lines FR, fr, &c. and so GR and gr grow so much the longer, by how much the more remote they are from the vertex E; that on the contrary the lines QG and qg must necessarily so much the more decrease and grow shorter, and consequently the right line CQ approach so much nearer and nearer to the curve EG.

V. But that they can never meet or coincide altho' produ­ced ad infinitum will thus appear; if it were possible there could be any concourse or meeting, so that the point G and Q or g and q could any where coincide, it would follow from Consect. 2. that as the □ DFE to the square FG so the square CF to the square FQ i e▪ to the same square FG; and so that the □ DFE would be = □ CF; which is absurd by Prop. 9 lib. 1. so that now it is evident that the lines COQ and CPR drawn according to Consect. 1. are really Asymptotes ( i. e. they will never Apoll. Prop. 1. lib. 2. coincide ( viz. with the curve of the hyperbola) as Apollonius has named them.

[Page 196]VI. Having drawn the right lines from G and g parallel to both the asymptotes, viz. GS and gs and likewise GT and gt, the rectangles TGS and tgs will be Apollon. Prop. 12. lib. 2. e­qual among themselves. For by reason of the similitude of the ▵ ▵ TQG and tqg, first, TG will be to QG as tg to qg; and, by reason of the equality of the □ □ QGR and qgr, secondly, QG will be to gr reci­procally as qg to GR, by Prop 19▪ lib. 1. and by reason of the similitude of the ▵ ▵ SGR and sgr, thirdly, as gr to gs so GR to GS, wherefore (since in two series

1. 2. 3. as TG to QG to gr to gs so tg to qg to GR to GS) you'l have ex aequo or by proportion of equality as TG to gs so tg to GS, by Prop. 24 lib. 1. Therefore, by Prop. 17. of the same, the □ of TG into GS=□ of tg into gs. Q. E. D.

SCHOLIƲM.

HEnce, lastly, we have a new genesis of the hyperbola in Plano about its given diameters from the speculations of De Witt Elem. Curv. lib. I. cap. 2. prop. 3. De Witt; if, viz. having drawn the lines AB and EF cross one another at pleasure (Fig. 132) to the angle BCF you conform the move­able angle BCD ( acd being to be delineated in the opposite hyperbola equal to the contiguous ACD) one of whose legs is conceived to be indefinitely extended, but the o­ther CD of any arbitrary length; and to the end of it D ap­ply the slit of a moveable ruler GD about the point G at any arbitrary interval GD (but yet parallel to the leg CB in this first station) and so carrying together along with it the moveable angle BCD about the line ECF, but so that the leg CD may always remain fast to it, and the other CB be inter­sected in its progress by the ruler GDH, e. g. in b or β This point of intersection, thus continually moved on, will describe the curve bGβ, which we thus prove to be an hyperbola: Be­cause the ruler GDH turning about the pole G, and carried from D e. g. to d or δ cuts the leg of the moveable angle CB [Page 197] brought to the situation cb or γβ, and in the mean while remain­ing always parallel to it self; and having drawn from the points of intersection b and β and G the lines GI, bK and βη parallel to the ruler CF, because e. g. in the second station, having ta­ken the common quantity cD from the equal ones CD and cd, the remainders C c and D d are equal, and by reason of the si­militude of the ▵ ▵ dcb and dDG,

  • as D d
  • i. e. C c
  • or bK

to DG,

  • so dc
  • i. e. DC
  • or GI

to cb; the rectangle of K b into bc will be = □ of DG into GI, by Prop. 18. lib. 1. and in like manner, when in the third station having added the common line Dγ to the equal ones CD and γδ, the whole lines Dδ and Cγ are equal, and, by reason of the similitude of the ▵ ▵ βγδ and GDδ

  • as Dδ
  • i. e. Cγ
  • or βη

is to DG so

  • is γδ
  • i. e. DC
  • or GI

to γβ; the □ of ηβ into βγ=□ of DG into GI, by the same 18. Prop. Wherefore the three points b, G, β, (and so all the o­thers that may be determined the same way) are in the hyper­bola, whose asymptotes are CB and CF and its centre C, &c. by the present Prop. Consect. 6. Q. E. D.

You may also determine innumerable points of this curve separately without the motion we have now prescrib'd, viz. as the point a in the opposite hyperbola, if thro' any assumed point c in the asymptote CE you draw a parallel to the other asymptote CA, and having made cd equal to CD, from G thro' d draw G da, and so in others.

CHAP. II. Of Parabolical, Hyperbolical and Elliptical Spaces.

Proposition X.

THE Archim. de Quadratur. Parab. Prop. 17. and 24. Parabolick Space ( i. e. in Fig. 133. that comprehended under the right line GH and the parabola GEH) is to a cir­cumscribing Parallelogram GK, as 4 to 6 (or 2 to 3) but to an inscribed ▵ GEH as 4 to 3.

Demonstration.

Suppose FH divided first into two then into four equal parts, and draw parallel to the ax EF the lines ef, ef, &c. dividing also EF into four parts, the first fg will be 3, the second 2, the third 1, by Prop. 34. lib. 1. but as ef is to ge so is ge to he, by Consect. 1. Prop. 4. Therefore he in the diameter EF is = o, in the first ef it is = ¼ (for as ef, 4, to ge, 1, so ge, 1, to he, ¼) in the second ef a portion of he is = [...], in the third to [...], and so the portions eh in the trilinear figure E hHK make a series in a duplicate arithmetical progression, viz. 1, 4, 9, 16: After the same manner, if the parts F f, &c. are bisected, you'l find the portions eh in the external trilinear figure to make this series of numbers ⅛, [...], [...], [...], [...], [...], [...], [...], and so onwards. Wherefore since the portions eh or the indivisibles of the trili­near space circumscribed about the parabola are always in a duplicate arithmetical progression: the sum of them all will be to the sum of as many indivisibles of the parallelogram FK, [...]qual to the line KH, i. e. the trilinear space it self to this pa­ra [...]lelogram as 1 to 3, by Consect. 10. Prop. 21. lib. 1. Wherefore the semi-parabola [...]E hH will be as 2, and the ▵ FEH as 1 ½; therefore the whole parabola as 4, and the whole ▵ GEH as 3, and the whole parallelogram GK as 6. Q E. D.

CONSECTARY I.

IT is evident Archim. Prop. 19. with the Coroll. that in the first division, the second line fh (i. e. that drawn from the middle of the base FH) is three such parts whereof FE is 4; for eh is [...] i. e. 1, therefore fh is 3.

CONSECTARY II.

IT is also evident, that this demonstration will hold of any parabolick segment.

Proposition XI.

THE Elliptical Space Archim. lib. de Conoid. &c. Prop. 5. comprehended by the Ellipsis DAEB (Fig. 127.) is to a circle described on the transverse ax DE, as the Axis Rectus or conjugate diameter AB to the transverse ax DE.

Demonstration.

THis is in the first place evident from the genesis of the ellipse we deduced in Schol. 1. Prop. 7. for in that de­duction we shewed that FO, i. e. HN was to NI as AB to DE: Which since it is true of all the other indivisibles or or­dinates HN and IN ad infinitum; it is manifest that the planes themselves constituted of these indivisibles will have the same reason among themselves, as the Axis Rectus AB to the trans­verse DE. Q. E. D.

CONSECTARY I.

THerefore the quadrature of the ellipse will be evident, if that of the circle be demonstrated.

CONSECTARY II.

SInce a circle described on the least diameter AB will be to one described on the greater diameter DE, as AB to a third proportional by Prop. 35. lib. 1. it follows by vertue of the present Prop. that the ellipse is a mean proportional between the greater and lesser circle, i e. as the ellipse is to the great­er circle so is the lesser circle to it, viz. as AB to DE.

CONSECTARY III.

HEnce you may have a double method of determining the area of an ellipse. 1. If having found the area of the greater circle, you should infer, as the greater diameter of the ellipsis to the less, so the area of the circle found to the area of the ellipse sought. 2. If having also found the area of the les­ser circle, you find a mean proportional between that and the area of the greater.

SCHOLIƲM.

WE may also shew the last part of the second Consect. thus, 1. If having described the circle E adbE ( Fig. 134.) about the least axis of the ellipse we conceive a regular hexagon to be inscribed, and an ellipse coinciding with one end E of its transverse ax, and with the other or opposite one D to be so elevated, that with the point d it may perpendicu­larly hang over the circle, and further from all the angles of the figure inscribed in the circle you erect▪ the perpendiculars gG, bB, &c. it is certain that the sides ED and E d of the triangle DE d will be cut by the parallel planes FG gf, &c. into proportional parts, and that those by reason of the simi­litude of the ▵ ▵ FDG and fdg, and so also the other rectan­gles will be among themselves as the intercepted parts of the lines ID and id, CI and ci, and in infinitum, (viz. of how many sides soever the inscribed figure consists:) Wherefore also all the parts of the ellipse taken together will be to all the parts of the circle taken together, i. e. the whole ellipse to the whole [Page] [Page]

Pag. 201.

135

136

137

138

139

140

141

[Page 201] circle as all the parts of the diameter ED or ab, i. e. as DE it self to AB. Q. E D.

CONSECTARY IV.

IT is also evident that both these demonstrations of the pre­sent Prop. will be also the same in any segments of the el­lipsis or circle.

Proposition XII.

ANY Hyperbolical space GEHG (Fig. 135) is to any Hyperbolick figure of equal heighth gEhg [whose Latus Rectum and Transversum are equal (as in the circle) and also equal to the Latus Transversum of the former DE, as the Axis Rectus (or conjugate) AB is to the Latus Transversum DE (as in the ellipsis.)

Demonstration.

By the Hypoth. and Prop. 7. and its second Consect. the □ F g is = □ DFE. Wherefore this □ DFE i e. the □ F g is to the □ FG as the Latus Transversam to the Latus Rectum of the hyperbola GEHG, by the same seventh Prop. i. e. (by Consect. 2 of the same) as the square of the Latus Transvers. DE to the square of the conjugate AB: Therefore the roots of these squares will be also proportional, viz. F g to FG as DE to AB; and consequently (since the same is true of any other ordinates ad infinitum) the whole hyperbola gE hg will be to the whole one GEHG as DE to AB. Q. E. D.

CONSECTARY I.

THerefore having found the quadrature of such an hyper­bola, whose Latus Rectum and Transversum are equal, you may have also the quadrature of any other hyperbola.

CONSECTARY II.

IT is evident that the same demonstration will hold in any other hyperbola's.

Proposition XIII.

ANY Parabolick segments upon the same base, and hyperbo­lical and elliptical ones described about the same conjugate (one whereof shall be a right one, the other a scalene) and con­stitued between the same parallels, are equal.

Demonstration.

I. It is evident of Parabola's; for both the right one GEHG, and the scalene one G EHG ( Fig. 136. n. 1.) (for the demon­stration of Prop. 10. will hold in both) is to a ▵ inscribed in them as 4 to 3. But the triangles GEH and G EH are equal, by Consect. 5. Def. 12. or Prop. 28. lib. 1. Therefore the Parabola's also.

Or thus, in the right parabola GEHG every thing is the same as in 1. and 4. Prop. of this Book, viz. EI= eb, EF= ib, the square IK= oecc, the □ FG= oicc. And because therefore in the scalene Parabola also the square FG remains = oicc, make F E= n, and find both the abscissa EI, and the □ answering to it IK.

1. For the abscissa; as FE to EI so F E to EI, per [...] Consect. 4 Prop. 34 lib. 1.

2. For the □ IK; as F E to EI so □ FG to □ IK, [...]. per Prop. 4. of this.

Therefore the □ IK=□IK and IK=IK, and this in any case ad infinitum: Therefore the one parabola is = to the other. Q. E. D.

[Page 203]II. The business is much after the same way evident of el­lipses and hyperbolas. For making all things in the ellipsis and right hyperbola ( n. 2. and 3. Fig. 136.) as in Prop. 2, 3, 5, 7. viz. the □ IK oecd− eecd in the ellipsis, oecd+ eecd in the hyperbola, the □ AB oocd by Consect. 2. Prop. 7. EI= eb, DE= ob, &c. if in oblique ones for the Latus Transversum DE you put n, and seek the Latus Rectum and abscissa EI, you may by means of these also have the square IK, by Prop. 2. and 3.

1. For the Latus Rectum. As n to √ oocd so √ oocd to [...] by Cons. 2.7▪

2. For EI the abscissa. As ob to eb so n to [...]= EI.

3. For the side RS □ deficient or exceeding, from Prop. 2. and 3. As n to [...] so [...] to [...]=RS.

Now the abscissa multiplyed by the Lat. Rect. The abscissa multiplyed by RS.
[...] by [...] gives □ oecd. [...] by [...] gives □ eecd.
The sum of these □ □ oecd + eecd in the hyperb. = □ IK by Prop. 2. evidently = □ IK The difference of these □ □ oecd− eecd gives in the ellipsis □ IK by Prop. 3. evidently = □ IK.

Wherefore the lines IK and IK, and the whole KL and KL will be equal; and since the same thing is evident after the same way of all other lines of this kind ad infinitum, the elliptical and hyperbolical segments will be so also. Q. E. D.

CHAP. III. Of Conoids and Spheroids.

Proposition XIV.

A Parablick Conoid Archimed. Prop 23. and 24. (al. 26. and 27.) is subduple of a Cylinder, and in sesquialteran reason (or as 1 ½) of a cone of the same base and altitude.

Demonstration.

Because in the parabola the □ AD ( Fig. 137.) is to the □ SH, as BD to BH, i. e. as 3 to 1, and so to the □ TI as BD to BI, i. e. as 3 to 2, by Prop. 4. of this; it is evident that these squares of SH and TI and AD and consequently of the whole lines also S h, T i, AC, and the circles answering to them will be in arithmetical progression, 1, 2, 3; and more­over if there are new Bisections in infinitum, as the abscissa's so also the squares and circles of the ordinates, by vertue of the aforesaid fourth Prop. will always be in arithmetical Progres­sion 1, 2, 3, 4, 5, 6, &c. It is evident that an infinite series of circles in the conoid, consider'd as its indivisibles, will be to a series of as many circles equal to the greatest AC, i. e. the co­noid to the cylinder AF as 1 to 2, or as 1 ½ to 3, by Consect. 9. Prop. 21. or Consect. 4. Prop. 16. lib. 1. but to the same cylinder AF the inscribed cone ABC is as 1 to 3, by Prop. 38. lib. 1. therefore the cylinder, conoid and cone are as 3, 1 [...] and 1. Q.E.D.

Proposition XV.

THE half of Archim. 29. and 30. ( al. 32. and 33.) any Spheroid, or any o­ther segment of it is in subsesquialteran pro­portion to the cylinder, and double of the cone having the same base and altitude.

Demonstration.

Having divided the altitude BD ( Fig. 138.) e. g. into three equal parts, because in the ellipse as well as in the circle the square of AD is to the square of SH as the □ GDB to the □ GHB, i. e. as 9 to 5, and so to the square TI as 9 to 8, by Consect. 1. Prop. 5. of this; and in like manner if you make new bisections, the squares (and consequently the circles) of the ordinates go on or decrease by a progression of odd num­bers, as 36, 35, 32, 27, 20, 11, and so ad infinitum, the bisections being continued on; as we have shewn in the sphere and circumscribed cylinder Prop. 39. lib. 1. and it will neces­sarily follow here also (by vertue of Consect. 12. Prop. 21.) that the whole cylinder will be to the inscribed segment of the spheroid, as 3 to 2; and since the same cylinder is to the cone ABC as 3 to 1, also the segment of the spheroid will be to the cone as 2 to 1. Q. E. D.

Proposition XVI.

AN hyperbolical Conoid Archi [...]. Prop. 27. and 28. (al. 30. and 31.) is to a cone of the same base and altitude, as the aggre­gate of the ax of the hyperbola that forms it and half the Latus Transversum, to the aggregate of the said axis and Latus Transversum.

Demonstration, containing also the Invention of this Proportion.

Make (in Fig. 139.) CE= a, EF= b, OE= c; then will CF= a+ b. Since therefore as CE to OE so CF to FQ [...] the □ EO will = cc and □ FQ= [...].

But as these squares so also are the circles of the lines EO and FQ to one another, by Prop. 32. lib. 1. and so the cone COP [Page 206] will be as [...], and the cone CQR as [...] ( viz. by multiplying the third part of the altitude CF by the base FQ:) Having therefore substracted the cone COP from the cone CQR, there will remain the truncated cone QOPR [...], and from this solid truncated cone having further substracted the hollow truncated cone, which the space EHRP produced in the genesis of the conoid (and which ac­cording to Consect. 2. Definit. 9. is as bcc) there will remain the hyperbolical conoid [...] i. e. (by substituting now the values of the ax or abscissa EF, and of half the Lat. Transv. EC, and of the conjugate diam. OP, &c. found in the de­monstrations of the preceding Chapter, viz. [...] for a, eb for b, and √ oocd for c or oocd for cc) the hyperbolical conoid will come out [...] i. e. [...]. But the cone GEH (multiplying the third part of EF into the □ GH, i. e. ⅓ eb into 4 oecd+4 eecd) is as [...]. Therefore the conoid is to the cone as 6 eebocd+4 e (powerof3) bcd to 4 eebocd+4 e (powerof3) bcd, i. e. (dividing on both sides by 4 eecd) as ½ ob+ eb to ob+ eb. Which was to be found and demon­strated.

SCHOLIƲM.

IF any one had rather proceed herein by indivisibles, as in the precedent Prop. having divided the ax EF ( Fig. 140.) again into three equal parts, and assuming the values of the lines determined in the hyperbola, viz. eb for the abscissa EF, ob for the transverse ax, [...] for the Latus Rectum, oecd+ eecd for the square of the semiordinate FG, &c. the lowest and greatest circle of the diameter HG will be as oecd+ eecd, and, if you make [Page 207] as the Latus Transv. to the Latus Rectum, so the □ D fE ob− [...] made of ob+⅔ eb into ⅔ eb (i. e. ⅔ oebb+ [...] eebb) to a fourth; there will come out ⅔ oecd+ [...] eecd for the second circle of the diam. hg; and by the same inference (as ob to [...] so ob+⅓ eb into ⅓ eb to a fourth) for the third circle of the diameter HG ⅓ oecd+ [...] eecd; so that these indivisibles [for which here and in the precedent also the partial circumscribed cylinders may be assumed] proceed in a double series of numbers, the first in a simple arithmetical progression 3, 2, 1, the latter in a du­plicate Arithmetical progression of squares 9, 4, 1; and the same if you make further new bisections, will necessarily hap­pen ad Infinitum, (the former numbers e. g. in the first bise­ction will be [...] ⅚ [...] oecd the latter [...] eecd, &c) it is manifest from the consectaries of Prop. 21. lib. 1. that the whole cylinder HK will in like manner be expressed by a double series of parts answering, in numbers to the indivisibles of the conoid made by any bisection, but in magnitude to the greatest of them all, and in the sum of its first series of parts will be to the sum of the first in the conoid, both being infinite, as 2 to 1 or 3 to 1 ½ oecd, by Consect. 9. of the said Prop. 21. and the sum of its latter to the sum of the former in the conoid will be as 3 to 1 eecd and so the whole cylinder to the whole conoid as 3 oecd+3 eecd to 1 ½ oecd + eecd i. e. (dividing by ecd) as 3 o+3 e to 1 ½ o+ e i. e. mul­tiplying both sides by b) as 3 ob+3 eb to 1 ½ ob+ eb; and consequently the cone (which is ⅓ of the cylinder) to the co­noid as ob+ eb to 1 ½ ob+ eb. Q.E.D.

CONSECTARY.

HEnce also appears the proportion of the hyperbolick co­noid to a cylinder of the same base and altitude, which we did not express in the Prop. viz. as the aggregate of the ax and half the Latus Transversum to triple the aggregate of the said ax and Latus Transversum.

CHAP. IV. Of Spiral Lines and Spaces.

Proposition XVII.

THE Archim. Prop. 24. de Spi­ral. first spiral space is subtriple of the first circle, i. e. as 1 to 3.

Demonstration.

Having divided the circumference of the circle into ( Fig. 141. n. 1.) three equal parts by lines drawn from the initial point, beginning from the first line BA, the line BC will be as 1, BD as 2, BA as 3, by Consect. 1. Def. 12. of this book, and consequently the sectors circumscribed about the spiral will be CB c as 1, DB d as 4, AB a as 9, by Prop. 32. lib. 1. and in like manner, if you make new bisections, the lines drawn from the point B to the spiral, will be 1, 2, 3, 4, 5, 6; but the circumscrib'd sectors, 1, 4, 9, 16, 25, 36; and so the circumscrib'd partial sectors ad infinitum will proceed in an order of squares, there being always as many sectors in the circle equal to the greatest of them. Therefore all the sectors that can be circumscrib'd ad infinitum about the spiral space, i. e. the spiral space it self (in which at last they end) to so many equal to the greatest, i. e. to the circle, is as 1 to 3, by Consect. 10. Prop. 21. lib. 1. Q. E. D.

CONSECTARY I.

SInce the first circle is to the second as 1 to 4 ( i. e. as 3 to 12) by Def. 12. of this, and Prop. 31. lib. 1. and the first spiral space to the first circle as 1 to 3 by the present Prop. the same spiral space will be to the second circle as 1 to 12; and to the third by a like inference as 1 to 27, to the fourth as 1 to 48, &c.

CONSECTARY II.

THE first spiral line is equal to half the circumference of the first circle. For the lines or radii of the sectors, and consequently their peripheries or arches proceed in a sim­ple arithmetical reason, as 1, 2, 3, 4, 5, 6, &c. while in the mean time the whole periphery of the circle contains so many arches equal to the greatest. Therefore the whole pe­riphery of the circle is to an infinite series of circumscrib'd ar­ches, i. e. to the spiral line it self, as 2 to 1, by Consect. 9. Prop. 21. lib. 1.

Proposition XVIII.

THE whole spiral Archim. Prop. 25. space comprehended under the second right line EA and the second spiral EGIA (see Fig. 141. n. 2.) is to the second circle as 7. to 12.

Demonstration.

For having divided the circumference of the circle first into three equal parts, there will be drawn to the second spiral four right lines BE, BG, BI and BA being as 3, 4, 5, 6, and but only three sectors circumscrib'd, viz GB g, IB i and AB a, which proceed according to the squares of the three latter lines, viz. 16, 25, 36, so that the sum is 77, while the sum of three equal to the greatest is 108, and so the one to the other (dividing both sides by 9) as 12 to 8 [...] ▪ Having moreover bisected the arches and parts of the line BE, so that that shall be 6, the second BF will be 7, and so the other five 8, 9, 10, 11, 12; and the sectors answering to them (excepting the first) 49, 64, 81, 100, 121, 144, so that their sum shall be 559, while the sum of six equal to the greatest, i. e. the whole circle is 864, and so one to the other (dividing both by 72) as 12 to 7 [...]. In the other bisection of the arches and the parts of the line BE, so that the one shall be 12, the second 13, &c. to the thirteenth BA which will be 24, the sum of twelve sectors will be found to be 4250; and the sum of [Page 210] as many equal to the greatest 6912, and so the one to the other (dividing both sides by 576) as 12 to 7 [...] s. [...]. Therefore the proportion will be

  • I. In the first case 12 to 7+1+½+ [...] viz. [...].
  • II. In the second case 12 to 7+½+¼+ [...] viz. [...].
  • III. In the third case 12 to 7+¼+⅛+ [...], &c.

The first and second fractions thus decreasing by ½ the latter by ¼. Wherefore the proportion of the second circle to the second spiral space will be as

12 to 7+1+½+ [...]
−½−¼− [...]
−¼ &c.−⅛ &c.− [...] &c.

By vertue of Consect. 3. and 8.=0=0 Prop. 21. lib. 1. i. e. as 12 to 7. Q. E. D.

CONSECTARY I.

BEcause the second circle is to the first spiral space as 12 to 1, by Consect. 1. of the preceding Prop. and to the se­cond spiral space as 12 to 7, by the present. it will be to the second space without the first ( viz. BCDEAIGE) as 12 to 6 i. e. as 2 to 1.

CONSECTARY II.

THerefore the second space separately to the first is as 6 to 1.

CONSECTARY III.

SInce in the trisection of both these circles, first and second, there arise six lines, and as many sectors, viz. three lines BC, BD, BE, i. e. 1, 2, 3, to which there answer three ar­ches in the same progression within the second circle, and also as many equal to its greatest; therefore the sum of all the un­equal arches will be 21, but the sum of the equal ones of both circles (each of which in the first are equivalent to 3, in the second to 6) will be 27. Wherefore the sum of both the Pe­ripheries to the sum of all the circumscrib'd arches will be as [Page 211] 27 to 21, i. e. (dividing both sides by 9) as 3 to 2 ⅓. More­over bisecting the arches of the circles and the parts of the line BA, there will arise six circumscribed unequal arches within the first circle, which are as 1, 2, 3, 4, 5, 6, and as many within the second 7, 8, 9, 10, 11, 12; the sum of all which is 78, while the sum of as many equal ones on both sides is 108. Wherefore the one will be to the other, i. e. the sum of both the peripheries to twelve circumscribed arches ta­ken together, is now as 108 to 78, i. e. (dividing both sides by 36) as 3 to 2 ⅙. And making yet another bisection, the proportion will be found to be as 3 to 2 [...], &c. and hence at length may be evidently inferr'd; that the sum of both the peripheries will be to the sum of all the arches circumscribible ad infinitum, i. e. to the whole helix as

3 to 2+⅓
−⅙
− [...] &c.=0. that is, as 3 to 2. Q. E. D.
CONSECTARY IV.

THerefore, since the periphery of the second circle is dou­ble of the first, that alone will be equal to the whole spiral.

CONSECTARY V.

THerefore, if the periphery of the second circle be 2, the periphery of the first will be 1, and the first spiral line ½ by Consect. 2. of the anteced. Prop. wherefore the second spiral alone will be 1 ½, and so the periphery of the second circle alone will be to the second spiral alone as 2 to 1 ½ i. e. as 4 to 3; and to the first alone as 4 to 1.

SCHOLIƲM I.

BUT as Consect. 4. may be also deduced after another way, viz. by comparing only the arches of the second circle with the correspondent circumscripts, but considering them as taken twice (because that circle is twice turned round while the whole helix or spiral is described) and finding in the first [Page 212] trisection the proportion of double the second periphery to all the circumscripts as 12 to 7; and in the succeeding bisection as 12 to 6 ½; in the second bisection as 12 to 6 ¼, &c. and at length by inferring, that the second periphery is double of all the arches circumscribible about the whole helix ad infinitum, that is to the helix it self.

as 12 to 6+1
−½
−¼ &c.=0. i. e. as 12 to 6;

and consequently the simple second periphery will be to the whole helix as 6 to 6: Thus the 5. Consect. may be separate­ly had after the same manner, if instead of the first trisection, you only bisect; ( vid. Fig. 141. n. 3.) for so in the first bi­section the arches circumscribed about the second spiral line would be separately two semi-circles D d, 3 and A a, 4, (for as the line BC is one, BE, 2, BD, 3, BA, 4; so the arch described by the radius BD is 3 and described by the radius BA=4,) and their sum 7; while the sum of two equal to the greatest is 8. In the second bisection (when BE is 4) BF and its arch is made 5, the arch BD 6, the arch BG 7, the arch BA 8, the sum 26; while the sum of so many quadrants equal to the greatest is 32. Thus in the third bisection the sum of eight Octants circumscrib'd about the second helix will be found to be 100, the sum of so many = to the greatest 128, &c. Wherefore the periphery of the second circle in the first case will be to the arches circumscrib'd about the se­cond helix as 4 to 3+½; in the second as 4 to 3+¼; in the third as 4 to 3+⅛, &c. and so to all the arches circumscri­bible in infinitum, i. e. to the second helix it self as

4 to 3+½
−¼
−⅛ &c. = 0. i. e. as 4 to 3. Q. E. D.

By the same method you may easily find the proportion of the third circle to the third spiral space, and of that periphery either to the whole spiral, or separately to the third, as will be evident to any one who trys.

[Page 213]I. For the third spiral space.

( Fig. 142)

BC 1 BF 4 BI 7 49
BD 2 BG 5 BK 8 64
BE 3 BH 6 BA 9 81

are the three first sectors circumscrib'd about the parts of the third helix. The sum of these three sectors is 194; and the sum of so many equal to the greatest 243. Therefore the first proportion of the one sum to the other will be as 243 to 194, i. e. (dividing both sides by 9) as 27 to 21 [...].

In the first bisection there will be seven lines:

BH 12, BL 13 169 Sectors circumscribed about the parts of the third helix.
  BI 14 196
  BM 15 225
  BK 16 256
  BN 17 289
  BA 18 324
  Sum 1459

; while in the mean time the sum of as many equal to the greatest is 944, and so the second pro­portion as 1944 to 1459 i. e. (dividing both sides by 72) as 27 to 20 [...].

In the second Bisection there will be thirteen lines, viz. BH 24, the rest 25, 26, &c. but the sum of the sectors, i. e. of the square numbers answering to the twelve latter will be found to be 11306; while in the mean time the sum of as ma­ny equal to the greatest will be 15552, so that you will have the third proportion of this sum to the other, viz. as 15552 to 11306, i. e. (dividing both sides by 576) as 27 to 19 [...].

Therefore the I. proportion will be as 27 to 19+2+ [...] i. e.
to 19+2+½+ [...]
II. — as 27 to 19+1+ [...] i. e.
to 19+1+¼+ [...]
III. — as 27 to 19+ [...]— i. e.
to 19+½+⅛+ [...].

Therefore the proportion of the third circle to the third spi­ral space will be

as 27 to 19+2+½+ [...]
−1−¼− [...] i. e. as 27 to 19.
−½ &c. −⅛ &c. − [...] &c.
= 0. = 0. = 0. Q. E. D.

II. For the third spiral line.

If instead of the first trisection (as less commodious for the end proposed) you make use here also, as before, of bisecti­on in the same figure, there will come out six lines from the point B to the helix, viz. B m, 1, BE, 2, B n, 3, BH, 4, B o, 5, BA, 6; to which there answer as many semicircular arches in the same progression, and to the greatest of the two as many equal to 2, 4, 6; so that the sum of the unequal ones is 21, and of the equal ones 24, and so the proportion of three peripheries together to all the circumscripts together will be as 24 to 21 (and dividing both by 6) as 4 to 3 ½. In the se­cond bisection the twelve unequal lines and arches make the sum 78, and as many equal to the greatest of the four will give the sum 96; so that the second proportion will be 96 to 78, i. e. (dividing both sides by 24) 4 to 3 ½. In the third bisection the proportion will come out as 384 to 300, i. e. (dividing both sides by 96) as 4 to 3 ½, &c. Therefore the proportion of the three circles together to the whole Helix will be as 4 to

3+½
−¼
−⅛ &c.=0. i. e. as 4 to 3 or 12 to 9.

Q. E. D.

CONSECTARY VI.

NOW, if the periphery of the first circle be made 2, the second will bee 4, the th [...]rd 6, and consequently the sum 12; it will be manifest that the third periphery separately will be to the whole helix as 6 to 9, i. e. as 2 to 3.

CONSECTARY VII.

AND because the second periphery (which is 4) is equal to the first and second helix together, by the above Con­sect. 4. the remaining third spiral will be 5, and so the pro­portion of the third periphery to it as 6 to 5.

CONSECTARY VIII.

WHerefore the proportions of each of the peripheries to their correspondent spirals will be in a progression of ordinal numbers, viz. so that the latter of every two will denote the periphery of a circle, and the former an inscribed spiral; and consequently the spiral lines will be in an arith­metical progression of odd numbers, and the peripheries of the circles in a progression of even ones.

  • 1—The first Spiral,
  • 2—The first Periphery,
  • 3—The second Spiral,
  • 4—The second Periphery,
  • 5—The third Spiral,
  • 6 &c.—The third Periphery, &c.
SHCOLIƲM II.

THE seventh Consectary may also be easily deduced sepa­rately this way: In the first bisection the line BA and its periphery is 6, the line B o and its periphery 5, the sum of the circumscribed Peripheries 11; the sum of as many equal to the greatest 12. Therefore the periphery of the third circle will be to the two circumscripts as 12 to 11, i. e. as 6 to 5 ½. In the second bisection the four circumscribed quadrants will be 12, 11, 10, 9, their sum 42; and the sum of four equal to the greatest, i. e. the periphery of the third circle 48. Therefore the proportion is now as 48 to 42, i. e. (dividing both sides by 8) as 6 to 5 ¼. Thus you will have the third proportion as 192 to 164, i. e. (dividing both sides by 32) as 6 to 5 ⅛. Wherefore the proportion of the third periphery to the third helix or spiral is [Page 216]

as 6 to 5+½
−¼
−⅛ &c.
=0. i. e. as 6 to 5. Q. E. D.
CONSECTARY IX.

AS Consect. 8. supplies us with a rule to determine the pro­portion of every spiral of every order to the periphery of the correspondent circle, viz. if the number of the order be doubled for the periphery of the circle, and the next antece­dent odd number be taken for the spiral line; so what we have hitherto demonstrated supplys also another rule, to define the proportion of the spiral space in any order to its circle. For since the circles are in a progression of Squares 1, 4, 9, 16, &c. but the first circle is to the first space as 3 to 1 ( i. e. a, 1 to ⅓) by Prop. 17. and the second to the second as 12 to 7 ( i. e. as 4 to 2 ⅓) by Prop. 18. the third to the third as 27 to 19 ( i. e. as 9 to 6 ⅓) by Schol. 1. of this. And contem­plating both these series one by another,

  • Of the circles, 1, 4, 9.
  • Of the spaces, ⅓, 2 ⅓, 6 ⅓.

We see the numbers of the spaces are produced, if from the spuare numbers of the circles you substract their roots, and add to the remainder ⅓. Wherefore, if, e. g. we were to de­termine the proportion of the fourth circle to the fourth spiral space; the square of 4 viz. 16 would give the circle; hence substracting the root 4, there wiill remain 12, and adding ⅓ you would have the fourth spiral space 12 ⅓; and in like man­ner the spiral space 20 ⅓ would answer to the circle 25, &c. And that this is certain is hence evident, that if we mul­tiply these numbers 16 and 12 ⅓, also 25 and 20 ⅓ by 3, that we may have those proportions in whole numbers, 48 and 37, 75 and 61, these are those very numbers Archimedes had hinted at in the Coroll. of Prop. 25.

Pag. 217.

142

143

144

145

146

CONSECTARY X.

NAY what we have now said, is that very Coroll. com­prehending also that 25th. Proposition, viz. that a spi­ral space of any order is to its correspondent circle, as the re­ctangle of the semidiameters of this and the preceding circle together with a third part of the square of the difference be­tween both semi-diameters to the square of the greatest semi­diameter. For, if e. g. the proportion of the third spiral space to the third circle be required, since the semidiameter of this third circle is as 3, and the semidiameter of the second precedent one is 2, and so the difference 1; the rectangle of 2 into 3 i. e. 6, together with ⅓ of the square of the diffe­rence will define the third spiral space 6 ⅓; since the third cir­cle may be defined by the square of the semidiameter of the greater, viz. by 9, and so in the rest; as the numbers we have found shew, or further that may be found according to given Rules which may be here seen in the following Table.

Orders. I II III IV V VI VII VIII XI X
Circles. 1 4 9 16 25 36 49 64 81 100
The whole spaces, the preced. ones being included. ⅓ 2 ⅓ 6 ⅓ 12 ⅓ 20 ⅓ 30 ⅓ 42 ⅓ 56 ⅓ 72 ⅓ 90 ⅓
Separate spaces the preced. ones be­ing excluded. ⅓ 2 4 6 8 10 12 14 16 18
CONSECTARY XI.

OUT of which table it is obvious to sight, that the se­cond space excluding the first is sextuple of the first, as we have already deduced in Consect. 2. Prop. 18. and the third separate space double of the second, and the fourth triple of the same second, and the fifth quadruple, and so onwards.

SCHOLIƲM III.

AND this shall suffice for spirals, which comprehends not only the chief Theorems of Archimedes of spiral spaces, but also the chief of spiral lines (whereof Archimedes has left nothing.) If any should have a mind to carry on our me­thod further, he may easily demonstrate after the same way what remains in Archimedes, and what Dr. Wallis in his A­rithmetick of Infinites from Prop. 5. to the 38, and what o­thers have done on this Argument.

CHAP. V. Of the Conchoid, Cissoid, Cycloid, Quadratrix, &c.

Proposition XIX.

THE first conchoid of Nicomedes Bbb (Fig. 110) on both sides of the perpendicular cDb approaches nearer always to the directrix or horizontal line AE, and yet will never coincide with it, altho' it be conceived to be produced on both sides ad infinitum.

Demonstration.

For since only D b is perpendicular to AE, and all the rest ab are so much the more inclined to it by how much the more remote they are from the middle one D b, and all in the mean while are equal both to it and to one another, by Def. 13. it is evident that the points b and B will come so much the near­er to AE, by how much the farther they recede from the mid­dle line D b. And yet because the lines BAC and bac are all right ones, whose points A, a, are in a right line AE it is e­qually as impossible that the point b or B, which is always in the conchoid should ever touch this right line, as it is impos­sible that the point C should be in it, by vertue of the afore­cited Def. Q. E. D.

Proposition XX.

YET no other right line can be drawn between the dire­ctrix AE and the conchoid, but what will cut it if pro­duced.

Demonstration.

For if such a right line be made parallel to AE, as GH, and you make, as DI to IC so D b to a fourth, which will be greater than IC, as D b is greater than DI, and consequently, if making that an interval you draw the circular arch from C, it will necessarily cut the line GH e. g. in G. Drawing there­fore C aG. you'l have as DI to IC so aG to GC, i. e. to that fourth proportional before found, by vertue of Prop. 34. lib. 1. but as DI to IC, so was also D b to the same fourth by Con­struc. Therefore aG and D b, which have both the same pro­portion to the same quantity, are equal; and consequently the point G is in the conchoid by virtue of Def. 13. and conse­quently the right line GH being produced will cut that pro­duced also, on both sides, by the same reason. Much more will it cut it on either side if it be not parallel to the directrix AE, which is very obvious. Therefore no right line can be drawn between the conchoid, &c. Q. E. D.

CONSECTARY.

HEnce, besides orher Problems, that may be very easily solved, which requires, having any rectilinear angle gi­ven ABC ( Fig 143.) and a point without it, from that point to draw a right line DEF, so that part of it EF, which is intercepted between the legs of the angle, shall be equal to a given line Z. For if you draw the perpendicular DGH from the given point D through the nearest leg of the angle BC, and make GH equal to the given line Z, and from the center C at the interval GH describe the conchoid IHK, which will be necessarily cut by the other leg of the angle by vertue of the present Prop. e. g. in F, the line DF being drawn will [Page 220] give the intercepted part = GH by the nature of the conchoid, and consequently = to the given line Z.

SCHOLIƲM.

BY means of this Consectary Nicomedes solves that noble Problem of finding two mean proportionals, after this way, which we will here shew from Eutocius, but drawn into a compendium, and somewhat changed as to the order. Let two given lines AB and BC ( Fig. 144.) between which you are to find two mean proportionals, be joined together at right angles, and divide both into two parts in D and E, and ha­ving compleated the rectangle ABCL, from L thro' D draw LG to BC prolonged; so that after this way GB may become = AL or BC: Having let fall a perpendicular from E cut off from C at the interval CF=AD the part EF, and having drawn FG make CH parallel to it; and lastly thro' the legs of the angle KCH draw the right line FHK, so that the part HK shall be equal to the line CF, by the preceding Consect. and also draw the right line KM from K thro' L to the con­tinued line BA: All which being done, CK and AM will be two mean proportionals between AB and BC; which after our way we thus demonstrate: By reason of the similitude of the ▵ ▵ MAL and LCK

MA is to LC or AB as AL or BC to CK

b− eb− c− ec and moreover,

as MA to AD so GC to CK i. e. FH to HK

b−½ eb−2 c− ec by reason of GF and CH be­ing parallel, by Consect. 4. Prop. 34. lib. 1. therefore since HK is = AD=½ eb, FH will be = A= b, and consequent­ly MD=FK, viz. both b+½ eb, and the square of both = bb+ eb+¼ eebb=□EF+EK by vertue of the Pythag. Theor. Now if to these equal quantities you add the equal □□ DX and EC=¼ cc, their sum, viz. □MD+□DX i. e. □ MX will be bb+ ebb+¼ eebb+¼ cc, equal to the sum of these, viz. □EF+□EC i. e. □ CF (by the Pythag. Theor. or EX by Construct.) +□KX; whence these two things now follow: 1. That the lines MX and KX are equal. 2. If from those equal sums you take away the common quan­tities [Page 221] ¼ eebb+¼ cc, the remainders will be equal, viz. bb+ ebb = ecc+ eecc; and (since the part taken away, viz. bb is ma­nifestly to the other part taken away, viz. ecc as the remain­der ebb to the remainder eecc, and the whole with the parts taken away and the remainders are in the same proportion by Prop. 26. lib. 1.) separately also bb will = ecc and ebb= eecc. But from the latter equation it follows that

as eb to ec so ec to b by vertue of the 19. Prop. lib. 1. AB to CK so CK to MA

and by the same reason it follows from the former Equation as ec to b so b to c

CK to MA so MA to BC i. e. CK and MA are two mean Proportionals between AB and BC. Q. E. D.

From which deduction you have also manifest the foun­dation of that mechanical way, which Hiero Alexandrinus makes use of in Eutocius, lib. 2. of the Sphere and Cylinder, and which Swenterus has put into his practical Geometry lib. 1. Tract. 1. Prop. 23. when, viz having joined in the form of a rectangle the given right lines AB and BC ( Fig. 145.) and continued them at the other ends, he so long moves the ruler in L, having a moveable center, backwards and forwards, 'till XK and XM by help of a pair of compasses are found equal. To which, another way of Philo's is not unlike, and flows from the same fountain, wherein, having made on AC a semicircle, the moveable ruler in L is so long moved back­warks and forwards, until LM and NK are found equal: Which seems to Eutocius to be more accommodated to practice, and easier to be perform'd by help of a ruler divided into small equal particles.

Proposition XXI.

IF from any point of the other diameter in the generating cir­cle e. g. from G (Fig. 111. n. 1.) you draw a perpendicu­lar GE thro' the cissoid of Diocles. the lines CG, GE, GD, and GH will be continual proportionals.

Demonstration.

For since GE and IF, as right sines, and also GD and IC as versed sines of equal arches by the Hypoth. are equal; you'l have as ID to IF ( i. e. CG to GE) so IF to IC ( i. e. GE to GD) per n. 3. Schol. 2. Prop. 34. lib. 1. But GD is to GH as ID to IF ( i. e. as GE to GD) by the forecited Prop. 34. lib. 1. Therefore CG to GE, GE to GD, and GD to GH, are all in the same continual proportion. Q. E. D.

CONSECTARY.

HEnce it was easie for Diocles to find two mean proportio­nals x and y between two given right lines V and Z; ( Fig. 146) for he made (having first described his curve DHB) as V to Z so CL to LK, and having drawn CKH to the curve, and thro' H the perpendicular GE, he had between CG and GH two mean proportionals GE and GD by vertue of the pre­sent Prop. when in the mean while CG the first would be to GH the last, as CL to LK, i. e. as the first V to the last Z given by vertue of the Constr. Therefore nothing remain'd but to make, 1. as CG to GE so V to X; and lastly, as GE to GD so x to y.

SCHOLIƲM.

IT may not be amiss to mention here another way of find­ing two mean proportionals between any two given lines by the help of two Parabola's, which Menechmus formerly made use of, viz. by joining at right angles the given lines AB and BC ( Fig. 147.) and prolonging them as occasion shall require thro' E and D; and then describing a Parabola about BE as its axis, so made that BC shall be its Latus Rectum, and in like manner describing another Parabola about BD as its ax, that shall have AB for its Latus Rectum, and that shall cut the for­mer in F: Which being done, the semiordinate FE (or BD which is equal to it) being drawn to the point of Intersection F, will be the two mean proportionals sought. For by ver­tue of the fourth Consectary of Prop. 1. of this Book, DF or [Page 223] BE is a mean proportional between AB and BD, and in like manner EF or BD is a mean proportional between BE and BC, and consequently as AB to BE so BE to BD, and as BE to BD so BD to BC; Q. E. D.

To this way of Menechmus that of Des Cartes is not unlike, which he gives us p. m. 91. except only that instead of two Parabola's, he makes use only of one and a circle in room of the other: In imitation of whom Renatus Franciscus Slusius has since shewn infinite methods of doing the same thing by help of a circle, and either infinite Ellipse's or Hyperbola's, in his ingenious Treatise which he thence names his Mesolabium.

Proposition XXII.

ANY semiordinate of the Cycloid as BF (Fig. 148.) or bf is equal to its corresponding Sine in the generating circle as BD, bd, together with the arch of that sine AD or Ad.

Demonstration.

For the motion of the point A describing the semi-cycloid AFE, by Def. 11. is compounded of the motion of the orb (or wheel) B along the semi-circle ADC, and of the motion of the centre along the right line BC equal to CE, and conse­quently to the semi circle it self, or motion of the orb, There­fore as the point A moving to E by the motion of the orb (or wheel) moved or was carried from the diameter AC thro' the whole semi circle ADC 'till it came to AC again, and by the motion of the Centre passes thro' the whole space BG or CE, which is equal to the semicircular arch; thus the same A, when come to F, will have describ'd the quadrant AD, by moving from the diameter AC the quantity of the sine BD, and moreover by the motion of its centre (which is equal to the motion of the Orb) moves from AC the space of DF: And so the semiordinate BF will be equal to the arch AD and to its sine BD taken together; and in like manner the semiordinate bf will be equal to the arch A d and its sine bd, &c. Q E. D.

CONSECTARY I.

HEnce may be easily assign'd by help of the cycloid a right line equal to the semi-periphery or any given arch AD or A d; viz. CE double, or taken twice for the whole circum­ference, and single for the semi-periphery or half circumfe­rence, DF for the quadrant A d, df for the arch AD, &c.

CONSECTARY II.

WHerefore the quadrature of the circle may be geome­trically obtain'd according to Consect. 2. of Def. 15. lib. 1.

CONSECTARY III.

IF you take B e, be, double of the sines BD, bd, &c. so that all the indivisibles bd taken together, may be to all the in­divisibles be taken together as BD to B e, a curve described thro' the points e will be an ellipsis by Prop. 11. and the curviline­ar space ADC eA will be equal to the semicircle ACDA.

CONSECTARY IV.

AND since DF ( i. e. D e+ eF) is equal to the quadrant DA, by vertue of the present Prop. BD+FG will be also equal to the quadrant (because the whole BG or CE is = to a semicircle) and consequently eF and FG will be equal; in like manner since df both above and below is equal to the arch dA, below bd+ fg will be = to the remaining Arch dC: and above bd+ ef (i. e. df) will be equal to the equal arch dA. Therefore ef above and fg below are equal, and (since the same may be shewn of all the indivisibles of the same sort throughout) the trilinear figure FGE will be equal to the tri­linear eFA.

Proposition XXIII.

THE cycloidal space is triple of the generating circle i. e. the semi-cycloidal space AECA is triple of the semi-circle ADCA.

Demonstration.

Since the parallelogram BCEG is equal to the whole circle by Consect. 2. of Def. 15. lib. 1. i. e. to the semi-ellipse A eCA, by the present construction the Trapezium C eGE will be equal to the quadrant of the ellipse or the semi-circle. But the trili­near space FEG is = to the trilinear space FA e, by the fourth Consect. of the preced. therefore also the trilinear space A eCEA is equal to the semi-circle. Therefore the whole cycloidal space is equal to the three semi-circles. Q. E. D.

Or thus.

Since the whole parallelogram AE is equal to two circles and the semi ellipsis A eCA to one; the remaining space A eCEC to one circle, and its half A eGC to a semi-circle. But the tri­linear space A eF is equal to the trilinear space FGE by Consect. 4. of the preced. Therefore the one being substituted in the other's place the trilinear space AFEC will be equal to the se­mi-circle: Therefore the remainder of the Parallelogram, i. e. the cycloidal space AFECA will be equal to three semi-circles. Q. E. D.

SCHOLIƲM.

TO these short Demonstrations, which we confess we owe for the most part to Hon. Faber, we will subjoin another somewhat more prolix, but yet not unpleasant, which we find in Carolus Renaldinus, lib. 1. de Resol. & Compos. Math. p. 299. But here we will give it the Reader more plain, and free from all Scruples, and likewise much easier. It is perform'd in these inferences, 1. That the right lined Parallelogram A baB ( Fig. 149 n. 1) is equal to the curvilinear space A bdaBDA. [Page 226] 2. That as that is divided into two equal parts by its right li­ned diagonal A a, so likewise is this by the semi-cycloid A aa, so that the right lined triangle A aB is equal to the curvilinear space A aaBDA. 3. Therefore the one as well as the other is equal to the generating circle; and consequently, 4 If to this curvilinear space there be added the semi-circle ADBA the semi-cycloidal space A aaBA will be equal to three semi-cir­cles. The first is evident, while if you take from the right lined parallelogram the semi-circle ADBA on the one side, and on the other add the semi-circle abda, there will arise the curvi­linear Parallelogram. The third is evident from Consect. 2. Def. 15. lib. 1. Because the line B a is equal to the semi-peri­phery, which multiplyed by the diameter BC gives the area of the circle. The fourth is self evident; and so there remains only the second to be demonstrated, viz. That the curvilinear paral­lelogram is divided into two equal parts by the cycloid, i. e. that the external trilinear Figure A adbA is eqaal to the inter­nal one A aaBDA; which may be thus shewn: Having divi­ded the base B a into three equal parts, and drawn thro' them three semi-circles, and moreover the transverse right lines D d and E e thro' the intersections of the sem-icircles and the cy­cloid; it is certain from the genesis of the cycloid, by vertue of the Cons. of Def. 11. that as the right line a1 is a third part of the whole aB, so the arch 1 a is a third part of the ge­nerating periphery, and by the same reason the arch 2 a two thirds, and so the remaining arch a11 also ⅓; insomuch that the first arch 1 a, and the last a11, and consequently their right sines af, ag, and likewise their versed ones f1, g11, are equal, and so the curvilined partial Parallelograms, both above and below, all upon equal bases, and of the same height ( viz. the two linear ones ae21 and da 11. 1.) are equal among themselves, and so likewise the two pricked or pointed ones D a 2B and aeb1. Wherefore if now the base B a (n. 2) be con­ceived to be divided into six equal parts, and having drawn semi-circles, and transverse lines thro' their intersections with the cycloid, the arches and [...] will be [...] and [...] and [...] of a semi-circle, [Page] [Page]

Pag. 227.

147

148

149

150

[Page 227] and so the versed sines of each, i. e. the altitudes of the corre­sponding parallelograms will be equal, and consequently the parallelograms of the internal and external trilinear space that are alike noted or signed will be equal to each other. Now this inscription of curvilinear parallelograms always respective­ly equal both in number and magnitude, since it may be con­tinued in both the trilinear figures ad infinitum; it will evi­dently follow, that the trilinear figures themselves, whose in­finite inscripts are always equal, will be likewise equal to one another.

Proposition XXIV.

THE base of the quadratrix AE (Fig. 150) and the se­midiameter of the generating quadrant AD and the qua­drant it self BD are in continual Proportion.

Demonstration.

For the quadrant DB is to the radius DA as the arch IB to the perpendicular H e by Consect. 1. Def. 16 and I b is to H e as A b to A e by Prop. 34. lib. 1. But the arch IB (if it be con­ceiv'd to be less and less ad infinitum) will at length coincide with I b, as ending in the same moment in the point B, where­in H e will end in the point E, and so A e will end in AE and A b in AB. Therefore at length DB will be to DA as IB ( i. e. I b) to H e, i. e, as A b to A e, i. e. as AB (or DA) to AE. Q. E. D.

SCHOLIƲM I.

CLavius about the end of the sixth Book of Euclid, and o­thers, demonstrate this indirectly by a deduction ad Ab­surdum, or concluding the opposite much after this manner: If DA or AB is not to AE as DB to DA, suppose it to be so to the greater A f or the less A e. In the first case therefore, because AB is to A f as DB to DA per Hypoth. i. e. as K f to A f the quadrant K f and the radius AB or DA will be equal. But as BD is to IB so is K f to H f by reason of the similitude of the arches; and as BD to IB so also DA (or =K f) to the sine H e, by [Page 228] Consect. 1. Def 16. Therefore the sine H e and the arch H f (to which the same K f bears the same proportion) will be e­qual; which is absurd. In the latter case, because AB would be to A e as DB to DA by the hypoth. i. e. as L e to A e, the quadrant L e and the radius AB or DA would be again equal. But as BD is to IB so is L e to M e by reason of the similitude of the arches; and as BD to IB so also is DA ( ie=L e) to H e by Consect. 1. Def. 16. Therefore the tangent H e and the arch M e (to which the same L e bears the same proporti­on) will be equal, which is again absurd. Wherefore BD is to DA, not as DA to a greater A f or a less A e; therefore as DA to AE. Q. E. D.

CONSECTARY I.

WHerefore it is evident from what we have deduced, if by means of the base of the quadratrix AE you draw a quadrant, the side of the quadratrix DA will be equal to it, and consequently double of the semi periphery, and quadruple of the whole periphery.

CONSECTARY II.

IT is evident also that you may obtain a right line equal to the quadrant DB of any given circle, if, having described a quadratrix, you make as AE to AD so AD to a third equal to the quadrant DB: Which third proportional taken four times will be equal to the whole periphery.

CONSECTARY III.

YOU may also obtain a right line equal to any less arch, if you make, as DA to H e so a third proportional found ( i. e. the quadrant DB) to a fourth, by vertue of Cons. 1. Def. 16.

CONSECTARY IV.

THE quadrature of the circle therefore, by vertue of Cons. 2. Def. 15. lib. 1. as likewise the trisection of an angle, by vertue of Consect. 2. Def. 16. lib 2. may be Geometrical­ly obtain'd, if the Quadratrix might be number'd among Geo­metrical Curves.

SHCOLIƲM II.

CLavius was also of this Opinion in the book afore menti­oned, who thought that if the quadratrix be excluded out of the number of geometrical curves, by the same reason you may also exclude the ellipse, parabola, and hyperbola, since they as well as this are commonly described thro' innu­merable points. But by that great Man's leave, we may de­ny this consequence, by the same reason as Des Cartes has de­ny'd the converse of it in his Geom. p 18. and 19. by vertue of which he suspects the ancients took the conick sections, &c. for mechanick or non-geometrick lines, because they did the spiral, quadratrix, &c. for such. But this is the difference between the description of the quadratrix and the conick secti­ons thro' points, that all and every of the points of the conick sections, relating to any given point of the axis, may be geo­metrically determin'd; but all the points of the quadra­trix promiscuously related to any point of the generating qua­drant, cannot be geometrically determin'd, but only those which respect some certain point, from which the quadrant may be divided into two arches of known proportion. For if, e g. in the quadrant BD the point X be given at pleasure, it will be impossible by Clavius's Rule to define a point of the quadra­trix answering to it, because the proportion of the arches DX and BX is unknown, and consequently neither can a propor­tional section of the right line AD be made: Not to mention that the last point E (which is the primary and most necessary one to the quadrature) even by Clavius's own confession cannot be geometrically defined. We may pass the like judgment on Archimedes's spiral and such like curves, which are conceiv'd to be described by two motions independent on one another; as [Page 230] will be manifest to any one who compares the genesis of the spiral with that of the quadratrix and what we have hitherto said. Whence neither will Monantholius's trisection of a gi­ven angle (which he essays) by means of a spiral be enough geometrical; which in his Book de Puncto, Cap. 7. p. 24. he attempts to perform thus: To the centre of a described spi­ral and its first helical or spiral line BA ( Fig. 151.) he applies the angle ABC equal to the given one abc; then having drawn circles thro' F and A where the legs of the angle cut the spi­ral, he divides the intermediate space DA into three equal parts in 1, and K: And then thro' these points he draws circles cut­ing the helix in L and M; and lastly having drawn BLN, BMO, he easily demonstrates from the genesis of the spiral that the arches AO, ON, NC are equal. And so after the same manner not only any angle or arch, but the whole peri­phery may be geometrically divided into as many parts as you please; only supposing that this spiral line may be numbred among geometrical ones; as we have heretofore hinted that the cycloid, conchoid, cissoid, and logarithmical curve, &c. might be; and we have above sixteen Years ago declared our opinion for it in our German Edition of Archimedes; and now are therein confirm d by those celebrated Mathematicians Leib­nitz, Craige, &c. who number lines of this kind, altho' they cannot be expressed by our common equations, among geome­trical ones, notwithstanding the contrary opinion of Des Car­tes, &c. because they admit of equations of an indefinite or transcendent degree, and are capable of a Calculus as well as others, tho' it be of a nature and kind different from that commonly used. See the Acta Erud. Lips. ann. 84. p. 234. and ann. 86. p. 292. and 294.

CHAP. VI. The Conclusion, or Epilogue of the whole Work.

NOW we may at length understand what Honoratus Fa­bri delivers concerning the distribution of figurate mag­nitudes into certain Classes, in his Synopsis Geom. p. 57. and the following.

1. The first Classe contains elementary figures, or equal in­divisibles, such as, 1. All Parallelograms, as the Square, Ob­long, Rhombus and Rhomboid, the elements whereof are e­qual right lines, as in Def. 12. lib. 1. 2. Convex or con­cave Surfaces, the elements whereof are curve lines moved thro' right lines by a parallel motion; among which are chiefly reckon'd cylindrical surfaces, whereof see Def. 16. lib. 1. about the end. 3. Parallelepipeds, and among them the cube, whose indivisibles are squares, or other Parallelograms. 4. Prisms made by the motion of a Triangle, Trapezium, or any Polygonous Body, along a right line, all the indivisibles whereof are consequently similar and equal to the generating plane.

2. The second Classe contains Figures whose Elements de­crease in a simple arithmetical Progression; such are, 1. Tri­angles, as is evident from Prop. 37. lib. 1. 2. The circle, and its Sectors, as resolvible into concentrick Peripheries ac­cording to Cons. 1. and 3. of the aforecited Prop. 3. The Cylinder as resolvible into concentrick cylindrick Surfaces, as its indivisibles. 4. The Surface of a Cone, whose elements are circular Peripheries, and also of the Pyramid whose indivi­sibles are similar angular Peripheries every where increasing in arithmetical Progression. 5. The Parabolick Conoid, whose indivi­sibles are Circles according to the proportion of the abscissa's in arithmetical Progression, by vertue of Prop. 14. lib. 2. &c.

3. The third Classe contains elementary Figures increasing in duplicate arithmetical Progression; such are, 1. The Py­ramid and Cone, the first whereof may be resolv'd into an­gular Planes, the second into circular ones increasing accord­ing to a series of square numbers; as is evident from Prop. 38. [Page 232] lib. 1. and its Consectary. 2. The trilinear parabolick Space, as defin'd Prop. 10. lib. 2. by the letters E h HK. 3. The Sphere, as far as it may be resolved into spherical concentrick Surfaces, every one whereof may be consider'd as a base, tak­ing the semidiameter for the altitude. 4. The Cone, as re­solvible into parallel conical Surfaces describ'd by the parallel indivisibles of the Triangle. 5. The remainder of a Cy­linder after an Hemisphere of the same base and altitude is tak­en out, according to Schol. 1. of Prop. 39. lib. 1.

4. The fourth Classe would comprehend all magnitudes re­solvible into elements or indivisibles increasing in triplicate, quadruplicate, &c. Arithmetical Progression; such we have not treated of, but may be found among Planes terminated by Curves of superior Genders; see Fabri's Synopsis, p. m. 67.

5. The fifth Classe is of those Magnitudes, whose indivisi­bles decrease, proceeding from a square number by odd num­bers, as, 36, 35, 32, 27, 20, 11, &c. such are, first, an Hemisphere, as is evident from Prop. 39. lib. 1. 2 An Hemi­spheroid, as in Prop. 15 lib 2. 3. A Semi-parabola, as may be gather'd from the demonstration of Prop. [...]0 lib. 2. For since the indivisibles of the circumscrib'd trilinear figure eb are found in a duplicate arithmetical Progression, ¼, [...], [...], [...], the indivisibles of the semi-parabola will necessarily be [...], [...], [...], [...], &c.

6. We may make a sixth Classe of those Magnitudes whose indivisibles decrease in a like Progression, not of the numbers themselves descending by odd steps from a given square, but of their roots, which are for the most part surd ones; such as is first, the Semi-circle, as is evident from Prop. 43. lib. 1. and by vertue of Prop. 5. lib. 2. and also the semi-ellipse, &c.

7. The seventh Classe comprehends those Magnitudes, whose Elements are in a Progression of a double series of num­bers, as in the Parabolick Conoid, as may be seen in the Scho­lium of Prop. 16. lib. 2.

But, to omit the other Classes of Magnitudes of a superiour Gender, the consider [...]tion whereof these Elements either have not touch'd on, or only by the by; (which any one who pleases may see in Faber s Synopsis, especially those which he compre­hends under the sixth and seventh Classes, p. 70 and the fol­lowing) about those we have here particularly noted, there re­main [Page 233] only two things to be taken notice of. 1. That since in the first Classe we place Parallelograms and Cylinders, in the second Triangles, in the third Pyramids and Cones, in the fifth Hemispheres, in the sixth semi-circles, &c. We may with Hon. Faber call the first Classe, that of Cylindrical or Parallelogrammatick Figures; the second, the Classe of Tri­angular Figures; the third, of Pyramidals; the fifth, of He­mispherical Figures; the sixth of semicircular ones, &c. 2. That having ranged or reduced after this manner homoge­neous Figures, or those of like condition, to a few Classes, their dimension, and consequently almost the whole business of measuring may be very compendiously reduc'd to a few Rules; whereof we will here give the Reader a short Speci­men, in the following

CONSECTARYS.

I. THE dimension of Parallelogrammatick Figures, i. e. of those of the first Classe, may be had, by multiply­ing the whole base by the whole altitude: See lib. 1. Def. 12. Cons. 7. Def. 18. Cons. 6. Def. 16. Cons 3. and 4.

II. The dimension of Triangular Figures, i. e. of those of the second Classe, may be had by the multiplication of the whole Base by half the Altitude, or of half the Base by the whole altitude; [see lib. 1. Def. 12. Consect. 8. Def. 15. Con­sect. 2. Def. 18. Consect. 4. lib. 2. Prop. 14.] and their Pro­portion is to their respective circumscribing Parallelograms, as 1 to 2; [see besides the Prop. already cited, lib. 1. Prop. 37. and its first Consect.]

III. The dimension of Pyramidals, i. e. of magnitudes of the third Classis, may be obtain'd by the multiplication of the Base by the third part of the of Altitude; [see lib. 1. Def. 17. Consect. 3. and 4. and Def. 20. Consect. 1. &c.] and their Proportion to the corresponding Figures of the Classe of the same Base and Altitude is as 1 to 3. [see besides the Prop. al­ready cited Prop. 38. lib. 1. and its Cons. Prop. 39. and Schol. 1. lib. 2. Prop. 10. &c.

[Page 234]IV. The Proportion of Hemispherical Magnitudes, i. e. of the fifth Classis to corresponding ones of the first Classe of the same Base and Altitude is as 2 to 3; [see lib. 1. Prop. 39. lib. 2. Prop. 10. and 15.] and so their dimension may be had by multiplying theit Base by ⅔ of their Altitude.

V. The Proportion of semi-circular Magnitudes, i. e. of those of the sixth Classe to so many corresponding ones of the first Classe of the same Base and Altitude cannot be expressed by whole numbers or by a small fraction [see lib. 1. Prop. 43. and lib. 2. Prop. 11.] and consequently their exact numeral dimension cannot be had.

AN INTRODUCTION TO S …

AN INTRODUCTION TO SPECIOUS ANALYSIS, OR, The New Geometry, chiefly according to the Method of Des Cartes, But much facilitated by later Inventions, &c.

By J. CHRIST. STƲRMIƲS.

THE PREFACE TO THE READER.

SINCE those ingenious Mathematicians of this present Age, which is now drawing to a Conclusion, Vieta, Ougthred, Harriot, Cartes, Schooten, Beaune, Van Hudde, Heuraet, de Witte, and Slusius, and several other Famous Men coeval with them, have by their Endeavours improved the Algebra of the Ancients, raised it to vastly an higher pitch, and brought it from Numbers to universal Symbols, and not only found the excellent uses of it in Geometry themselves, but also communicated them to others; almost all Countries have furnished us with some excellent Persons, who treading in the footsteps of their Predecessors, have endeavoured to advance it yet further: And even our Times are not without those of the high­est rank, as Wallis, Baker, Renaldinus, Mengolius, Hug­geas, Malbranch, Leibnitz, Craan, and several others, who endeavour to promote this Science, deservedly reputed the very Apex of Humane Reason, and carry it more and more towards its utmost Perfection by dayly augmenting it with new and curious Inventions. But in the mean time, while these ingenious Men wholly busie themselves in promoting it, there are few found who condescend to explain the first Principles of it, and shew a ready way to young Beginners to arrive at the knowledge of those In­ventions. It is not long since a certain Friend of mine, who for some time has publickly and successfully taught these Sciences, [Page] complain'd to me by his Letters, of the want of a good Guide to Specious Analysis, whereby he might instill the Principles of that admirable Art into his Auditors. To whose Desires being not just then at leisure to satisfie, I immediately after projected this In­troduction, which at length you see finished, in the Form we now present you with it, wherein we have all along consulted to suit the Endeavours of young Beginners, as far as possible; as we thought our selves engaged by the Duties of our Professorship to do, & have compriz'd the Precepts of the Art in six or seven Pa­ges, and accommodated Examples of every kind to illustrate them. [...]herein if I have but indifferently accomplish'd my De­sign▪ I shall not think my Labour lost. We here add it to our Mathesis Enucleata, both as being properly a part of it, and more especially, because this Introduction presupposes the Reader to be acquainted with the first Principles of Specious Computation, which we have therein laid down. And so we commit our Endeavours to the Perusal and Censure of the Can­did Reader.

INTRODUCTION TO SPECIOUS ANALYSIS.

THE Analytick Art, or Specious Analysis, is solely subvervient to finding of Theorems, and resolving Problems, by leading us from certain Data or given Quantities, into the knowledge of unknown and sought ones, by a Chain of certain and infallible Consequen­ces: This admirable Artifice may be reduced to four Primary Heads, viz. Denomination, Reduction, Equation, and Effe­ction (if the Problem be a Geometrical one) or Construction.

I. DENOMINATION.

BY Denomination is understood a preparatory imposition of Names peculiar to each Quantity, whereby every one of the Quantities given or sought, are denoted by one or more peculiar Letters of the Alphabet at pleasure, but with this (arbitrarious) difference, that known or given quantities are mark'd by the former Letters of the Alphabet, a, b, c, &c. and the unknown or sought ones by the latter, z, y, x, &c. But although this imposition of Names, is, as we have said, altogether arbitrarious, yet there often happens not a little faci­lity to the Solution it self, by its being chosen as accommodate as possible to the conditions of the quantities given and sought; which any one w [...]ll learn better by Use than Precepts: As we find that both Theorems may be demonstrated, and Problems resolved e. g. by an extraordinary Compendium, if we denote any reason of two given Homogeneous quantities by a and e a, b and i b, d and o d, &c. (v z. by expressing the Names of the Reasons by e, and i, and o, &c.) and continued proporti­onality [Page 2] by a, ea, e (powerof2) a, e (powerof3) a, &c. and discontinued or discrete by b i b, c i c, d i d, or after the like manner, as we have done in our Math. Enucl. Lib. 1. Cap. 2, 3, 4, 7. and Lib. 2. Cap. 1. &c.

II. EQƲATION.

HAving thus given each quantity its Name, and making no further distinction between the quantities given and those sought, but treating them all promiscuously, and as already known, you must carefully search into and discuss all the Cir­cumstances of the Question, and making various Comparisons of the quantities, by adding, substracting, multiplying, and dividing them, &c. 'till at length, which is the chief aim and design of it, you can express one and the same quantity two ways, which is that we call an Equation: And you must find as many of these Equations, or Equalities of literal quantities, (as expressing the same thing) as there are several unknown quantities in the Question, independent on each other, and consequently denominated by so many different Letters, z, y, x, &c. But if so many Equations cannot be found, after hav­ing exhausted all the Circumstances of the Questions by one or two Equations; that is a sign the other unknown quanti­ties may be assumed at pleasure: Which the Examples we shall hereafter bring will more fully shew.

But as here also (as likewise in all this Art) Ingenuity and Use do more than Rules and Precepts; yet we will here shew the principal Fountains, for the sake of young Beginners, whence Equations, according to circumstances obvious in the Questi­on, are usually had. These are partly Axioms self evident, E. g.

That the whole is equal to all its parts taken together.

That those quantities which are equal to one third, are equal among themselves.

That the Products or Rectangles under the Parts or Segments, are equal to the Product of the whole.

Partly some universal Theorems that are certain and already demonstrated, as,

Three (α) continual Proportionals being proposed, the Rect­angle of the Extremes is equal to the Square of the mean.

(β) Four being proposed, whether in continued or disconti­nued Proportion, the Product or Rectangle of the Extremes is equal to that of the Means.

And several others such like, which we have demonstrated in Cap. 2, 3, and 4. Lib. 1. of our Mathesis Enucleat. partly in the last place, some particular Geometrical Theorems al­ready demonstrated, as e. g. that common Pythagorick one.

That in rightangled Triangles (γ) the Square of the Hypo­thenusa is equal to the two Squares of the sides.

That the Square of the Tangent of a (δ) Circle is equal to the Rectangle of the Secant and that Segment of it that falls without the Circle; the first whereof, we have demon­strated, Lib▪ 1. Math. Enuc. Def. 13. Schol. and also Prop. 34. Consect. 8. also Prop 44. after various ways; to which may be numbred Prop. 34. with Schol. 11. n. 3. Prop. 37. and following, Prop. 45. and 46. also the 48. and several others in Lib. 1. Math. Enucl. and likewise Lib. 2 Prop. 1, 2, 3, and se­veral following. And as for Examples both of Denominati­on, and Equations found after various ways, you may see them hereafter follow, and some we will here give you by way of Anticipation.

III. REDƲCTION.

An Equation thus found must be reduc'd, i. e. those two e­qual quantities, which for the most part are very much com­pounded of the quantities given and sought together, must be reduc'd to such a form, by adding or substracting something to or from each part, or multiplying and dividing by the same, &c. that the unknown or sought quantity alone, or its Square or Cube or Biquadrate, &c. may be found on the one side, and on the other the quantity express'd by meer given or known Letters, or affected with the unknown and sought ones; such are these Forms which follow, distinguish'd by their Names prefix'd to them,

A simple Equation, z= b, or y= [...].

A pure Quadratick, [...].

A pure Cubick, [...].

An affected Quadratick, [...].

An affected Cubick, [...] &c.

A Biquadratick, [...] &c.

To one of which, or some other like them, when yo [...] reduc'd your Equation first found, there are Rules in rea [...] whereby the Value of the unknown or sought quantity z or x, may be either expressed in Numbers, if the Quest [...] an Arithmetical one, or geometrically determin'd if it Geometrical one: Which is that we call Effection or Co [...] ction.

Thus therefore the whole, or at least the chief busine [...] Analyticks, is conversant in finding a convenient or fit Eq [...] on: For Reduction is very easie, and consisting only in Operations and mere Axioms, as e. g.

If to equal quantities you add or substract equal ones Aggregates or Remainders will be equal;

If equal quantities are multiplyed or divided by the s [...] the Products or Quotients thence arising will be equal, [...] [Page]

Pag. 5

Fig. I.

Fig. II.

Fig. III.

Fig. IV.

Fig. V.

VI

VII

VIII

IX

X

XI

XII

XIII

XIV

XV

XVI

XVII

IV. EFFECTION, or CONSTRƲCTION.

1. In simple Equations.

[...]Uppose z= b, the quantity b is sought.

If z be = [...].... or or x= [...].... or [...] or [...] make

  • as c to b so a to z,
  • as b to a so a to x,
  • as h−1 to b+ g so f to y,
  • as h+1 to b− g so f to z,

&c. every where according to n. 2.

Schol. 2. Prop. 34. Lib. 1. Math. Enuc.

If z be = [...], the Resolution of it into Proportionals [...] be more difficult, because neither of the Letters are found [...]ce in the Numerator. That therefore you may have e. g. twice, you must make as k to n so m to a fourth Proportio­ [...]l which call p; then will, by vertue of Prop. 18. Lib. 1. [...]= mn, and the proposed Equation be changed into this [...], to be now constructed from the 2 d. Case.

Or if [...], find a mean Proportional between k [...] l, which call p; and between m and n, which call q, [...]ording to n. 3. of the afore-cited Schol.; and the propo­sed Equation, by virtue of Prop. 17. will be in this form: [...]. Make therefore in the right-angled ▵ ( Fig. 1.) B= p and BC= q; and the □ AC by vertue of the Py­ [...]gorick Theorem, = pp+ qq: Which since it must be di­ [...]ed by r− s, make further, by Prop. 18 as r− s to the [...], so is [...] to y, according to the afore­ [...]ed.

[Page 6]4 In like manner if x be = [...], make 1st. as b to m so n to a fourth which call k; and so putting bk for mn, the Equation will be reduced to the second Case under this form: [...]

Or thus: Find a mean Proportional between b and g, which call p, and between m and n, which call q; and the proposed Equation will be in this Form: [...] Make therefore (in Fig. 2.) AB= p, and having on this described a Semi-circle, apply BC= q; then will, by vertue of Schol. 5. Prop. 34. □ AC= pp− qq: Which since it must be divided by c+ d, make farther, as c + d to [...] so [...] to x; all from the same Foundations, whence you have the Construction of the third Case.

5. If z be = [...]; make first as f to a, so a to a third Proportional m, and you'l have (putting fm for aa) [...], i. e. [...]. Make secondly as f to m, so b to a fourth n, and by putting fn for mb you'l have z= [...], i. e. [...], wherefore thirdly you'l have as g to n so c to z.

6. If y be = [...] make first as m to n, so l to a fourth, which call n, and by putting now mn for hl, you'l have [...] i. e. [...]= y. Therefore you'l now have secondly, as m to n, so l to y by Case 2: So that the Construction of the fifth and sixth Cases is nothing but reiterations of the Rule of three, ac­cording to what we have often inculcated, N. 2. and 3. Schol. 2. Prop. 34.

2. In simple Quadratick Equations.

1.
  • IF xx= ab
  • or y (powerof2)=1 c
  • or z (powerof2)=¾ dd

you'l have

  • [...]
  • [...]
  • [...]

that is to a mean propor­tional between

  • a and b
  • 1 and c
  • ¾ d and d

and so the Construction will be had from n. 3. Schol. 2. Prop. 34. (see Fig. 3.)

  • 2. If y (powerof2)= fg+ kl
  • or x (powerof2)= fg− kl

you'l have

  • [...]
  • [...]

make there­fore on the one side the Right-angled Triangle ABC ( Fig. 4.) whose side

AB is = to a mean Proportional between f and g,

BC is = to a mean Proportional between k and l;

On the other a Right-angled ▵ ( Fig. 5.) whose side AB is = to a mean Proportional between f and g, and the side BC = to a mean Proportional between k and l; and on the one hand the Hypothenusa, on the other the side

  • AC will be the va­lue of y
  • AC will be the va­lue of x

sought.

And all by vertue of the Pythag. Theor. and according to Schol. 5. Prop. 34. or the Consectarys of Prop. 44. See Fig 4. and 5.

3. If z (powerof2)= [...] i. e. [...] extracting the Roots on both sides, z will = [...], and so be the second case of simple Equations.

4. If y (powerof2) be = [...], make first as l to f, so g to a fourth which call n, and by putting ln for fg, you'l have y (powerof2)= [...] i. e. [...]. Make Secondly, as m to n, so h to a fourth, which call p; and by putting mp for nh, you'l have y (powerof2)= [Page 8] [...] i. e. pk, and so the first case of the present Equations.

5. If [...] in the first place the Rectangles fg and lm being turned into Squares, and collected into one Sum, make them = nn. Then (since cc and cd are multiplyed by qb+ bd) in like manner qb and bd added make pp; and you'l have [...]. Thirdly (since pp is already multiplyed by cc+ cd) having added cc+ cd into one Sum that they may e. g. make rr; x (powerof2) will = [...], and so be the third case of the present Equations.

3. In affected quadratick Equations.

1. IF zz be = az+ bb, then will [...]; which may be thus in short demonstrated a priori; Since z (powerof2)− az= bb per Hypoth. and that first quantity if it be added to ¼ aa, it becomes an exact Square, the root where­of is z−½ a; therefore [...], and consequently [...]; and lastly [...]; which last Root is a false one and less than nothing, but yet gives you the propo­sed Equation back again as well as the former; as will be evi­dent to any one who trys, viz. having transferr'd ½ a on the other side, and so the two equal quantities z−½ a and [...] being squared. For here will come out ¼ aa+ bb as well as if the radical Sign were affected with the Sign + be­cause − by − gives +. Therefore [...], and taking away on both sides ¼ aa, zz− az= bb i. e. zz= az+ bb.

The value therefore of this Root will be had geometrically, by making (in Fig. 6.) CD=½ a and DE= b, that the Hy­pothenusa CE may be [...]; and moreover, drawing out on both sides CD, and at the interval CE describing a Se­mi-circle [Page 9] AEB: This being done, AD will be the value sought of the true Root z, and DB of the false one.

2. If y (powerof2) be =− ay+ bb, then will [...]; which again may thus appear: Since y (powerof2)+ ay is = bb per Hy­poth. adding to both sides ¼ aa, the first quantity will be an exact Square, and [...]. Therefore the Roots will be also equal, viz. [...], and consequently [...]; which is a false Root.

The value of these Roots may be had geometrically, viz. of the true Root DB in Fig. 6. or BE in Fig. 7. and of the false one in the first AD, in the second AE.

3. If xx is = ax− bb, you'l have [...] or [...].

Which may be demonstrated after the same way a priori, as the former Cases, viz. Since x (powerof2)− ax is =− bb, adding on both sides ¼ aa, the former quantity will be an exact Square, viz. [...]. Therefore the Root of the one x−½ a = to the Root of the other, viz. [...], and adding on both sides ½ a, [...] which is one of the true Roots. Or [...] which in this case is al­so a true one. But the value of each may be obtain'd by making ( Fig. 8.) CB=½ a and by erecting BD perpendicu­larly = b, and making the Semi-circle BEA, and drawing DE parallel to CB, and letting fall the Perpendicular EF: For thus CF will be [...] and consequently AF, [...], and FB [...].

Or, with Cartes, making ( Fig. 9) CB=½ a and BD= b, drawing DF parallel to CB, that FD may be one root and ED the other; as is manifest from the precedent Construction, and its Rule. See also another Deduction from Cartes's Con­structions, Schol. 1 Prop. 47. Lib. 1. Math. Enucl.

NB. 1. The ingenious Schooten has before shew'd this Me­thod of demonstrating, and also of finding out these Rules in his Comment on the Geometry of Des Cartes, p. m. 163. and moreover deduces another ingenious Method for all the three Cases of these Equations, by taking away the second term in [Page 10] the Equation, p. 290, and the following, where we may make only this Remark concerning the third Case; that perhaps the Rule might be better deduc'd, if we make x=½ a− z rather than x= z−½ a.

NB. 2. If any one has a mind to see the new Constructions of affected quadratick Equations of the Abbot Catelan, he may find them in Acta Erud. Lips. Ann. 1682. p. 86. and in the 27th Journal des Scavans. 1 Dec. 1681.

IV. For Cubick and Biquadratick Equations both simple and affected, and also for all before mentioned, and consequently universally for all not exceeding the fourth Dimension.

THE value of the unknown Quantity or Root may for any Case be determined by one general Rule, found out by Mr. Thomas Baker an English-man, occasioned by what Des Cartes had taught concerning this matter, Lib. 3. Geom. p. 85, and the following; but now very much perfected by this Rule, and made more simple. Now that this Rule may be the better comprehended by Learners, we will premise these following things.

1. That all Equations occurring under those Forms which we have before shewn in the Article of Reduction, or the like, must always for this purpose be so changed as to have all the terms or parts of the Equation both known and unknown, af­fected and not affected, brought over to one side promiscu­ously, and so on the other there will stand o or nought, as e. g. let [...], or [...]; or [...]; or [...]; or [...] &c. which also was usual to Des Cartes in Lib. 3.

[Page 11]2. In all Equations the known quantity or Co-efficient of the second Term we will generally denote by the Letter p, that of the third Term by the Letter q, of the fourth by r, and the fifth (or absolute Number) by S; according to Cartes, but with some little alteration: So that hence the Equations we have before been treating of, and all others like them (every where denoting the unknown quantity by x) may all be reduced to these forms: [...], &c. &c.

3. These and the like Equations may either occur whole, or with all their Terms, as here, or depriv'd of one or more of them, as the following Examples will shew, where we will always put an Asterisk in the place of the deficient Term.

[...]

4. The unknown quantity in any Equation has N. This is an error, as has been prov'd by Dr. Wallis. as many diverse Roots or Values, as the Equation has Dimensions; which Des Cartes shews, Lib. 3. Geom. p. 69. at the same time evidently demonstrating this, viz. that some of those Roots may be false ones, i. e. less than nothing: Which from him we here suppose.

When therefore Des Cartes in his Construction of Cubick and biquadratick Equations, p. 85, and the following, requires as a necessary Condition, the ejection of the second Term in the given Equation, unless it were already wanting, and so was obliged to shew a way to eject it, with several other Prepa­rations; and afterwards, when by help of his Rule delivered p. 91. he had found a way of finding two mean Proportionals, and [Page 12] dividing any given Angle into three equal parts, then he use [...] it for solving other solid Problems, or finding two mean Pro­portionals, or trisecting an Angle. But the general Rule of Baker has no need of these methods or helps, neither of the Ejection of the second Term, nor any other Preparation, but immediately shews us a way, by the help of a Circle and Para­bola to find all the Roots of any given Equation, both true and false, whether the Equation want any term or not, and howsoever affected, after the way we will now, and perhaps a little more distinctly, shew.

1. It supposes with Cartes a Parabola NAM to be already described, (See Fig. 10. and 11.) whose Latus Rectum shall be L or 1, and its Axe ay; which Des Cartes only making use of, and never thinking of the other Diameters, was forced to take away the second Term of the Equation, &c. Baker therefore (strangely perfecting the Cartesian Geometry by this one thought) if the quantity p or second Term be in the E­quation applys to the Ax ay (or draws an ordinate to it) BA= [...] i. e. he erects at top of the Ax a on the right hand the perpendicular aE= [...] and from E draws EA y parallel to the Ax ay; whereby he obtains the Diameter A y sought.

2. Having made this Preparation, the whole business de­pends on this, to find the Center of the Circle to be described through the Parabola, which (by vertue of some arbitrary suppositions in the beginning) he always seeks on the left side of the Ax or Diameter, by help of two Lines [...] or b, and DH or d; viz. by placing the former upon the Ax from a to D, if p be wanting in the Equation, or upon the Diameter A y from A to D if p be there; and letting fall from the point D the latter to aD or AD perpendicularly towards the left hand.

3. He shews how to find the quantity of either of these Line (which is here very requisite) in any given Equation, by [...] certain general Rule (which he calls the Central Rule, because it alone helps to find the Center H) comprehended in these terms:

Directions for the Bookbinder.

The Eight half-sheet Plates that are to be folded in, and the single Leaf mark'd Page 89, are to be placed in those Pages of the Introduction to Specious Analysis which the figures at the top of them direct to.

A SYNOPSIS of Mr. Baker's CLAVIS, to be annexed to Page 13, of the Introduction to the Specious Analysis.
Of Aequations. Of Aequations. Of Central Rules.
Class. I. 1. [...] [...].
[...] [...] [...]
[...] [...] [...]
[...] [...] [...]
[...] [...] [...]
[...] [...] [...]
[...] [...] [...]
[...] [...] [...]
[...] [...] [...]
[...] [...] [...]
[...] [...] [...]
[...] [...] [...]
[...] [...] [...]
[...] [...] [...]
[...] [...] [...]
[...] [...] [...]
[...] [...] [...]
[...] [...] [...]
[...] [...] [...]
[...] [...] [...]
[...] [...] [...]

N B. The Sign ∽ denotes a dubious Case, viz. That either the Antecedent must be Substracted from the Consequent, or the Consequent from the Antecedent, according as the matter will bear.

[Page 13] [...]

This Rule as it stands here whole, only answers to those [...]ations wherein are all the Terms p, q, and r; and in the [...]n time may also be easily accommodated to all other Cases, [...] observing these things. 1. Whatever Term, or Quantities [...]r, be wanting in the proposed Equation, that must also be [...]ctively omitted, or put out of the general Central Rule, that [...] [...]emaining quantities may determine the special or particular [...]ral Rule. 2. As for what belongs to the Signs, viz. whether [...] ∽ (which latter Sign denotes a dubious Case, either that the [...] must be substracted from the latter, or contrary-wise, as [...] matter will bear) must be put in the Central Rule, he [...] (α) that in the Rule you'l always have [...], unless [...] in the proposed Equation p and r are affected with diverse [...]s: (β) By what Sign soever in the proposed Equation it [...]ens that the quantity q is marked with, it must be noted [...] the contrary one (altho' involv'd with other quantities) [...]e Rule; as may be seen in the application of the Rule to [...]ecial Cases done by the Author himself for the sake of Be­ [...]ers, and is exhibited in the Synopsis hereunto adjoyning, [...]h yet we have thought fit to give at the end of this Trea­ [...] much more contract as to the Central Rules, in a short [...]pendium by way of Appendix.

By these Rules therefore, the quantities of the Lines aD [...]D and DH will be so determined, that the parts in the [...] marked with the Sign + (taken either aggregately or [...]) will be put downwards from a to A towards y, and on [...]ft hand of D; but the negative Parts, or those affected [...] the Sign—, will be cut off, on the one part above, on [...]her on the right hand: Which being done the Center H [...]e found.

From the Center H thro' the Vertex of the Ax a (if the [...]D is found in the Ax) or in the other Case thro' the [...] of the Diameter A, you must draw a Circle which by [...]g or touching the Parabola will deermine the Roots [Page 14] sought, if the Equation be not a Biquadratick i. e. has not the quantity S; otherwise another Point L or Z must be found, ( vid. Fig. 12. and 13.) and a Circle described on the Radius HL or HZ, according to Des Cartes p. 86, and following, of his Geometry.

7. Viz. If you have−S, you must take on the Line H a or HA produced, on the one side AI=L or 1, and on the other AK = [...], and describing a Semi-circle on IK, draw AL perpendicular to AH, to obtain the point L. (see Fig. 12.) But if you have +S, then in another Semicircle described on AH, apply the Line AZ = to AL found, thereby to obtain Point Z, (see Fig. 13.)

8. A Circle therefore described from H through a or A, if S be wanting, but thro' L if there be−S, and thro' Z if +S, may touch or cut the Parabola either in 1, 2, 3 or 4 Points; from which if you let fall Perpendiculars to the Ax or Diame­ter, you will obtain all the Roots of your Equation both true and false.

9. And, 1. If in the Equation p be wanting and− r be there, the true Roots will be on the left side of the Ax, as NO, and the false ones as MO on the right side. 2. But if there be in the Equation p and− p, the true Roots will fall on the left side of the Diameter, and the false ones on the right; but if + p, on the contrary the true will be on the right hand and the false on the left.

10. But if the Circle neither touches nor cuts the Parabola in any point, it is a sign that the Equation is impossible, and has no Root either true or false, but only imaginary ones. All which, how they may be found out, and that they are undoub­tedly true, are demonstrated a posteriori, in an easie and plain way by the Author, wherefore we shall not give the Demon­strations of them here; but remit the Reader, after he has made a little progress in this Art, to the Author himself.

11. Wherefore now, (omitting also in this place the Do­ctrine of the Composition of the plain and solid Geometrical Loci, or Places, which would serve for a Complement of the Analytick Art) we will shew the Practice of these Rules al­ready delivered, premonishing only this from Mr. Baker, if the Latus Rectum be made Unity, that L in the Central Rules [Page 15] and all its Powers may be omitted, and so the Rules exhibited more compendiously, as we have already done in our Synop­sis, and may be seen from the form of a general Central Rule hereunto annexed.

[...]

To which Premonition of Baker we may also add this, if any given Line in the Problem it self be taken for Unity, which may be often very commodiously done [as a in the former Problem p. 91. Geom. Cartes, and the Line NO in the lat­ter, and a again in the Equation p. 83. the last line] and then the same Line also may be taken for the Latus Rectum of the Parabola to be described, if we have a mind to make use of this Compendium for abbreviating the Central Rules. For otherwise if we would construct all Problems, as Baker rightly asserts we may, by only one Parabola, we shall fall often into very tedious Prolixities.

SOME EXAMPLES OF SPECIOUS ANALYSIS, In each kind of Equations.

I. In Simple Equations.

PROBLEM I.

HAving the sum of any two sides given for forming a Triangle ABC, to find each of the sides, and form the Triangle.

Supppose e. g. three Lines given in Fig. 14. the first =AB+AC in the Triangle sought, the second =AB+BC, the third =BC+AC, to find each of the sides e. g. to find AB, which being known, the rest will be so also.

SOLƲTION.

1. Denomination. Make AC+AB= a; AB+BC= b; BC+AC= c; AB= x; then will AC= a− x, and BC= b− x, and so the Denomination be compleat.

2. Equation. Now if the values of the two last Lines BC and AC be added into one Sum, which we had before given; you'l have this Equation a+ b−2 x= c.

3. Reduction. By adding on both sides 2 x, you'l have [...]; and substracting from both sides [...]; and dividing both sides by [...].

[Page 17]4. The Effection or Geometrical Construction, which the Equation thus reduced will help us to

Join AE= a and ED= b in one Line AD, and from this backwards cut off DF= c; and divide AF which remains into two equal parts in B, and you'l have AB the first side of the Triangle to be formed; and BE will give the other side AC, which substracted from ED, will leave GD= to the third side BC; of which you may now form the Triangle ABC.

5. A general Rule for Arithmetical Cases. Add the two former Sums, and from the Aggregate substract the third Sum; half the Remainder will give the side AB common to the two former Sums. For an Example take this Question: There are three Towns of ancient Hetruria, viz. Forum Cassii (which the Letter A denotes in ▵ ABC) Sudertum (B) and Volsinii (C) which are at this distance one from another; if you go from Volsinii to Forum Cassii and thence to Sudertum, you mst go 330 Furlongs; from Forum Cassii to Sudertum and thence to Volsinii there are 306 Furlongs; lastly, from Sudertum to Volsinii and thence to Forum Cassii 272 Furlongs. How far is each Town distant from each other.

PROBLEM II.

IN a right-angled Triangle ABC, having given the Base AB, and the difference of the Perpendicular AC and the Hypothenusa BC to find the Perpendicular and Hypothenusa, and form the Triangle.

Make e. g. the Base AB ( Fig. 15.) and the difference of the Perpendicular and Hypothenusa BD, to find the Perpen­dicular AC; which being known, the Hypothenusa AC will be known also, if the given difference be added to the found Perpendicular.

SOLƲTION.

1. Denomination. Make AB= a, BD= b, AC= x; then will BC= x+ b.

2. Equation by the Pythagorick Theorem, [...], viz. the two Squares of the Sides to the Square of the Hypothenusa.

[Page 18]3. Reduction. Substracting from both sides xx, you'l have [...]; and moreover by substracting also [...]; and dividing by [...].

4. Effection or Geometrical Construction. Having described upon the given Base AB a Semi-circle, apply therein the given difference BD, and draw AD, whose Square is = aa− bb. Since this must be divided by 2 b, make, as AE=2 b to AD = [...], so AD = [...] to AC the Per­pendicular sought. To which if you add CF=BD, you will have AF= to the Hypothenusa sought BC; which will come of course together with the whole Triangle sought, if the found Perpendicular AC be erected at right Angles on the given Base AB.

5. The Rule for Arithmetical Cases. From the square of the given Base substract the square of the given difference, and divide the Remainder by the double difference; and you'l have the Perpendicular sought. E. g. suppose the Base = 20 foot, and the difference between the Perpendicular and Hypothenu­sa 10.

PROBLEM III.

IN the right angled Triangle ABC, having given the side AC and the sum of the other side AB and the Hypothenusa BC, to find the other side and the Hypothenusa separately, and form the Triangle. Suppose the given side (that is to be) AC (Fig. 16.) and the sum of the other sides AD, to find the side AB, which being known the Hypothenusa BC will be known also.

SOLƲTION.

1. Denomination. Make AC= a, AD= b, AB= x, then will BC= b− x.

2. Equation. [...] and substract­ing xx.

[Page 19]3. Reduction. [...]; and adding [...]; and substracting [...]; and dividing by [...].

4. The Effection or Construction is like the former, and so will be manifest only by inspecting that Scheme.

5. The Arithmetical Rule. From the square of the given sum substract the square of the given side, and divide the Re­mainder by double the given sum; and youll have the other side, and substracting that from the given sum, you have the Hypothenusa also. E. g. let one side be 15, and the sum of the other two 45.

PROBLEM IV.

HAving given the Perpendiculars and sum of the Bases of two right-angled Triangles having equal Hypothenuses, to find the Bases separately, and form the Triangles. Sup­pose e. g. to form the Triangle ABC (see Fig. 17.) you have given the Perpendicular AB, and for the other ▵ ADC, the Perpendicular CD, and the given sum of the Bases BE, to find the Bases singly, viz. the less for the greatest Perpendicular, and the greater AD for the less Perpendicular.

SOLƲTION.

1. Denomination. Make AB= a, CD= b, the sum BE= c; make the lesser Base BC= x; the greater AD will = c− x.

2. Equation. Since the Hypothenuses of the two Trian­gles are supposed equal, the two □ □AB+BC, i. e. xx+ aa will be = to the two □□AD+CD, i. e. [...].

3. Reduction. By taking away therefore xx and adding 2 cx, aa+2 cx will = bb+ cc; and further taking away from both sides [...]; and dividing both sides by [...].

[Page 20]4. Geometrical Construction. Join the Lines b and c CD and BE at right Angles, ( n. 2.) and the square of Hypothenusa DE will = bb+ cc. Upon this Hypothen [...] having described a Semi-circle, apply therein the Line AD, [...] the square of AE will = bb+ cc− aa. Which, since it m [...] be further divided by 2 c, make ( n. 3.) as BF=2 c to [...] = [...], so BG to BC, the lesser Base sought.

5. The Arithmetical Rule. From the sum of the squa [...] of the lesser Perpendicular and the sum of the Bases substract [...] square of the greater Perpendicular, and the Remainder di [...] ded by the double sum of the Bases, will give the lesser Ba [...] E. g. make AB 76, CD 57, and BE 114.

PROBLEM V.

HAving given the Perpendiculars of two right-angled T [...] angles standing on the same given Base, to find the Se [...] ments of the Hypothenuses. E. g. suppose the common give Base be AB ( Fig. 18.) and the Perpendicular of one Triang [...] AD of the other BC; to find the segments of the Hypothen [...] ses, cutting one another geometrically.

SOLƲTION.

If a Geometrical Solution be required there is no need [...] any Analysis; for having erected perpendicularly on the co [...] mon given Base AB the given Perpendiculars AD and B [...] the Hypothenuses AC, BD being drawn immediately, exhib [...] their Segments EA, EB, EC, ED. But if it be to be do arithmetically by a general Rule, then an Analysis will be n [...] cessary.

1. Denomination. Make the common Basis AB= a, B [...] = b, AD= c; and, having found AF, all the rest may had (for as AB to BC so AF to FE; which being given, y [...] have also GD and HC, and consequently also DE, CE, & [...] make AF= x, then will BF or HE= a− x.

2. The Equation from FE found twice. 1. As AB to BC so AF to FE. [...] [Page]

XVIII

XIX

XX

XXI

XXII

XXIII

XXIII

XXIV

XXV

XXVI

XXVII

XXVIII

As BA to AD so BF to FE. [...].

Therefore [...].

Reduction. Multiplying both sides by a you'l have bx, c− cx; and adding on both sides cx, bx+ cx= ac dividing both sides by [...].

The Arithmetical Rule. Multiply the commom base [...]e least Perpendicular, and divide the Product by the sum [...]e Perpendiculars; and you'l have the lesser segment of the [...] which being given you'l have all the rest. E. g. sup­ [...] AB=10, BC=9, AD=6.

PROBLEM VI.

O inscribe a Rhombus in a given Oblong, i. e. having the sides of the Oblong AB and BC given, (Fig. 19.) to find Segment BF or DE, which being cut off, the remainder FC [...]E will be the side of the Rhomb sought.

SOLƲTION.

Make AB= a, BC= b, BF= x: FC or FA will be = x (so far the Denomination.) Therefore the square of FA, [...]ch is bb−2 bx+ xx will be = aa+ xx, viz. to the two [...]res of AB and BF (so far the Equation;) and substracting both sides [...]; and [...], and [...]; and [...]. (so far the Reduction.)

The Geometrical Construction. Having described a semi­ [...]le on BC ( n 2.) apply CD or AB, and the □ BD will be bb− aa. Which since it must be divided by 2 b, make, [...]E=2 b to BD = [...], so BD to BF sought, and [...]e cut off the side of the Oblong BC ( n. 1.)

The Arithmetical Rule. From the square of the greater side substract the square of the lesser, and divide the remainder by double the greater side, and the quotient will give the Seg­ment BF sought. E. g. suppose AB=4, and BC=8.

PROBLEM VII.

TO inscribe the greatest square possible in a given Triangle, i. e. having given the heighth of the Triangle CD ( Fig. 20) and the Base AB, to find a portion of the altitude CE, which being cut off there shall remain ED=FG.

SOLƲTION.

Make the base AB= a; the altitude CD= b, CE= x; then will ED or FG= b− x.

By reason of the similitude of the Triangles ABC and FGC you'l have as AB to CD so FG to CE. [...]

Therefore the Rectangles of the means and extremes will be equal, i. e. ax= bb− bx; and adding on both sides bx, ax+ bx= bb, and dividing by [...].

Construction. Upon the side of the Triangle CB produced, make CH= b, and HI= a, so that the whole Line shall be a+ b. And having joined ID and parallel to it HE drawn from H, the part CE will be cut off, which is that sought.

For as CI to CD so CH to CE, [...] according to the second case of simple Effections.

Arithmetical Rule. Square the given heighth of the Tri­angle, and divide the Product by the sum of the base and al­titude; and the quote is the part to be cut off CE. E. g. sup­pose CD=10, and AB=15.

PROBLEM VIII.

IN an acute-angled Triangle having all the sides given, to find the Perpendicular that shall fall from the Vertex on the Base, i. e. having given AB, AC, BC (Fig. 21.) to find AD [Page 23] or BD (for having found the one you may easily find the o­ther) Coroll. Prop. 13. Lib. 2. Eucl.

SOLƲTION.

If there be only required a Geometrical Construction of this Problem, there will be no need of any Analysis; for having formed a Triangle ABC of the three given sides, you need only let fall the perpendicular AD from the Vertex A, which would determine the Segment BD. To find the general A­rithmetical Rule, which is the general Corollary of Euclid, or if any one for exercise sake had rather determine the Per­pendicular DA by the segment of the Base BD, than the latter by the former, the Analysis will proceed thus:

Make AB= a, BC= b, AC= c, BD= x; then will CD be = b− x. Wherefore by the Pythagorick Theorem □AD= aa− xx, and by the same reason the same □ AD = [...]. Therefore [...]; and by adding on both sides [...]; and by transferring [...], and dividing by [...].

The Arithmetical Rule. Substract the square of the lesser side from the Sum of the □□ of the base and greater side, and the remainder divided by double the base will give its greater segment: If the □ of the greatest side be substracted from the sum of the other squares, &c. you will have the less segment CD.

Geometrical Construction Having described a semi circle upon AB ( n. 2.) apply therein AC, and the □ BC will = aa− ccx, and continuing AC to B, till C B be = CB ( n. 1.) the □ of BB will = aa− cc+ bb; which since it is to be divided by 2 b make as BE=2 b to BB = [...], so BF= BB to BD the segment sought = BD n 1.

PROBLEM IX.

IN an obtuse angled Triangle having the thr [...]e sides given to find the Perpendicular let fall from the Vertex to the Base being continued: i. e. Having given AB, BC, AC (Fig. 22. n. 1.) to find AD or CD (for the one being found the other will readily be so also) Coroll Prop. 12. Lib. 2. Eucl.

SOLƲTION.

What we premonished about the former Problem, we un­derstand to be premonish'd here also. For the rest make here also AB= a, BC= b, AC= c, CD= x; then will BD= b+ x: Wherefore by the Pythagorick Theorem the □ AD will = cc− xx, and by the same Theorem the same □ AD = [...].

Therefore [...]; and adding to both sides [...]; and transposing cc and [...]; and dividing by [...].

The Arithmetical Rule. Substract from the square of the greater side the sum of the squares of the base and lesser side; and the Remainder divided by double the base will give its continuation to the Perpendicular.

Geometrical Construction from the Equation reduc'd: Having described a semi-circle upon AB ( n. 2.) apply therein AC, and the □ of CB drawn will [...]; which since it must be divided by 2 b make, as CF=2 b to CE [...], so CE to CD the segment sought, n 1.

PROBLEM X. Commonly ascribed to Archimedes.

THE Diameter AB of a given semi-circle (Fig. 22 n. 1.) being any how divided in L, and from L erecting a Per­pendicular LX, and upon the segments LA and LB having de­scribed two other semi-circles, whose semi-diameters are also gi­ven as well as that of the great Circle CB; to find the Radii FM and Vy of the little Circles that are to be so described, that they shall touch the Perpendicular LX, the Cavity of the greater semi-circle, and the Convexities of the less.

SOLƲTION.
I. For the Radius FM.

1. Denomination. Make CB= a, EB= b then will CE = a− b; for which for brevities sake put c. And let FM or FN or FK = x: Therefore EF will be = b+ x, and CF (sub­stracting FK from CK) = a− x. Wherefore now you'l have at least the names of the three sides in the ▵ CFE, so that according to Poblem 8. the Segment of the base GE may be determined (which indeed is determined already, as being =LE−LG or MF i. e. b− x) for which in the mean time we will put y; and now will CG= c− y.

2. For the Equation. If the □ GE= y be substracted from the □ EF = [...], you'l have the square of the Perpendicular FG = [...]; and, if □ CG = [...] be substracted from the □ CF = [...], you'l have the same □ of the Perpendicular FG = [...]. Therefore [...].

3. Reduction. And taking from both sides the quantities xx and yy, [...]; and adding 2 ax and cc, but taking away aa from both sides, [...]; and adding 2 ax and cc, and taking away from each side [...]; and dividing by [...]. but the same y or EG is = EL−MF i. e. b− x. There­fore [...]; which is a new and more principal Equation: And multiply­ing both sides by 2 c (you have a new Reduction) [...]; and adding 2 cx, and transpo­sing the others, [...]; and dividing by [...].

The Geometrical Construction of this first Case. Add the de­terminate ( n 2.) quantity 2 bc into one sum with the quantity aa, as in the beginning, n. 3. Then from this sum substract successively the quantities bb and cc, and there will come out (the same n 3) FH, whose □ [...]: Which since it must be divided by [...], make (the same n. 3) as FI = [...] to FH = [...], so FH to FM the Radius sought of the little Circle to be de­scribed. This quantity FM being thus found, place it from L to G ( n. 1.) and from G erect a Perpendicular, which be­ing cut off at the interval CF (which may be had, if from CB or CK you cut off FK=FM) or from E at the interval EF (which is composed of the Radii EN and FN) gives the Center of the little Circle to be described.

The Arithmetical Rule. Add twice the □ CEB to the square of the greatest semi-diameter CB, and from the sum substract the Aggregate of the □□ CE and EB; divide the remainder by the sum of all the three Diameters, (AB, AL and LB) i. e. by double the greatest AB; and you'l have the Radius FM, &c. For Example sake let a be = 12, b=4; c will be = 8, and x will be produced = 2 ⅔.

II. For the Radius Vy by help of the obtuse-angled ▵ DVC.

1. Denomination. CA= a as above, DA or DL= b, and putting x again for the sought V y or VK, CV will be = a − x, DL or DR= b, and consequently DV= b+ x, and DC= a− b, for which for brevity's sake we will put c. Now you'l have at least in Denomination in the ▵ CVD the three sides, so that aecording to Problem 9. the segment CW may be determined, for which in the mean while we will put y; then will DW= c+ y, which is the same as DL−WL or V y i. e. b− x.

2. For the Equation. If the □ CW= yy substract it from the □ CV = [...], and you'l have the □ of the Perpendicular VW, [...]; and if the □ DW = [...], substract it from the □ DV = [...], and you'l have the same □ of the Perpendicular VW = [...] Therefore [...].

[Page 27]3. Reduction. Therefore taking from both sides xx and yy, [...]; and adding 2 cy and 2 ax, [...]; and substract­ing [...], and dividing by 2 c, [...]

But if you add to the same y or CW DC= c, you'l have DW = [...] i. e. reducing this c to the same Denomination, [...] = DW. But the same DW =DL−WL= b− x. Therefore [...], and multiplying [...], and adding 2 cx, and transposing the rest, [...], and dividing by [...] just as above in the first Case.

4. The Geometrical Construction therefore will be the same as there. See Fig. 23. n. 4. and 5.

5. The Arithmetical Rule is also the same, but the given quantities in this Example, which the figure of the Problem will shew, thus vary, while a remains 12, b will be 8, and c 4, from which data (or given quantities) there will not­withstanding come out again, for x or the Radius V y 2 ⅔.

II. Some Examples of simple or pure Quadratick Equations.

PROBLEM I.

TO make a Square equal to a given Rectangle; i. e. having given the sides of the Rectangle, to find the side of an equal Square, Eucl. Prop. 14. Lib. 2. Suppose e. g. the gi­ven sides of the Oblong to be AB and BC ( Fig. 24.) to find the Line BD whose square shall be equal to that Rectangle.

SOLƲTION.

Make AB= a and BC= b, and the side of the square sought = x, and the Equation will be ab= xx; and extra­cting the root on both sides [...].

Geometrical Construction. Join AB and BC in one right line, and describing a semi-circle upon the whole AC, from the common juncture B erect the Perpendicular BD which will be the side of the square sought, according to Case 1. of the Effection of pure quadraticks.

Arithmetical Rule. Multiply the given sides of the Oblong by one another, and the square root extracted out of the Pro­duct will be the side of the square sought.

PROBLEM II.

THE square of the Hypothenusa in a right-angled ▵ being given, as also the difference of the other two squares to find the sides. E. g. If the Hypothenusa be BC ( Fig. 25.) and the difference of the squares of both the legs, and conse­quently its Leg also BE given (for the squares being given the sides are also given geometrically) to find the sides of the right-angled ▵ which shall have these conditions; or more plainly, to find one side e. g. the lesser which being found, the other, or the greater, will be found also.

SOLƲTION.

Let the □ of the given Hypothenusa = aa, and the square by which the two other differ = bb. Let the less side = x, and its □= xx. Wherefore the greater will be xx+ bb. And since the sum of these is = to the □ of the Hypothenusa, you'l have [...]; and substracting [...]; and dividing by [...]. Therefore [...].

Geometrical Construction. Having described a semi-circle on BC, and applyed therein BE, the □ EC will = aa− bb; and having described another semi circle upon EC divided into two Quadrants the □ DC will be [...], and so DC = [...] or the side sought; which being also transferr'd upon the other semi-circle describ'd on BC, viz. from C to A gives the other side AB and the whole ▵ sought.

The Arithmetical Rule. From the square of the Hypothe­nusa substract the given difference, and the square root extra­cted out of half the remainder gives the lesser side of the ▵ sought.

PROBLEM III.

HAving an equilateral ▵ ABC given (Fig. 26. n. 1.) to find the Center and Semi diameter of a Circle that shall circumscribe it. i. e. Find the BD the side of an Hexagon that may be inscribed in it. For if we consider the thing as already done; it will be manifest that the side of the Hexagon BD will fall perpendicularly on the side of the ▵ AB, as making an angle in a semi-circle, so having bisected the Hypothenusa DA you'l have E the Center sought.

SOLƲTION.

Make the side of the Triangle AB= a, BD= x, then will AD=2 x. Since therefore the square BD i. e. xx being sub­stracted out of the square AD i. e. 4 xx, there remains the square AB 3 xx, you'l have the Equation [...]; and dividing by 3 [...]; therefore [...]

The Geometrical Construction. Having produced AB ( n. 2.) to F a third part of it, the square of a mean proportional BD between BF and BA will be ⅓ aa or [...], and so the Line BD = [...]. Therefore the Hypothenusa DA being divided in two in E, or at the interval BD, making the intersection from B and A, you'l have the Centre sought.

The Arithmetical Rule. Divide the square of the given side into three equal parts, and the square Root of a third part will give the semi-diameter AE or BE sought, by the intersection of two of which you'l have the Centre.

PROBLEM IV.

HAving given, in a right-angled Parallelogram, the Dia­gonal, or for a right-angled ▵, the Hypothenusa and the proportion of the sides, to find the sides separately and construct the Parallelogram or ▵. Suppose e. g. the given Diagonal to be AB (Fig. 27. n 1.) and the given reason of the sides as AD to DE, to find the sides.

SOLƲTION.

Make AB= a, the reason of AD to DE as b to c; make the lesser side = x, then will the greater be [...].

For the Equation, the □□ of the sides are [...], □ AB; and multiplying both sides by bb, [...]; and dividing by bb + cc, [...]. Therefore [...] i. e. Extracting the roots as far as possible [...]

Another Solution.

Call the name of the given reason e, so that assuming any line for unity, the value of e may also be expressed by a right line, which shall be equal e. g. to DE above. Wherefore be­cause we make the less side x, the greater will be ex, and so [...] i. e. dividing by [...] i. e. [...]

The Geometrical Construction. The last Equation above being reduc'd to this proportion as the [...] to b so a to x, make ( n. 2) AD and DE at right angles, and AE will be = [...], and continuing AE and AD make as AE to AD so AB to AC the lesser side sought. Having therefore drawn BC which determines the lesser side AC, the greater side and so the ▵ ABC will be already formed, and may be easily compleated into a Rectangle. In the other Solution the last Equation agrees with the precedent (for it gives us this pro­portion as [...] to 1 so a to x in which 1 is = b, and ee = cc by what we have supposed) and so the Construction will be the same.

The Arithmetical Rule may be more commodiously expres­ed by this last Equation under the last form but one, after this way, divide the □ of the Diagonal by the □ of the name of the Reason lessen'd by unity, and the root extracted out of the Remainder is the lesser side sought.

PROBLEM V. (Which is in Pappus Alexandrinus, and in Cartes's Geometry, p. 83. in a Biquadratick affected E­quation, and p. 84. he gives us thereon a very remarkable Note.)

HAving given the Square AD (Fig. 28) and a right line BN, you are to produce the side AC to E, so that EF drawn from E towards B shall be equal to BN.

It will be evident, if you imagine a semi-circle to pass thro' the points B and E, that the most commodious way will be to find the line DG, that you may have the Diameter BG; upon which having afterwards described a semi circle, there will be need of no other operation to satisfie the question, than to produce the side AC 'till it occur to the prescrib'd Peri­phery.

SOLƲTION. (As found by Van Schooten, p. 316. in his Com­ment on Cartes's Geometry, which we will here give somewhat more distinct.)

1. Denomination. Make BD or DC= a, BN or FE= c, BF= y, and DG= x; the Perpendicular EH will be = a, and EG=BF, viz y (because the ▵ EHG is similar to ▵ BDF, by n. 3 Schol 2 Prop. 34. Lib. 1. Math [...]s. Enucl. and BD in the one = to EH in the other) and BG= a+ x, BE= y+ c; and BH will have its Denomination, if you m [...]ke (by reason of the similarity of the ▵ ▵ BFD and [...]EH) [Page 33] as BF to BD so BE to BH [...]. and you'l have also HG = [...], i. e. having reduc'd them all to the same Denomination, [...], i. e. [...]. Having therefore na­med all the lines you have occasion for, you must find two E­quations, because there are assumed two unknown quantities, viz. x and y.

2. For the first Equation and its Reduction. By reason of the similiarity of the ▵ ▵ BGE and BEH, as BG to GE so BE to EH [...]: Therefore the Rectan­gle of the Extremes will be = to the Rectangle of the means, i. e. [...]; and taking from both sides [...].

3. For the second Equation and its Reduction. Since BH, HE and HG [...] are continual proportionals, the Rectangle of the extreams are equal to the square of the mean, i. e. [...]; and multiplying both sides by yy, and dividing by [...], and taking away ayy and transposing the rest, [...]; and dividing by [...]; i. e. dividing actually as far as may be by [...].

4. The comparison of these two Equations thus reduc'd▪ gives a third new one, in which there will be only one un­known quantity, viz. [...]; and adding to both sides cy, [Page 34] [...]; and multiplying by [...]; i. e. [...] and dividing both sides by [...]; and adding [...]. Therefore [...].

5. The Geometrical Construction, which is the same Pap [...] prescribes in Cartes, viz. having prolong'd the side of [...] square BA to N, so that BN shall be = to a given right li [...] since BA is = a, and BN= c, the Hypothenusa DN will [...]. Having therefore made DG=DN, a [...] describ'd a semi-circle upon the whole line BG, if AC be pr [...] longed until it occur to the Periphery in E, you'l have do that which was requir'd.

PROBLEM VI. (Which Van Schooten has in his Comment, p. m. 150, and following.)

HAving given a right line AB, from the ends of it A an [...] B (Fig. 29.) to inflect two right lines AC and B [...] which shall contain an angle ACB = to the given one D, an [...] whose squares shall be in a given proportion to the Triang [...] ACB, viz. as 4 d to a.

Viz. You must determine the point C, which the two rig [...] lines AH and HC or EH and HC will do, assuming the mi [...] dle point E in the line AB. Wherefore here will be two u [...] known quantities HE and HC, and consequently two Equ [...] tions to be found in the Solution; one whereof the giv [...] proportion in the Question supplies us with, and the other [...] have from the similar Triangles AIC and GFD, which repr [...] sent equal angles.

SOLƲTION.

Denomination. Make AE, half AB= a, HE= x a [...] HC= y; therefore AH will be = a− x and HB= a+ [...] whence the Denomination of the squares AC and BC is ea [...] had; viz. the one [...], and the other, a [...] [Page]

XXIX

XXX

XXXI

XXXII

XXXIII

XXXIV

XXXV

XXXVI

XXXVII

[Page 35] xx+ yy, so that the sum of the squares is 2 aa+2 xx+

And the ▵ ACB will be = ay: And since the ▵ ▵ [...] and AIC are similar, and the sides of the former FD and [...]rbitrary, so that for FD we may put b and for FC, c; [...]he sides of the latter are determined by the similitude of ▵▵ ABI and HDB, as being right-angled ones, and ha­ [...] the common angle B; they will be obtain'd by making the Hypothen. BC to the Hypoth. AB so the base HB to [...]ase BI, i. e.

[...], whence substracting BC= e, there remains CI = [...].

For the first Equation, by virtue of the Problem as [...] to ay. And the Rectangle of the [...]mes is = to the Rectangle of the means, i. e. [...].

For the other Equation, since as DF to FG so CI to AI [...] the Rectangle of the extreams will again be = to the Re­ [...]gle of the means, i. e. [...]; multiplying both sides by [...]; [...]h is the second Equation.

The Reduction of both Equations. [...] first was [...]. [...]refore dividing by [...]. substracting [...]. [...]he latter Equation was [...], i. e. [...]ituting again the value ee, which was [...]; and by transposition, [...]; and dividing by [...], or [...], [Page 36] or (putting 2 f for [...]) [...].

Wherefore we have the value of xx twice expressed, but by quantities partly unknown, because y is found on both sides. Wherefore now we must make a new comparison of their values, whence you'l have this new

5. Third Equation, in which there is only one of the un­known quantities: [...]; and adding on both sides both yy and aa, [...]; or dividing by 2, [...]; and transposing [...]; and dividing by [...]; which is the value of the quantity y in known terms.

But this value in one of the precedent Equations, viz. in this [...], being substituted for y and its square for yy, will give [...]; i. e. all being reduced to the same denomination, [...]; and [...].

6. The Geometrical Construction, which Schooten gives us p. 153. Having made the angle KAB ( n. 2. Fig. 29.) e­qual to the given one D, erect from A, AL perpendicular to KA, meeting the Perpendicular EM in L; and from the Centre L, at the interval of the given right line d, describe a Circle that shall cut KA and EL produced to K and M. Then assuming EN=KA, join MA, and from N draw NH pa­rallel to it, which shall meet AB in H. Afterwards, having described from L, at the interval LA, the segment of a Circle ACB, draw from H, HC perpendicular to AB meeting the circumference in C, and join AC, CB.

NB. The reason of this elegant Construction, which the Author conceal'd, for the sake of Learners we will here shew. 1. Therefore, he reduc'd the last Equation (extracting the root, as well as it could bear, both of Numerator and Deno­minator) to this: [...] multipl. by [...], so that after this way the Construction would be reduc'd to this proportion, as d+ f to a so [...]. 2. He made the angle KAE = to the given one D, and the angle KAL a right one, so that having described the segment of a Circle from L the inscribed angle will also be made equal to the gi­ven one, according to the 33. Lib. 3. Eucl. 3. By doing this, EL expresses the quantity f, since by reason of the simi­larity of the ▵ ▵ KOA s. GFD, n. 1. and AEL (for the angles LAE and AKO are equal, because each makes a right one with the same third Angle KAO) you have as KO to OA so AE to EL i. e. [...]

4. Making now LM and LK = d you had EM= d+ f, and AK = [...] (for the □ AL is = to aa+ ff, which being substracted from □ LK = [...].)

5. Wherefore there now remains nothing to construct the last Equation above, but to make EN=AK, and to draw HN parallel to AM; for thus was the whole proportion as EM to EA so EN to EH [...] to x. Q. e. f.

For the point H being determined, a perpendicular HC thence erected in the segment already described defines the Point C, which answers the Question.

PROBLEM VII.

HAving given the four sides of a Quadrangle to be inscribed in a Circle, to find the Diagonals and their Segments, and so to construct the Quadrangle, and inscricbe it in the Circle. As e. g. suppose the given sides are AB, BC, CD, DA ( Fig. 30 n. 1.) which now we suppose to be joined in [...]o a quadr [...]n­gle [Page 38] inscrib'd in the Circle, the Diagonals also AC and BD being drawn ( n. 3.) to find first the segments of the diagonals A e, B e, &c. which being had, the Construction is ready.

SOLƲTION.

Denomination. Make AB= a, AC= b, CD= c, DA = d, A e= x [for this segment alone being found, the rest will be found also, as will be evident from the process.] Since therefore the vertical angles at e are equal, and likewise the angles in the same segment BCA, BDA, also DAC, DBC, &c. are equal, the Triangles A eD and B eC, also A eB and C eD are similar: wherefore it will follow that,

  • 1. As DA to A e so CB to B e [...]
  • 2. As AB to B e so CD to C e [...]
  • 3. As AB to A e so CD to D e [...]

Therefore the whole Diagonal AC will be = [...] and BD = [...].

2. The Equation. But now by Prop. 48. Lib. 1. Math. Enucl. the Rectangle of the Diagonals is equal to the two Rectangles of the opposite sides.

Diag. AC, [...]

Diag. BD, [...]

Therefore the □ of the Diagonals [...].

3. Reduction, i. e. taking (a) for unity [...]; i. e. the quantities on the left hand being reduc'd to the same denomination.

[...]; and multiplying both sides by dd, [...]; and dividing both sides by d, [...]; and then dividing both sides by [...]; i. e. in the present case, where b by chance happens to be = a, [...]

Therefore [...] or in our case [...]

4. The Geometrical Construction, which, by supposing a (and in the present case also b) to be unity, ought to deter­mine, [Page 40] 1. The quantities cd, [...], cc, and their aggregate with unity. 2. The aggregate of cd and dd. 3. To divide the one by the other. And, 4. To extract the root out of the quotient, or also to extract the roots first out of each quantity, and divide them by one another; which may all of them be separately done in so many separate Diagrams, but more ele­gantly connected together after the following or some such like way. 1. Join AD and DC ( n. 2.) into one line, and having described a semi-circle thereupon, erect the Perpendicular DE; and the line AE drawn will = [...]. 2. Making the angle CAG at pleasure, make AF=AB, and draw CG pa­rallel to the line DF; so FG will be = [...]. Now if, 3. in the vertical angle you make AH=CD, the line HI drawn parallel to DF will cut off AI= cd. 4. In AK erected = to AB, if you take AL=AH or CD, and draw LM paral­lel to KH, you'l have AM= cc. 5. Having prolonged AG to N and AH to O, so that GN shall be =AI+AM and AO=AB or AK, and having described a semi-circle upon the whole line NO, a perpendicular erected AP will be = [...] and so, 6. if AQ be made =AE and AR=AF or AB, and you draw a line RS from R pa­rallel to PQ; AS will be = x, i. e. the segment sought A e of the Diagonal AC; which being given, by force of the first Inference premis'd in the Denomination above, by drawing DS and, having made DT=BC, TV parallel to it; you'l have also the other segment B e=SV and by their Intersection on the line AB ( n. 3.) the point e, thro' which the Diagonals must be drawn which will be terminated by the other given sides, and thence you'l have the quadrilateral figure ABCD sought, to be circumscribed about the Circle, according to Con­sect. 6 Defin. 8. Mathes. Enucl.

NB. Unless we had here consulted the Learner's ease, the artifice of this Construction might be proposed after a more short and occult way, thus: Make DE a mean proportional between AD and DC, and draw AE. Then having made any angle CAG, make AF=AB, and at this Interval de­scribe [Page 41] the circle FROK, and draw CG parallel to DF. More­over in the opposite vertical angle, having made AH=CD, draw HI parallel to DF, and having erected the perpendicular AK, and thence the abscissa AL=AH, make LM parallel to HK, and thence having prolonged AH to O, and GN being made equal to AI+AM, make AP a meam proportional be­tween AO and AN, cutting the hidden circle in R; and last­ly having made AQ=AE, if RS be drawn parallel to QP, you'l have AS the value of x sought, &c.

III. Some Examples of Affected Quadratick Equations.

PROBLEM I.

HAving given, to make a right angled Triangle ABC, the differences of the lesser and greater side, and of the greater, and the Hypothenusa, to find the sides separately and form the Triangle. E. g. Having given the right line DB ( Fig. 31.) for the difference of the perpendicular and base, and CE for the difference of the base and Hypothenusa, to find the perpendicular AC, which being found, you'l have al­so, by what we have supposed, the base AB, and the hypothenu­sa BC.

SOLƲTION.

Make the difference DB= a, CE= b; put x for the perpendicular; the base, which is greater than that will be x+ a and the Hypothenusa x+ a+ b. Therefore by vertue of the Pythagorick Theorem,

[...]; and substracting from both sides [...].

Wherefore by the first case of affected quadratick Equati­ons [...].

Construction. Find a mean proportional AK between AH=2 b and AI= a (n. 2.) ( Fig. 31.) and having made both AF and AG= b, place the Hypothenusa KF from AL, and [Page 42] cut off GC equal to the hypothenusa GL; thus you'l have AC the perpendicular of the Triangle sought, and adding DB you'l also have the base AB, and from thence having drawn the hy­pothenusa BC, it Will be found to differ by the excess required CE.

The Arithmetical Rule. Join twice the product of the differences multiplyed by one another, to twice the square of the difference of the base and the hypothenusa; and if the square root of this sum being extracted be added to the afore­said difference, you'l have the perpendicular sought. Suppose e. g. both the differences of CE and DB=10.

PROBLEM II.

IN a right-angled ▵ having given the Hypothenusa and sum of the sides, to find the sides. E. g. If the Hypothenusa BC be given ( Fig. 32.) and the sum of the sides CAB, to find the sides AB and AC separately, to form the Triangle.

SOLƲTION.

Make the Hypothenusa BC= a, the sum of the sides = b. Make one side e. g. AB= x, then will the other side AC be = b− x. Therefore

[...]; and adding 2 bx, and taking a­way [...]; and dividing by [...].

Therefore according to Case 1. of affected Quadraticks. [...] i. e. [...] or [...].

The Geometrical Construction. Having described a semi-circle upon BD=BC so a apply therein the equal lines BE and DE, and having described another semi-circle on BE apply therein BF=½ b, to be prolonged farther out. Lastly, if another little semi circle be described at the [...]nterval EF, the whole line AB [Page 43] will be the true root or the side sought, and GB the false root, &c.

The Arithmetical Rule. From half the square of the Hy­pothenusa substract the fourth part of the square of the given sum, and the root extracted out of the remainder, if it be ad­ded to half the sum, will give one side of the Triangle; and substracted from the given sum, will give also the other Suppose e. g. BC to be 20, and the sum of the sides 28.

PROBLEM III.

HAving given again in the same ▵ the Hypothenusa, as above, and the difference of the sides DB (Fig. 33.) to find the sides.

SOLƲTION.

Make the less side x, the difference of the sides = b; the greater side will be x+ b. Let the Hypothenusa be = a. Therefore, [...]; and taking away [...]; and dividing by [...].

Therefore by case 2, [...] i. e. [...].

The Geometrical Construction. Having described a semi-circle upon BD=BC or a, apply therein the equal lines BC and DC, and having described another semi-circle on DC apply in it DF =½ b, and if at the same interval you cut off FA from FC, the remainder AC will be the lesser side sought, &c.

The Arithmetical Rule. From half the square of the Hy­pothenusa substract the square of half the difference, and if you take half the difference from the root extracted out of the re­mainder, you'l have the lesser side of the Triangle required, and by adding to it the given difference you'l have also the greater. E. g. Let the Hypothenusa be 20, and the difference of the sides 4.

PROBLEM IV.

HAving given the Area of a right-angled Parallelogram, and the difference of the sides to find the sides. E. g. If the Area is = to the square of the given line DF, and the difference of the sides ED ( Fig. 34.) to find the sides of the rectangle.

SOLƲTION.

Make the given Area = aa, the difference of the sides = b, the lesser side x; then the greater will be x+ b. Therefore the Area xx+ bx= aa; and substracting bx xx=− bx+ aa.

Therefore according to case 2, [...].

The Geometrical Construction. Join at right angles AG= a, and GH=½ b, and having drawn AH and prolong'd it, describe the little Circle at the interval GH: so you'l have AE the lesser side, and AD the greater of the Rectangle sought, &c.

The Arithmetical Rule. Add the given Area and the square of half the difference, and having the sum, substract and add the difference from or to the root extracted, and so you'l have the greater and less sides of the rectangle.

PROBLEM V.

HAving given for a right-angled Triangle the difference of both the Legs from the Hypothenusa, to find the sides and so the whole Triangle. E. g. Suppose the difference of the less side to be BD and of the greater DE ( n. 1. Fig. 35.) to find the sides themselves, and so make the Triangle.

SOLƲTION.

For BD put a, for DE, b. Let the greater side be x; the Hypothenusa will be x+ b; therefore the lesser side will be x+ b− a. Now the □□ of the sides are = to the □ of the Hypothenusa, i. e. 2 xx−2 ax+2 bx+ bb−2 ab+ aa [Page 45] = xx+2 bx+ bb; and taking away aa, [...]; and adding 2 ax and 2 ab, and taking away [...]. Therefore [...] i. e. [...].

The Geometrical Construction. Between the given diffe­rences BD and DE▪ find ( n 2.) a mean proportional DF▪ and join to it at right angles the equal line FG, and cut off DH equal to DG; and so you'l have BH the greater side of the triangle sought. This being prolong'd to C, so that HC shall be = b, having described a semi-circle upon the whole line BC apply therein BA=BH; and having drawn AC, the Triangle sought ABC, will be formed.

The Arithmetical Rule. If the square root extracted from the double rectangle of the differences be added to the greater difference, you'l have the greater side sought, &c.

PROBLEM VI.

HAving given, to make two unequal Rectangles, but of equal heighth, the sum of their Bases with the Area of [...]her (viz. the greater,) and the proportion of the sides of the [...]her (viz the least,) to find the sides separately. E. g. Let the sum of the bases be AB ( n. 1. Fig. 36) and the square of the line BC= to the Area of the greater rectangle; and let the sides of the lesser rectangle be to one another as CD to DE: To find the sides of both the rectangles; i. e. to find the com­mon altitude, which being found the other sides will be easily obtain'd from the Data; or to find the base of the greater which, with the same ease, will discover the rest.

SOLƲTION.

Make AB= a, and the Area of the greater rectangle = bb; [...]d the proportion of the altitude to the base in the lesser, as c [...]d; to find e. g. the greater base which call x. Therefore [...] common altitude will be = [...], and the base of the lesser [...]ctangle= a− x.

Wherefore you'l have for the Equation, as c to d so [...] to a− x.

Therefore ac− cx= [...]; and multipl. by x, acx− cxx= bbd; and adding cxx and taking away bbd, acx− bbd= cxx. Now that you may conveniently divide both sides by c, make first as c to b so d to a fourth which call f, and then put cf for bd, and you'l have [...]; and dividing by [...]; and so according to case 3. [...] or [...].

The Geometrical Construction. Find first the quantity f [...] (num. 2. Fig. 36.) according to the following proportion, as c to b so d to f; and a mean proportional between b and f will be = [...]. Then having at the interval ½ a described a semi circle ( n. 3.) upon the given line AB, and erected BD= [...], and having made EF= to it, CF will be [...] ▪ To which AC being added will give x for one value and FB for the other. And for the common altitude, which we called [...], make as x to b, so b to a fourth, i. e. as AF to FH so FH to FG; which will be the altitude of both rectangles A g and B g which may now easily be constructed.

The Arithmetical Rule might easily be had from this Equa­tion reduced; but you may have it more commodiously from this other

SOLƲTION.

Let the Denomination remain the same as above, only her [...] put x for the common altitude, and express the reason of th [...] lesser base of the rectangle to this altitude by e, and that bas [...] will be = ex: Therefore the base of the greater Rectangl [...] will be = a− ex. Having now multiplyed the com [...]mon altitude by each base, the area of the greater rectangl [...] will be ax− exx, and hence you'l have the Equation

ax− exx= bb; and adding exx, and taking away bb, ax− bb= exx; and dividing by [...]. Therefore by case 3.

[...].

Wherefore now this will be the Arithmetical Rule. If from the fourth part of the square of the sum of the bases divided by the □ of the name of the reason you substract the given area divided by the same name of the reason, and if the root extracted [...]ut of the remainder be added to or substracted from half the sum of the bases divided by the same name of the reason; this sum or [...]emainder will give the altitude of the given Rectangles, and [...]hat multiplyed by the name of the reason one of the bases: And that being substracted from the given sum of the bases [...]ill give the other base. For Example, let the sum of the ba­ [...] be 16, the area of one of the rectangles 30, the [...]ame of the reason which the common altitude has to the base [...]f the other rectangle = 2. There will come out the com­ [...]on altitude, on the one side 5, on the other 3, &c.

PROBLEM VII.

HAving given the Perpendicular of a right-angled Trian­gle let fall from the right angle, and its Base, to find the [...]ments of the Base, and so to form the Triangle.

E g. If the base of the right-angled Triangle you are to [...]m be AB ( Fig. 37.) and the length of the perpendicular [...] or BF; to find the segments of the base, and so the point [...], from which you are to make the perpendicular CD, to form [...]e Tiangle ABG.

SOLƲTION.

Let the given base be = a; and the given perpendicular [...]b: Then will one of the segments of the base be = x; [...]d the other = a− x, and b a mean proportional between [...]e said segments, i. e.

x to b as b to a− x;

[...]herefore ax− xx= bb; and by adding xx and taking a­ [...]y bb, [Page 48] ax− bb= xx. Therefore by case 3. [...].

The Geometrical Construction. Having described a semi-circle upon the given line AB, if you erect the perpendicular BE, and from the point G (which is determined by EG pa­rallel to AB) let fall GD equal to it, you will have the two segments sought, AD = [...] and DB = [...], which Construction, it cannot be denyed, but it may be evident to any attentive person even without the A­nalysis.

But that case may by the by be taken notice of wherein the given perpendicular would not be BE but BF. For in this case the perpendicular BF being erected upon AB, the paral­lel FG would not cut the semi-circle; which is an infallible sign that the Problem in this case is impossible, where the perpendicular is supposed to be greater than half the base; which is inconsistent with a right angle.

The Arithmetical Rule. From the square of half the base take the square of the given perpendicular, and add or sub­stract the square root extracted out of the remainder, to or from half the base; and on the one hand the sum will give the greater segment, and on the other the difference will give the less.

PROBLEM VIII.

HAving given the perpendicular of a right-angled Triangle that is to be let fall from the right angle, and the diffe­rence of the segments of the base, to find the segments, and de­scribe the Triangle.

E. g. If the perpendicular is, as above, BE, and the diffe­rence of the segments AH ( n. 1. Fig. 38.) to find the seg­ments AD and DB, from whose common term you are to e­rect a perpendicular DG or DC to form the Triangle.

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SOLƲTION.

Make the lesser segment = x, and the difference of the seg­ [...]ts = a, the greater segment will be x+ a Make the gi­ [...] perpendicular as before = b: Therefore you'l have as x+ a to b so b to x; and consequently, xx+ ax= bb; and substracting ax, xx= bb− ax. Wherefore according to case 2. [...].

The Geometrical Construction. Make HD=½ a, DG equal perpendicular to b; HG will be = [...] = HB or [...], viz having drawn a semi-circle from H thro' G. [...]erefore DB is the less segment, and AD the greater; and [...]ing drawn AG and BG, or on the other side, (making the [...]pendicular DC=DG) having drawn AC and BC, the [...]iangle will be constructed. Or with Cartes, make ( n. 2.) [...]=½ a and EB= b, and having described a Circle from [...]hro' E draw BHA; and so you'l have the two segments [...]ght AB the greater and DB the lesser.

The Arithmetical Rule. Join the squares of the half diffe­ [...]ce and perpendicular into one sum, and then having extra­ [...] the root substract half the difference from it; and the re­ [...]nder will be the lesser segment sought; and having added [...] difference you'l have also the greater.

PROBLEM IX.

HAving given for a rigbt-angled Triangle one segment of the base and the side adjacent to the other segment, to find rest and construct the Triangle.

As if the lesser segment of the base DB be given ( Fig. 39.1.) and the side AC adjacent to the other segment; to find greater segment of the base, which being found the rest easily obtain'd, and consequently the whole Triangle.

SOLƲTION.

Make the greater segment = b, the given side = c, the segment sought = x. Now if we suppose the triangle ABC to be already found, it is evident, 1. If from the square of AC you substract the □ AD, you'l have the □CD= cc− xx. 2. The same □ CD may also be otherwise hence obtain'd, be­cause, the angle at C being a right one, CD is a mean pro­portional between BD and DA, i. e. between b and x; whence the rectangle of the extremes bx is = □ of the mean CD. Wherefore now it follows, 3. that cc− xx= bx; and adding xx, cc= bx+ xx; and substracting bx,− bx+ cc= xx. Therefore according to case 2. x=−½ b+√¼ bb+ cc.

Geometrical Construction. Join EF=½ b (n. 2. Fig. 39.) and FA= c at right angles, and having described a Circle from E thro' F draw AEB; so you'l have DA the greater segment and DB the less; having erected therefore a perpendicular from D, and described a semi-circle upon AB, you'l have C the vertex of the triangle sought, whence you are to draw the sides AC and BC.

The Arithmetical Rule. Join the □ of half the given seg­ment, and the □ of the given side into one sum; and having extracted the root of it, if you thence take half the given seg­ment, you'l have the segment sought.

PROBLEM X.

HAving given in an oblique angled Triangle the perpendi­cular height, and the difference of the segments of the base, and the difference of the other sides, to find the sides and form the triangle.

As, if the altitude CD be given ( n. 1. Fig. 40.) and also the difference of the segments of the base EB, and the diffe­rence of the sides FB (as is evident from the triangle ABC ( n. 2.) conceived to be so formed beforehand) and you are to determine the base it self and both sides, &c.

SOLƲTION.

Make the given perpendicular CD= a (see n. 2. Fig. 40.) EB= b, FB= c: For the lesser segment of the base AD put x, and the greater will be x+ b. It is now evident that you may obtain the □ CB by the addition of the □ □ DC and BD, [...], and the □ AC by the addition of the □ □ AD and DC, viz. aa+ xx: So that the side AC will be = [...], and the side BC = [...]. But since also this same side BC may be obtain'd by adding the difference c to the side AC, so that it shall be = [...]: you'l have this Equation, [...]; and squaring both sides, [...]; and substracting from both sides [...]; and again squaring it, [...]; and substracting from both sides 4 ccxx (because c is less than b) and transposing the rest, [...] and dividing by [...] i. e. dividing the affected quantities by 4 both above and un­derneath, [...] and actually dividing the former part by [...].

Therefore according to the second case, [...]; or reducing ¼ bb to the same denomination with the rest [...]; and leaving out those quantities that destroy one another [...].

The Arithmetical Rule. Multiply four times the square of FB by the square of the perpendicular CD, and add to it the product of the square of EB into the □ FB, and from the sum substract the biquadrate of FB; and the remainder will be the first thing found. Then substract four times the square of FB from four times the square of EB; and the remainder will be the second thing found. Lastly, divide the first thing found by the second, and from the quotient take half after having extracted the root: Thus you'l have the lesser segment of the base sought, &c. E. g. In numbers you may put 2 for FB, 4 for EB, 12 for CD.

As for the Geometrical Construction, the quantity of the last Equation contain'd under the radical sign will help us to this proportion,

as [...] so cc to a fourth; or divi­ding all by 4,

as [...] so ¼ cc to a fourth, which is ¼ of the quantity under the radical sign. Assuming there­fore the quantity c for unity, make ( n. 3.) IK= c IN and KL= b; NO will be = bb, and substracting OP= cc (i. e. to unity) there will remain NP= bb− cc. In like manner IS and KM= a, ST will = aa; to which if you add SX=¼NO, and take thence XV=¼ unity; TV will be = [...]. Wherefore if you make NR equal to this TV, and PQ=¼ of unity or cc, since NP is = bb− cc; by the rule of proportion there will come out DR¼ of that quantity, which is under the radical sign. Therefore this being taken four times will give DZ for the whole quantity; to which if [Page 53] you join D y= to unity, and, having described a semi-circle upon the whole line YZ, erect the perpendicular DE; this will be the root of the said quantity, and taking hence more­over EF=½ b, you'l have DE or DA the less segment of the base sought. Therefore adding GB= b to DG, DB will be the greater segment, and, having let fall the perpendicular DC= a, BC and AC will be the sides sought. Q. E. F.

IV. Some Examples of Affected Biquadratick Equa­tions, but like Affected Quadratick ones.

PROBLEM I.

TO find a square ABCD (such as in the mean while we'll suppose n. 1. to be in Fig. 41.) from which having taken away another square AEFG, which shall be half the former, [...]here will be left the Rectangle GC whose Area is given. E. g. Suppose the given area equal to the square of the given line LM, to find the true sides of the squares AB and AE, answer­ [...]ng to these supposed ones, n. 1.

SOLƲTION.

Make the area of the rectangle that is to remain = bb, and GB= x; BC or AB will be = [...], and substracting hence GB, the remaining side of the lesser square AG= [...]− x, [...]. [...]. Since therefore the square of this is supposed [...]o be half of the square of AB, this will be the Equation: [...]; [...]nd multiplying by xx, [...] [...]nd multiplying by 2, [...]; [Page 54] and substracting 2 b (powerof4), and adding [...]; and dividing by 2, [...].

NB. The same Equation may be obtain'd, if, putting x for GB or FH, and having found the □ of AG or GF as a­bove, you infer [...].

This last Equation, tho' it be a biquadratick, yet may be rightly esteem'd only a quadratick one, because there is neither x (powerof3) nor single x in it, and so you may substitute this for it, [...], viz. by supposing y= xx. Whence according to the third case of affected quadraticks, y will = [...] i. e. [...] or = [...].

Therefore [...].

Geometrical Construction. Now if the given line b be assu­med for unity, bb and b (powerof4) will be = = to the same line Therefore, if between LM as unity, and MN=½ b viz [...] you find a mean proportional MO ( n. 2. Fig. 41.) that wil [...] be = [...], which being substracted from LM, and added to it, will give the two values of the quantity y. Moreover there­fore by extracting its roots, i. e. by finding other mean pro­portionals LR and LS between the quantities found LP and LQ and unity ( n. 3.) they will be the two values of the quantity x sought; the first whereof LR will satisfie the que­stion, and the other LS be impossible. Wherefore to form [Page 55] the square it self, since its side will be = [...]; by making ( n 4.) as x to b so b to a fourth, it will be obtain'd: And this may be further prov'd, if finding a mean proportional BK between BI=LR and the side of the □ BC, it be equal to the given quantity LM.

Arithmetical Rule. From the given area or the square of the given line LM substract the root of half the biquadrate of the same line; thus you will have the value of the □ FC, viz. xx: Therefore extracting further the square root of [...]is, it will be the value of x sought.

PROBLEM II.

TO find another square ABCD (Fig. 42. n. 1.) out of the middle whereof if you take another square EFGH, which [...]all be a fourth part of the former, the area of the rectangle [...]K intercepted between BC and FG prolonged, shall be equal [...] the square of a given line LM; i. e. having these given to [...]nd the segment BI, and consequently also the side BC or [...]B.

SOLƲTION.

Make the area of the given rectangle, or the square of LM [...] to bb, and the side sought of the rectangle BI= x; the o­ [...]er side BC will be = [...], and having substracted out of it [...] and GK ( i. e. 2 x) the side of the lesser square FG will [...] [...]−2 x, i. e. [...]; whose square since it is the [...]orth part of the greater square by the Hypothesis, you'l have [...]; [...] multiplying both sides by [...]; [...] taking away 4 b (powerof4), and adding [...]; [Page 56] and dividing by 4, [...].

Therefore according to the third case of affected quadratick [...] quations, [...] i. e. [...].

Therefore [...].

Geometrical Construction. If the given line b be taken [...] unity, b (powerof4) and bb will be equal to it. Therefore if betwe [...] LM as unity, and MN=3 ⅓ b, you find a mean proporti [...] nal MO ( n. 2. Fig. 42.) 'twill be [...]; which subst [...] cted from MQ=2 b, or added to it, will give two values the quantity xx, viz. PQ and IQ; the first whereof will [...] only a true one, and of use here. Now therefore a mea [...] proportional QR found between PQ and unity will expre [...] the quantity sought x.

Therefore for forming the square it self, since its side AB= [...], you may proceed as in the former Constructon, ( vi [...] n. 3.)

PROBLEM III.

HAving given the base of a right-angled Triangle, and mean proportional between the Hypothenusa and Perpe [...] dicular, to find the Triangle. As if the given base be A ( Fig 43.) and the mean proportional between AC and B [...] be CD; to find the perpendicular BC, and Hypothenu [...] AC.

SOLƲTION.

Make the given base = a, and the mean proportional = the perpendicular BC= x, then will the Hypothenusa be [...] the Hypoth, − [...].

Pag. 56.

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[...]erefore [...]; [...]d multiplying both sides by [...]; [...]d substracting [...]. [...]erefore by the second case of affected quadraticks [...], and [...].

Or thus.

Make the Hypothenusa AC= x, then will the perpendi­ [...]ar be BC= [...]. Therefore [...]; [...]d multiplying by [...]. [...]erefore by case 1. [...] and [...].

Geometrical Construction; the first for the latter Equation. [...] be put for unity, the line AB will be also = aa, and king, as a to b so b to a third, i. e. as LM to MN so LO OP, and you'l have bb. Having erected the perpendicu­ [...] AQ=OP upon AM, and drawn MQ or M n equal to [...], and consequently A n will be = [...], [...] the value of xx. Moreover a mean proportional AC [...]nd between A n and AR unity will be the value of x, i. e. Hypothenusa sought; which being found, you may easily [...]mplete the Triangle ABC.

2. In the case of the former Equation, making every thing before AK would be the value of the quantity xx, i. e. [...]. Therefore a mean proportional RT [...]nd between RS=AK and AR unity will be the value of x, [Page 58] i. e. the perpendicular sought, and so AT the Hypothenusa of the Triangle sought.

Arithmetical Rule. In the first Solution add the biquadrate of the given mean proportional to the biquadrate of half the given base; and having extracted the square root of the sum, take from it half the square of the given base; the root of the remainder will give the perpendicular of the triangle sought, and the root of the sum will give the hypothenusa of it.

PROBLEM IV.

HAving the Hypothenusa of a right-angled Triangle given, and a mean proportional between the sides to find the Tri­angle. As if the hypothenusa be AC ( Fig. 44.) and a mean proportional between the sides BD, to find the sides AB and BC.

SOLƲTION.

Make the given Hypothenusa = a, and the mean propor­tional = b, and the perpendicular BC= x; the basis AB by the hypoth. will be [...]. Therefore [...]; and multiplying by xx, [...]; and substracting b (powerof4), [...].

Therefore by the third case, [...] and [...].

Geometrical Construction. If a be put for unity, AC will be also = aa, and by making as AC to CG ( a to b) so AF to GH ( b to a third) this third will be GH= bb. Assuming therefore OC=½ aa=OB the radius of a semi-circle, and ha­ving erected CD= bb=BE parallel to it, EO will be [Page 59] [...], and consquently EC = [...], and EA [...], viz. the double value of the quantity xx. Therefore for the double value of x, you must extract the roots out of them, i. e. you must find the mean proporti­onals AL and AM between unity AC and AI=EC on the one side, and AK=AE on the other; Altho' these last may be more compendiously had, and the triangle it self immedi­ately constructed, if having found EC and EA, you draw CB and AB: For these will be those two last mean proportionals = = AL and AM; for by reason of the ▵ ▵ ABC, AEB, and BEC, BC is a mean proportional between AC and CE, and AB a mean proportional between the same AC and AE by the 8. Lib. 6. Eucl. which is Consect. 3. Schol. 2. Prop. 34. Lib. 1. Math. Enucl.

PROBLEM V.

HAving given the Area and Diagonal of a right-angled Parallelogram, to find the sides and so the Parallelogram. As if the given Area be = to the square of a given line BD ( Fig. 45.) and the Diagonal AC, to find the sides AB and BC.

SOLƲTION.

If for the given Area, or square of the line BD you put bb, and make the Diagonal AC= a, and put for the lesser side BC, x; the other side will be [...].

Therefore [...]; and multiplying by xx, [...]; and substracting b (powerof4) [...]. Which Equation, since it is the same with that of the preceding Problem (which is no wonder, since this fifth perfectly coincides with the fourth; for the Di­agonal AC is the hypothenusa, and BD, whose square is = to the given area of the rectangle, is a mean proportional be­tween the sides AB and BC) and so will have the same Con­struction, [Page 60] (see Fig. 45.) and the same Arithmetical Rule, which may be easily formed from the last Equation of the preceding Problem.

PROBLEM VI.

HAving given the first of three proportional lines, and ano­ther whose square is equal to both the squares of the other two, to find those two proportionals. As if AC ( Fig. 46.) be the first of the three proportionals, and another line ED gi­ven, whose square equals the two squares of the others taken together; to find those two as second and third proportio­nals.

SOLƲTION.

If for AC you put a, and make the given line ED= c, and the second proportional = x, the third will be [...]. Where­fore the squares of the two last will be [...]+ xx= cc, □ ED by the hypoth. and multiplying both sides by [...]; and substracting [...]. Therefore [...] and [...].

Geometrical Construction. If a be put for unity, AC will also = aa and a (powerof4), and making as AC to CD ( a to c) so AF to DE ( c to a third) DE will be = cc. Now having made AK=DE, i. e. cc, a mean proportional AI found between AC and AK will be [...]. Therefore taking AO=½AC, viz. ½ aa, the hypothenusa OI will be = [...]. And OA=½ a being substracted from OI or OH equal to it will leave AH the value of xx; and the root of that being extra­cted, i. e. finding another mean proportional AG between AC and AH, it will be the value of x, i. e. the second of the proportionals sought, and since AC is the first given, AH will be the third. Q. E. F.

NB. This Construction may be abbreviated, and the first Operation, by which you find DE, to which afterwards AK is made equal, may be omitted. For since you make use of a mean proportional between CA and DE sought, which is = to the given line ED, and afterwards AI a mean pro­portional between AC and AK is sought; it is evident that AI will be equal to ED given, and consequently that they may be immediately joined at right angles at the beginning of the given line AC, and the rest may then be done as before.

Some Examples of Cubick and Biquadratick Equations, both simple and affected, whether reducible or not.

PROBLEM I.

BEtween two given right angles to find two mean proportio­nals. E. g. Suppose given AB the first and CD the fourth, ( Fig. 47. n. 1.) between these to find two mean pro­portionals.

SOLƲTION.

Make the first of the given quantities AB= a, the other CD= q, the first mean = x, then will the latter be [...], and consequently [...]= q; and multiplying by aa, x (powerof3)=** aaq.

The Central Rule will b [...]=AD [...]=DH. i e. according to a supposition we shall by and by make, ½ a=AD, and ½ q= DH.

Geometrical Construction. If AB or a be made unity, and also the Latus Rectum of your Parabola, and you describe, by means of this Latus Rectum, the Parabola, according to Schol. 1. Prop. 1. Lib. 2. Math. Enucl. [see n. 2. and 3. Fig 47.] [...]n which AB is the Latus Rectum; A1, A2, &c. the Abscis­ [...]a's; AI, AH, &c. the semiordinates; make moreover ( n. 4) AD=½ a, and having from D erected a perpendicular =½ q, describe a circle at the interval AH, cutting the para­bola [Page 62] in N: Which being done, a perpendicular to the Ax will be the root sought or the value of x, i. e. the first of the means, and consequently OA the other; since NO by the first property of the Parabola (see Prop. 1. Lib. 2. Mathes. Enucl.) is a mean proportional between the Latus Rectum AB, and the abscissa AO. And by this means there will come out, by Baker's Central Rule the very construction of Des Cartes, Ge­om. p. m. 91.

The Arithmetical Rule. Multiply the square of the first by the fourth given, and the cube root extracted out of the pro­duct, will express the first of the means sought.

PROBLEM II.

HAving given the solid Contents of a solid or an hollow Pa­rallelepiped, and the proportion of the sides, to find the sides. As, if the given capacity or solid contents be = to the cube of a certain given line IK ( Fig. 48. n. 1.) and the pro­portion of the heighth to the length be as AB to BC, and to the latitude as the same AB to BD; to find first the altitude, which being had, the other Dimensions will also be known, by the given proportions.

SOLƲTION.

Make IK= a, AB= b, BC= c, and BD= d; and lastly the heighth sought = x, then as b to c, so x to the length required [...]; and as b to d, so x to the latitude sought [...].

Multiplying therefore these three dimensions of the Parallelepi­ped together, you'l have its capacity or solid contents [...] = a (powerof3); and multiplying by bb, cdx (powerof3)= a (powerof3) bb; and dividing by cd, [...] i e. [...].

Therefore the Central Rule will be the samc as above, [...]=AD and [...]=DH, i. e. according to the supposi­tion which will by and by follow, [...]=AD and [...]=DH.

Geometrical Construction. If IK or a be made unity, and at the same time Latus Rectum, and by means of it you de­scribe a Parabola, after the way we have shewn, Fig. 47. n. 2. and 3. and shall always hereafter make use of; and then to prepare the quantity [...] (which in the Central Rule is the the quantity [...]) make ( n. 2.) IK−IM=BC−KL=BD−MN as a to c so d to e; so that for cd you may put ae, and afterwards divide by a both above and underneath; you'l have the quantity [...].

Therefore by further inferring as 2 e to bb, so aa to a fourth IO−IP=IT−IK−IQ, and you'l have the quantity DH, which will determine the centre, after AD is made equal [...]. Having therefore described from that centre a circle through the vertex of the Parabola A ( n. 3.) a semi­ordinate NO drawn from the intersection will be the altitude sought, which will easily give you the length and breadth by the reasons above shewn.

Another Solution.

Which will be more accommodated to the Arithmetical Rule.

Let the rest of our Positions or Data remain as above, but the name of the proportion which the altitude has to the length be = e, and of that which it has to the latitude = i, the [Page 64] length will be = to ex, and the latitude to ix. Wheref [...] multiplying the sides together, you'l have the whole solidity [...]; and dividing by ei, [...]. Therefore [...] Hence

The Arithmetical Rule. Multiply together the given na [...] of the reasons, and divide the given cube by the produ [...] which done, the cubick root extracted out of the quotient [...] be the altitude of the solid sought.

Another Geometrical Construction. Now if we would a [...] construct this Equation x (powerof3)= [...] geometrically, putting AB= [...] for unity, BC and BD will be the names of the reasons = [...] and i. Making therefore first IK−IM−KL−MN as a to e so i to a fourth f (Fig. 49. n 1.) af [...] be = ei, and the proposed Equation will have this form: [...] i. e. aa / f.

Therefore 2. making as f to a so a to a third IQ, that will be the value of [...] But hence 3 by extracting the cube root, i. e. by finding t [...] mean proportionals between unity b, viz. AB and the [...] found [...]Q; the first of them will give the root sought.

NB. The same Central Rule would come out according Baker [...]s Central Rule, which would have the same form of Equation, as the precedent Example x (powerof3)**− aa= o,

  • taking b for the Lat.
  • Rect. and unity,
    • [...]=AD and [...]=DH,
    • [...]=AD and [...] i. e. ½IQ=DH

(see Fig. 49. n. 2.)

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PROBLEM III.

HAving given the solid contents of a solid or hollow Paral­lelepiped, and the difference of the sides, to find the sides. [...] if the given capacity be equal to the cube of any given [...]e LM ( Fig. 50. n. 1.) and the difference whereby the [...]gth exceeds the breadth = NO, and the difference by which [...] breadth exceeds the altitude or depth = PQ, to find the [...]gth, breadth and depth.

SOLƲTION.

Make the side of the given cube = a, the excess of the [...]gth above the breadth NO= b, and of the breadth above [...] depth PQ= c, make the depth = x, the breadth = x+ c, [...]d the length = x+ b+ c. Multiplying therefore these [...]ree dimensions together.

[...]

[...] according to the forms of Baker and Cartes [...]

Wherefore the Central Rule contracted by the supposition [...]hich will hereafter follow will be this, [...] = AD and [...] = DH. [...] e. by virtue of the supposition just now mentioned (which [...]kes LM viz. a for unity and also for Lat. Rectum)

[...] = AD. And [...] = DH.

Or more short; [...] i. e. [...] = AD; and [...] (∽ [...] i. e. [...] = DH.

Geometrical Construction. If LM or a be taken for unity or Latus Rectum of the Parabola to be described, that being described ( Fig. 50. n. 3.) you are first of all to determine two quantities AD and DH; which may be done two ways; either by Baker's form of his Central Rule, or by ours immediately divided by the quantities of the last Equation.

1. For AD by our form, [...] = AD, you must make ( Fig. 50. n. 1.) as a to b so b to a third (AB to AC so BE to CF) which will be bb, and D (powerof2) the eighth part of this CF must be added to A (powerof2) the half of AB. And by Baker's form you must make, 1. as AB (= a, n. 2.) to AC (=½ p i. e. ½ b+ c) so BE (=¼ p i. e. ¼ b+½ c) to a fourth CF (which will be = [...] ▪) 2. Make moreover as AB to AG ( a to b) so BH to GI ( c to bc) and, as AB to AK ( a to c) so BH to KL ( c to cc) and the two quantities found GI and KL ( bc and cc) being added into one sum will give the quantity q=MN, the half whereof MO will express the quantity [...] in the [Page 67] Rule, and to be substracted from the former [...]. Actually therefore to determine the quantity AD not on the Ax, but on another Diameter of the described parabola n. 3. (because the quantity p is in the Equation) having made a perpendi­cular to the Ax aE= [...] i. e. to the line BE n. 2. and from E having drawn EA parallel to the ax, according to our form AD n. 1. transfer it only on the diameter of the parabola n. 3. from A to D, either by parts [...] i. e. ½LM from A to c, and [...] i. e. ⅛CF n. 1. from c to D: But according to Baker's form, first you must put [...]=½LM n. 1. from A to 1. Se­condly you must put from 1 to 2 the quantity [...]=CF n. 2. Thirdly you must put from 2 to 3 backwards the quantity to be substracted [...]=MO n. 2. which being done, the point D will be determined.

[It is evident by comparing these two ways of Construction, that we may join our forms not incommodiously to Baker 's; because by ours the quantity AD was obtained more compendiously than by Baker 's, which will also often happen hereafter. And where this Compendium cannot be had, there is another not inconside­rable one, that, if the quantities AD and DH determined ac­cording to both ways shall coincide, (which happens in the pre­sent case) we may be so much the more sure of the truth]

2. As for DH by our form, you must put it from D to e on a perpendicular erected from D on the left hand, the quantity [...] = BE n. 2. falling here upon the Ax. Then for the quantity [...] make n. 4. as a to ½ bb (LM to LN) so ⅛ b to [...] (MO to NP) and this quantity must be put from e to f in a perpendicular to the Diam. Thirdly, for the quantity [...] [Page 68] you must farther make n. 4. as a to ½ bb (DM to LN) so ¼ c to a fourth (MQ to NR) which must be put from f to g. Lastly the quantity [...] (which is = MO n. 2.) must be put backwards (because to be substracted) from g to H, which is the centre sought. In like manner by Baker's form, first [...]=BE is put from D to 1 even to the Ax. Secondly, for the quantity [...] make, n. 4. as a to [...] (LM to LS=CF n. 2.) so [...] to a fourth (MT=AC n. 2. to SV) and this SV is further put ( n. 3.) from 1 to 2. Thirdly, make in the same Fig. n. 4. for the quantity [...], as a to [...] (LM to MT) so [...] to a fourth (LX to XZ) and this XZ is put back­wads ( n. 3.) from 2 to 3, which precisely coincides with the point g. Lastly the remaining quantity [...] (=MO n. 2. and so by what we have said above, precisely coinciding with the interval g H) is put backwards from g to H the Centre sought.

[Hence it appears again that Baker 's form is more laborious than ours; tho' both accurately agree, and hereafter, for the most part, we shall use them both together, tho in' the work it self, rather in Figures, than in that tedious process of words, which we have here for once made use of, that it might be as an Example for the following Constructions.]

Having therefore found by one or both ways the Center H, and thence described a circle thro' the vertex of the parabola A, the intersection N will give the perpendicular to the dia­meter NO, the value of the quantity x sought.

PROBLEM IV.

TO divide a given angle NOP, or a given arch NQTP (Fig. 51. n. 1.) into three equal parts; i. e. having gi­ven or assumed at pleasure the Radius NO, and consequently also the Chord of the arch NP, to find NQ the subtense of the third part of the given arch.

SOLƲTION.

If NO be made unity, NP= q, and NQ be supposed = z; having drawn QS parallel to TO, you'l have three similar triangles NOQ, QNR and RQS. For since the an­gle QOP is double of the angle QON, and the same (as be­ing at the Center) double also of the angle at the Periphery QNR; it will be equal to the angle QON. But the angle at Q is common to both triangles: Therefore the whole are equi-angular, and consequently the legs NQ and NR equal, as also NO and QO; and by the like reason also PY and PT. Wherefore if RS should be added to RY, the line NP by this addition would be triple of the line NQ; and so would give the Equation, if RS was determined; which may be done by means of the ▵ QRS, similar to the two former NOQ and QNR; for the angle RQS is equal to the alternate one QOP=QNR, and the angle at R common to the tri­angles QNR and RQS, &c. Wherefore as NO to NQ so NQ to QR 1− z− z− zz and as NQ to QR so QR to RS z− zz− zz− [...]

Therefore according to what we have above said [...]; and substracting [...]; or [...]

Therefore the Central Rule will be (supposing also unity NO for the Latus Rectum,) [Page 70] [...] = AD i. e. by our ½+ [...] i. e. 2NO=AD form, and [...]=DH. and [...]=DH.

Geometrical Construction. Having described your parabola ( Fig. 51. n. 2.) take on its Ax (because the quantity p is want­ing in the Equation) AD=2NO, and from D having e­rected a perpendicular = [...]NP to H; that will be the Center, from which a circle described thro' A, by cutting the parabola in three places, will give the three roots of the Equation, viz. NO and no true ones, the first whereof will express the quantity sought NQ ( n. 1.) the latter the line NV, being the subtense of the third part of the compl. of the arch; and MO will express the false root, which is equal to the former taken together: All the same as in Cartes p. 91. but here some­what plainer and easier.

PROBLEM V.

HAving three sides given of a quadilateral Figure to be in­scribed in a circle, AB, BC, CD, (Fig. 52. n. 1.) to find the fourth side, which shall be the Diameter of the Cir­cle.

SOLƲTION.

If we consider the business as already done, and make AB= a, BC= b, CD= c, and AD= y; we shall have first in the right-angled ▵ ABD, □BD= yy− aa, and (since in the obtuse angled ▵ BCD the □ BD is = □□BC+CD+2▭ of BC into CE) if those two □□BC+CD ( i. e. bb+ cc) be substracted from □BD ( yy− aa) you'l have 2. yy− aa− bb− cc= to the two said rectangles of BC into CE. But these two rectangles may also be otherwise ob­tain'd, 3, if the segment CE be otherwise determined; which may be done by help of the similar ▵ ▵ ABD and CED (for the angles at B and E are right ones, and ECD and BAD equal, because each with the same third BCD makes two right [Page 71] ones; the one ECD by reason of its contiguity, the other at A by the 22. Eucl. Lib. 3.) viz. by inferring as DA to AB so DC to CE y− a− c− [...] for now multiplying CE= [...] by BC= b, the ▭ of BC into CE will be = [...] and two such [...].

Now therefore yy− aa− bb− cc= [...]; and multipl. by y,

  • y (powerof3)*− aa y=2 abc; i. e. by Baker's and Cartes's forms.
  • − bb y=2 abc; i. e. by Baker's and Cartes's forms.
  • − cc y=2 abc; i. e. by Baker's and Cartes's forms.
  • y (powerof3)*− aa y−2 abc= o.
  • − bb y−2 abc= o.
  • − cc y−2 abc= o.

Therefore the Central Rule will be (supposing the same quantity e. g. a for unity and Lat. Rect.) [...] = AD and [...]=DH, i. e. according to our form, [...] i. e. [...] = AD, and

Geometrical Construction, which, without any circumlocuti­on, from our form is founded on the foll. in Fig. 52. n. 2.

LM= a n. 3. A 1.=LM from n. 2.

MN and LP= b 1, 2=½PQ

PQ= bb 2, 2=½ST

LS and MO= c DH=PR

PR= bc MO and mo two false roots

ST= cc NO the true root; upon which having de­scribed a semi-circle the quadrilateral will be easily made. Ac­cording to Baker's form, for AD there would be n. 3. first [Page 72] A c=½LM, then cD= [...]=VX, half the line VZ, which is compounded of LM, PQ and St; but DH is = PR as above.

PROBLEM VI.

HAving given to form a right-angled Triangle the least side BA (Fig. 53. n. 1.) and the difference of the seg­ments of the base, to find the difference of the sides, and so form the Triangle. If we represent the business as already done, ha­ving given AB and EC to find FC.

SOLƲTION.

Make AB= a, and EC= b, and FC= x; then will BC= a+ x: Therefore the □AC=2 aa+2 ax+ xx and the line AC [...] and [...]. Now therefore ACE i. e. [...] multiplyed by b or √ bb i. e. [...] is = GCF [but GC is = 2 a+ x] i. e. 2 ax+ xx by Cons. 1. Prop. 47. Lib. 1. Mathes. E­nucl. and squaring both sides

[...]; and transposing all, [...]. − bb (p) (q) (r) (s)

Therefore (taking a for 1 and the Latus Rectum of the parabola) the Central Rule will be, [...] = AD and [...] = DH i. e. according to our form and Reduction, [...]

Geometrical Construction. First from our form, the com­pendiousness whereof will here appear, for it requires only one preparation n. 2. in which LM= a, MN and LO= b, OP= bb, which being premis'd, and the diameter A y (be­cause the quantity p is in the Equation) being drawn ( n. 3.) make AI=½LM, and 1, 2 or [...]D=½OP, and DH=LM. The rest therefore being also perform'd, which the quantity S occurring in the present Equation requires, according to the last precepts of our introduction, you'l have NO the va­lue of x sought; whence ( n. 4.) at the interval AB having describ'd a Circle, and made a right angle at B, if FC be made = to the found NO, you'l have the ▵ ABC required, and EC will be found at the beginning of the prescribed mag­nitude.

Now if you were to find the center H by Baker's form with­out our Reduction, 1. you must put of ( n. 3.) ½AB from A to c. 2. For the quantity [...] make as 1 to [...] so [...] to a fourth, which would be = 2AB, viz. 2DM, and to be set from c to d. 3. The quantity [...] (to obtain which you must substract OP ( n. 2.) from the quadruple of LM, and divide the re­mainder by 2) must be set off backwards from d to e, by thus determining the point D. 4. The quantity p, which here is precisely = LM, must be transferr'd from D to f on a perpen­dicular erected from D. 5. For the quantity [...] make as 1 to [...] (= 2AB) so [...] (also = [...]) to a fourth quadruple of LM; and this must further be produc'd from f to the point g (which here the paper wil not permit) 6. For the quanti­ty [...] make as 1 to [...] so [...] to a fourth, (which would be = to the quadruple of LM, but taking away OP) which must be set backwards from g to h. 7. Lastly the quantity [...] (=OP) set off from h backwards or towards the right hand to i will at length give the point H required.

Another Solution of the same Problem.

This Problem may be more easily solved, and will give a far more simple Equation, if you are to find not FD but AE. Make therefore ( n. 1. Fig. 53.) = x, and the rest as above; AC will be = x+ b, and its □ xx+2 bx+ bb; therefore the □BC= xx+2 bx+ bb− aa; therefore the □ of the tan­gent HC= xx+2 bbx+ bb−2 aa. But the rectangle ACE will be bb+ bx. Therefore by Prop. 47. Lib. 1. Math. E­nucl.

[...]; and turning all over to the right hand, [...] or [...]. Therefore by case 2 of affected quadraticks, [...].

The Geometrical Construction may be performed according to the rules of quadraticks Fig. 54. n. 1. as will be evident to any attentive Reader. Having therefore described a circle at the interval BA, whether it be done from any arbitrary cen­ter (see n. 4. Fig. 53.) or upon AE found in the pres. Fig. making an intersection at the said interval in B; and apply­ing AE, and producing it until EC become equal to the gi­ven quantity q, and at length having drawn BC you'l have the triangle right-angled at B, and also the difference of the sides FC. But to make it more short and elegant; having determined AE by a little circle ( Fig 54 n. 3.) prolong it to the opposite part of the circumference in C, and draw AB and CB; for as the radius of the little circle is ½ b, so EC is = b.

Now if you would construct the Equation above by Baker's rule (that its universality may also be confirm'd by an Exam­ple in quadraticks) viz.

[...];

The centtal rule will be (taking a for 1 and the L. R.) [...] = AD [Page]

LIV

LV

LVI

LVII

[Page 75] [...] = DH. i. e. by our Reduction, [...] = AD [...] i. e. [...] = DH.

The Construction therefore will consist in these: LM ( n. 2. [...]g. 54)= a, MO=¼ b, LP=½ b, therefore PR= [...]: [...]N=PR viz. [...], therefore Q= [...]. In Fig. 53. n. 3. [...]=¼ b, A1=½LM, 1, 2, or 1 D=PR; D1=LP MO, 1, 2, or 1 H=PQ Having drawn a circle from [...] thro' A you'l have the true root RO= to AE sought; [...]d MO the false root.

NB. Hence it is evident that one Problem may have seve­ [...]l Solutions and Constructions, some more easy and simple, [...]hers more compound and laborious; viz. according as the [...]known quantity is assumed more or less commodious to the [...]urpose: which may not be amiss here to note for the sake [...] Learners.

PROBLEM VII.

SUppose a right line BD (Fig. 55.) any how divided in A, to divide it again in C, so that the square of BA shall [...]e to the square of AC as AC to CD.

SOLƲTION.

Since the first segments BA and AD are given, call the first [...] and the other b, and call AC the quantity AC x; and CD [...]ill be = b− x. Now therefore since we suppose as the □ of AB to the □ AC so AC to CD [...]

aab− aax will = x (powerof3), and transferring the qu n [...]ities on [...]he left hand to the right, x (powerof3)* aax− aab= o.

Wherefore the Central Rule (taking a for 1 and the L. R.) will be ½− [...]=AD and [...]=DH. i. e. by our form [...] i. e. o=AD, that D may fall on the vertex of the parabola; and [...] i. e. [...]=DH. The Construction therefore will be very simple, as is evident from the fifty fifth Figure.

PROBLEM VIII.

THere is given AB the capital line of a horn work (which we represent (tho' rudely) n. 1. Fig. 56) and the Gorge AD, also part of the line of defence EF, to find the face BE, the flank DE, the curtain (or the chord) DF, also the angle of the Bastion ABE, &c. and so the whole delineation of the horn work. It is evident if you have the flank DE or the curtain DF, the rest will be had also. Suppose therefore, the capital line AB and the gorge AD, and part of the line of defence EF to be of the magnitudes denoted by the Letters a, b, c, on the right hand.

SOLƲTION.

Make AB= a, AD= b, and EF= c, and DF= x; then will AF= x+ b, and by reason of the similitude of the ▵▵BAF and EDE and ECB, as FA to AB so FD to DE [...]

But now □ □ DF+DE are = = □EF i. e. [...]; or giving the same denomination to all the quan­tities on the left hand, [...] [Page 77] and multiplying both sides by [...] and translating all the quantities on the right hand by the contrary signs to the left, [...]

Wherefore (putting a for 1 and L. Rectum) the Central Rule will be, [...] (because the quantity q is negative in the E­quation, for cc is greater than aa+ bb)=AD; and [...] = DH. or according to our Redu­ction, [...] = AD and [...] i. e. [...] i. e. [...] = DH.

Wherefore the Geometrical Construction requires no other preparatory determination by our form than of the quantities [...] for AD, [...] for DH at the center, and bbcc to determine the radius of the circle; which are exhibited by n. 2. Fig. 56. viz NP is = cc, RS= bcc, RV= bbcc, which are found by means of LM= a, LR= b, LN and MO= c, MQ=NP and MT=RS. Having therefore described a para­bola, n. 3. and drawn its diameter, transfer AD=½NP upon it (because the quantity p is in the Equation) and also ½RS from D upon H perpendicularly, and on the right hand, (because DH=− [...];) and so you'l have the center H; [Page 78] thro' which having drawn KAI so that AK shall be = to the quantity bbcc or S, i. e. RV, &c. a circle described at the interval HL will cut the parabola in M and N, and applying the magnitude NO it will be that of the Curtain sought; up­on which, n. 4. having laid down the circumference of the horned work by help of the given lines AB and AD, you'l have the line EF, of the magnitude which was above supposed. Now if any one has a mind to do the same thing by Baker's way; by laying down first the interval AB = [...] and then making bc= [...], and lastly, putting cd for the quantity [...]; he will fall upon the same point D, and in like manner may express the other parts of the Central Rule by the intervals D e, ef, fg, and setting back the last gh, he will fall upon the same center H: But this is done with a great deal more trou­ble and labour to determine so many quantities, and also is in more danger of erring by cutting off so many parts separately, as experience will shew; and thus we have by a new argument shewn the advantage of our Reduction.

Another Solution of the same Problem.

Things remaining as before (only assuming the given lines AB, AD and EF, n. 1. Fig. 56. one half less, that the Scheme may take up less room) make BE= x, as the first or chief unknown quantity; then will BF= x+ c, and its □ xx+2 cx + cc: And since as BE to BC=AD so BF to AF a fourth, which will be [...] and its square = [...]. Where­fore if this square be substracted from the square of BF, there will remain the square of BA, i. e.

[...]; i. e. all being reduced to the same denomination, [...]; [Page 79] or according to the forms of Cartes and Baker, [...]

Therefore the Central Rule (putting again a for 1 and the L. R.) will be [...] = AD and [...] = DH; or by our Reduction, [...] i. e. [...] = AD; and [...] i. e. [...]=DH.

Geometrical Construction. Having therefore described a para­bola ( Fig. 56. n. 5.) and drawn the diameter A y, make A1= a and 1, 2= [...], so you'l have the point D; make moreover D (powerof3) or 2, 3= c, and backwards 3, 4= [...] (we here omit to express the Geometrical determination of these quantities [...] and [...] as being very easy) and you'l have the point H, &c. and there will come out the quantity sought NO; which since it is equal to half BE n. 4. the business will be done; which Baker's form will also give, exactly the same, but after a more tedious process.

PROBLEM IX.

IN any Triangle ABC (the scheme whereof see n. 1. Fig. 57.) suppose given the Perpendicular AD, and the differences of the least side from the two others EC and FC to find all the three [Page 80] sides. i. e. Chiefly the least side AB which being found, the others will be so also.

SOLƲTION.

Make AD= a, EC= b, and FC= c, AB= x; then will BC= x+ b and its square be xx+2 bx+ bb, and AC= x+ c and its square be xx+2 cx+ cc; and BD will be [...], and DC [...]. But the □ BC may also be obtain'd otherwise, and the Equation also, if □□BD+DC+2▭ of BD by DC be added into one sum according to Prop. 4. Lib. 2. Eucl. viz.

[...], multiplyed by BD √ xx− aa, gives the rectangle of the segments [...] and this doubled i. e. multiplyed by √4, gives the quanti­ty which is contain'd under the radical sign in the Equa­tion]

Therefore turning all over on the left hand which are be­fore the sign √ to the right hand, prefixing to them the contrary signs, you'l have [...] and taking away the Vinculum on the left hand, and squaring on the right [...] [Page 81] and adding and substracting on both sides, as much as can be, [...] and transferring all to the left, [...] and dividing all by 3, [...]

Note, I sought this Equation also after two other ways; 1. By a comparison of the □ AC with the two squares AB+BC, after 2 □ □ CBD thence substracted, according to Prop. 13. Lib. 2. Eucl which is the 46. Lib. 1. Math. Enucl. and I form'd the same with the present. 2. By putting at the beginning y for x+ b and z for x+ c, and going on after the former me­thods, 'till you have this Equation, [...] in which▪ when afterwards I substituted the values answering the quantities yy and zz, &c. This same last Equation came out a little easier, but (NB) with all the contrary signs.

Now to form the Central Rule, and thence make the Geo­metrical Construction, we must determine first each of the quantities p, q, r and S, that we may know whether they are negative or positive; and you'l find ( n. 2. Fig. 57.) p=G2 positive, q=H (powerof4) negative, and K (powerof4)= S also negative; and that by help of the quantities LM= b or 1, MN and [Page 82] LO= a, OP= aa, MQ and LR= c, RS= cc, and also LT= cc, TV and LX= c (powerof3), XY= c (powerof4). Wherefore the form of the last Equation will be like this, x (powerof4)+ px (powerof3)− qxx− rx− S= o, and so the Central Rule (taking here b for 1 and the L. R.) [...] = AD and [...] = DH.

Wherefore, having now described a parabola (as may be seen n. 4.) having found the diameter A y transfer upon it first A b=½LM ( n. 2) and then bc= DP ( n. 3.) i. e. [...]; and thirdly cD= [...] i. e. ½H (powerof4) ( n. 2.) moreover from D to e put off MB ( n. 3.= [...], and from e to f put off DR= [...], and from f to g put off CF= [...]; and from g backwards to h put off half the quantity r, or I5 ( n. 2.) and having done the rest as usual, you'l have NO, the side required of the Trian­gle to be described; the description whereof will be now easy ( n. 5.) having all the three sides known. This may serve for an Examen, if having described a semi-circle AGB upon AB=NO, you apply the given line AD, and from B thro' D draw indefinitely BDC: Then at the interval AB having described the Arches AE and BF, add the given line EC to BE, for thus having joined the points A and C, FC ought to be equal to FC before given.

SCHOLIƲM.

WE have here omitted our Reduction, because it would be too tedious, and would express the quantities AD and DH (especially the latter) in terms too prolix. For AD would be = [...] ( viz. because [...] is [Page 83] found = [...] and [...] taken in it self = [...]; but here [where by vertue of the Central Rule it is taken positively, when it is in it self negative] un­der contrary signs it is = [...] or [...] or [...] or [...] yet more contractedly (because b is unity) AD = [...]; which parts may be expressed without any great difficulty on the Diameter A y, by its portions A1, 1, 2; 2, 3; 3D: But the other quantity DH, or the definition of the Center H, would also have some tedi­ousness, as because [...]

If you take away out of the quantities [...] and [...] (since this latter is to be substracted, and so left, as it is, under the sign −; but the other, also negative in it self, but here posi­tively expressed in the Central Rule, must have all the con­trary signs) I say, if you take out of these quantities those which destroy one another, and add the rest with the two for­mer quantities, they will be [...] = = DH. or a little more contracted (because b is 1) [...] = DH.

But now if any one has a mind to illustrate this by a numeral Example and try the truth, &c. of the quantities found; they may make e. g. a=12, b=1, and c=2; and they [Page 84] will easily find that in the last Equation the quantity p will be 4, and q−190, r−388, S−195: Secondly in the Central Rule of Baker they'l find [...]=½, [...]=2, and [...]= 95, and so the whole line AD=97 ½; and further [...]=1, [...]=4, [...]=190, and [...], and so the whole line DH=195−194 i. e.=1. Thirdly, likewise in our Reduction (if we proceed by each part corresponding to Baker's) [...], and [...]. The sum for AD 97 ½; but further [...]; and [...] to be sub­stracted; and so the sum for DH=195−194=1. Which same quantities will fourthly come out, if the quanti­ties AD and DH contracted, as they are expressed in letters a­bove, be resolved into numbers.

PROBLEM X.

YOU are to build a Fort on the given Polygons EAF (see Fig. 58. n. 1.) whose capital line AB shall equal the ag­gregate of the gorge and flank, and the squares of these added together shall be equal to the square of the given line GH, and the solid made by the multiplication of the square of the flank by the gorge, shall be equal to the cube of the given line IK.

LVIII

LIX

LX

SOLƲTION.

Make the Gorge AC= z, whose square zz substracted from bb the square of the given line b, will leave the square of the flank DC= bb− zz. Now this square being mul­tiplyed by the gorge AC or z will give bbz− z (powerof3)= g (powerof3), the cube of the given line IK; and adding to both sides z (powerof3), and substracting bbz, o= z (powerof3)*− bbz+ g (powerof3).

Therefore if we take g or IK for 1 and L (powerof3), g (powerof3) will be the line g, and

The Central Rule: [...] = AD. and [...]=DH. i. e. according to our Reduction, [...] = AD and [...] DH.

Geometrical Construction. Having described a Parabola ( n. 2. Fig. 58.) make on its ax A1=½IK and 1, 2, s. viz. 1D=½MN (from n. 1.) and DH=½IK. Then having de­scribed a circle from H, and found the true root NO upon the given angle EAF ( n 3.) make AC=NO, and having e­rected the perpendicular CD divide it by AD=GH ( n. 1.) and make AB=AC+CD; and the Fort will be drawn.

PROBLEM XI.

IN a right-angled Triangle ABC (which we denote by n. 1. Fig. 59) having given the greater side AC, and made the less side AB= to the segment CE, which shall cut off from the base BC a perpendicular let fall from the right angle A; to find these lines AB or CE, and consequently the whole trian­gle.

SOLƲTION.

Make AC= a, CE or AB= x. Therefore, 1. you'l have aa− xx=□AE. And because the ▵ ▵ BEA and CAE are similar, you'l have as AC to CE so BA to AE [...]

And so, 2. □AE= [...]. Therefore [...]= aa− xx; and multipl. by aa, x (powerof4)= a (powerof4)− aaxx; or, according to the form of Cartes and Baker, transposing all to the left, x (powerof4)*+ aaxx*− a (powerof4)= o i. e. x (powerof4)*+ qxx*− S= o.

Therefore (taking a for 1 and L. R.) the Central Rule will be [...] i. e. o=AD and [...] i. e. o=DH; that H may fall on the vertex A.

Geometrical Construction. Since a is assumed for unity and L, the quantity S also and Latus Rectum i. e. AI and AK and consequently the mean proportional AL and the radius HL will be = = to the given side AC, and conse­quently at that interval having described a circle, thro' the Pa­rabola rightly delineated, you'l have NO the value of the quantity x, i. e. of the lesser side AB. Having drawn there­fore NA, which is = to AC by Construction, if you draw to it the perpendicular AB cutting NO produc'd to B, you'l have the Triangle sought ABC, and AB (which will be a sign of a true Solution) will be found = NO or CE.

Another Construction. Since in the Equation above found there is neither x (powerof3) nor x, it may be look'd upon as a quadra­tick, and constructed after the same way, as several other like it among the Examples n. 4. viz. because [...]; according to case 2 of affected quadraticks, [Page 87] [...], and [...] i.e. [...] Wherefore ( n. 3. Fig. 59.) if AC be made = a and CD [...] a, the mean proportional CG will be = [...], and taking hence GF=½ a, there will remain FC = [...]: And now between this or CH equal to it, and unity AC, having found another mean proportional CE it will be the value of the quantity x sought=NO ( n. 2.)

PROBLEM XII.

IN a right-angled Triangle ABC (Fig. 60. n. 1.) there is given the Perpendicular BA, a segment of the Hypothenu­sa BD, and a segment of the Base EC, from C to the perpendi­cular DE let fall from the end of the segment BD; to find AE DC, and consequently the Base AC and the Hypothenusa BC, and so the whole Triangle.

SOLƲTION.

Make AB= a, BD= b, and EC= c, and DC= x; which being given, the rest cannot be wanting: Therefore xx− cc=□DE. But the same □ DE may be had, if you infer as BC to BA so DC to DE [...]

And then square the quantity DE, the square will be [...]; and multiplying both sides by [...] and substracting also [...] [...] [Page]

Pag. 89.

LXI

[Page 89] so PV= to it, that PX may be [...]; and lastly P y=½ q, that PZ may be [...]. These being thus prepar'd, if ( n. 4) A b be made = ½AB ( n. 1.) bc=RT, and cd backwards = ½ PQ we shall light on the point D; and, if we make D e=½BD, ef=PX ( n. 5. which interval was too big to be represented in the Paper) and from f you put backwards fg=PZ, and on from g beyond to the right hand gh=2RS; we shall light on the point H, &c. Which of these two methods is the shortest and fittest for practice, any one, never so little experienc'd, may here see; and first Learners may take notice if they would construct by Baker's form, in the Diagrams n. 3. n. 5. and the like, they must take care to make the angles PLN, SPT pretty large; which we have here represented the less to save charges in cutting on Copper.

PROBLEM XIII.

HAving given the Diameter of a Circle CD (n. 1. Fig. 61.) and the line BG, which falls on it perpendicularly, (which we have here only rudely delineated) to find the point A. from which a right line AC being drawn shall so cut the line BG in F, that AF, FG, GD shall be three continual Proportionals.

SOLƲTION.

If CF be found, the point A will be also had, and the se­ction of the line BG will be made. Make therefore CF= x, and (because the perpendicular BG is given, there will also be given the segments of the diameter CG and GD) make GD= b, and CG= c; then will BG=√ bc▪ and CD= b+ c. Since therefore the ▵ ▵ CAD and CGF are right-angled, and have the angle at C common, they will be simi­lar.

Therefore, as CF to CG so CD to CA [...]

Therefore AF = [...] and FG [...].

But by the Hypothesis, as AF to FG so FG to GD [...].

Therefore the rectangle of the extremes will be equal to that of the means, [...]; and multiplying by [...]; and transposing all to the left, [...]; i. e. by the Carte­sian form,

[...].

Therefore (taking b for 1 and L.) the Central Rule will be, [...] = AD and [...] = DH. or according to our Reduction, [...] i. e. [...] = AD and [...] i. e. [...] = DH.

Geometrical Construction. Having described upon the gi­ven line CD ( n 2.) a semi-circle, and apply'd in it the gi­ven perpendicular BG, as the figure shews, you'l have the seg­ments of the Diameter GD= b, and to the quantity p in Ba­ker's form, and CG= c, which ( n. 3. where LM= b, LO and MN= c) will give OP= cc and to the quantity q. Wherefore having describ'd a Parabola ( n. 4.) and the line VZ=2 ½ b, having cut off the fourth part of XZ, and the eighth of YZ (whereof the one will be = [...], viz. [...], and [Page 91] the other to [...]) if you transfer A1=XZ upon the diame­ter of the Parabola A y, and moreover 1, 2 or 1 D= to half OP ( n. 3.) and transversly D (powerof3)=YZ and backwards 3, 4=¼ OP, as also 4, 5=½CG ( n. 2.) you'l have the center H, and having describ'd a Circle at the interval HA, the root NO must be transferr'd from ( n. 2.) C to F, and continued to A the point sought. In Baker's Form (because the quantity p is = b or 1) [...] is = [...] and [...]= [...], and the quantity q or cc=OP, ( n. 3.) make therefore in the Diameter of the Para­bola A b=½GD, and bc=⅛GD, ( n. 2.) and lastly cd=½OP ( n. 3.) and you'l have the point D the same as before. Make moreover D e=¼GD and cf= [...]CG, and then back­wards fg=¼OP, and lastly gh=½GD, and you'l have the same center H, and the coincidence of the parts in both forms will be pleasant to observe; which otherwise seldom happens.

Other Solutions of the same Problem.

Carolus Renaldinus, from whose Treatise de Resol. & Com­pos. Math. Lib. 2. we have the present Problem, proceeds to solve it in another way, changing it plainly into another Pro­blem: viz. he observes, 1. That the angles FAD (see n. 1. of our 61. Fig.) and FGD, since both are right ones on the same common base FD, are in circle. Hence he infers, 2. (by vertue of the Coroll. of the 26. Prop. 3. Eucl.) that the □ □ DCG and ACF are equal, and consequently CD, CA, FC and CG are four continued proportionals. Then he observes, 3. That GD is the excess of the first of these proportionals above the fourth CG, and AF is the excess of the second AC above the third CF; and so, since 4. the rectangle of AF and GD is = to the square of the mean proportional FG (for AF, FG, GD, are supposed to be continual proportionals) and this □ FG is the excess, by which the square of the third CF ex­ceeds the square of the fourth CG; now the present Problem will be 5. reduc'd to this other: Having two right lines (CD and CG) given to find two such mean proportionals (AC and FC) that the ▭ of the excess of the first above the fourth ( viz. of FA into GD) shall be equal to the excess, [Page 92] by which the square of the third (FC) exceeds the square of the fourth (CG) viz. by the square FG.

Wherefore instead of the former he solves this latter Pro­blem, putting for CG, b, for GD, c, so that the first of the given lines CD shall be = b+ c, and the other GD= b; calling the first mean proportional AC, x; and thence deno­minating the latter [...] ( viz. multiplying the fourth by the first, and dividing the product by the second) and more­over he finds the excess of the first ( b+ c) above the fourth (b) to be c, and the excess of the second (x) above the third [...] to be [...] i. e. [...]; so that the □ □ of these two excesses is [...], and because the □ of the third FC is = [...], having substra­cted bb=□GC, there is given the □FG = [...] = ▭ of the excesses we just now found. So that now you'l have the Equation [...] &c.

We also endeavour'd to find another Solution, by finding an Equation from the line BD ( n. 1. Fig. 61.) as which might be twice obtain'd by means of the two right-angled Triangles FAD and FGD, since it is the hypothenusa of both. But here, besides the former denominations of our Solution we must first give a denomination to the line AD, by making as CF to FG so CD to AD [...], &c.

But whosoever shall prosecute this Solution of ours, or that of Renaldinus to the end, will find much more labour and dif­ficulty in either, than in the first we have given.

APPENDIX.

THE Invention of the special Central Rule for the case of Problem 1. Of Cubic [...] Equations, &c. which may serve for an Example for all the other special ones which be­long to our Synopsis, p. 354. (and from these special ones) to find a general one.

In Fig. 47. n. 4. make AD= b, DH= d; and so we shall have the □ of the radius HA= bb+ dd. But this □ HA or HN, may be also had otherwise, by putting,

2. For the quantity NO, as sought, the letter x, and by inferring from the known Property of the Parabola, as L to NO so NO to AO [...] and substracting AD= b from AO, you'l have DO or PN = [...]; whose □ is [...].

But you also have PH=DH−DP or NO, i. e. d− x; whose □ is therefore = dd−2 dx+ xx.

Wherefore adding the □ □ PN and PH, you'l have the □HN = [...].

Wherefore the Equation will now be readily had: (□ HN)—(□ HA) [...]; and taking from both sides [...]. and multiplying all by L (powerof2) and also dividing by [...].

+L (powerof2) x

[Page 94]3. And now farther comparing this Equation with a form like ours in Probl. 1. viz. with this, [...]; It is manifest, since in ours also the third term, viz. q is want­ing, that the correspondent one to it in the former −2 bL+L (powerof2) is equivalent to 0; and adding to both sides 2 bL, L (powerof2) will be made = 2 bL; and dividing by 2L, [...] will = b or AD. In like manner, since − r in our form answers to the quantity 2L (powerof2) d in the former; 2L (powerof2) d will = r, and, divid­ing by 2L (powerof2), DH or d will = [...]: Which is the other Mem­ber of the Central Rule to be found.

NB. The Analytick Art has this particularly to be admi­red in in, that it finds its own Rules by an Analysis: Where­of we have here an evident Example, and several others in the Resolution of affected Quadraticks, p. 345. and the follow­ing.

II. The Invention of the Central Rule, in the Case of Fig. 11. and the like.

1. □HA= bb+ dd as above.

2 Putting x for NO, as sought, we may infer from a new Property of the Parabola, which we have demonstrated Prop. 6. lib. 2

as L to NO so OR to AO, i. e. (putting a for BA or FO given; that NO−OF i. e. NF or OR shall be = x− a) as L to x so x− a to [...] = AO.

Therefore having substracted AD i. e. b from AO, you'l have DO or HP = [...]; whose □ is [Page] [...].

But PN also is = NO−PO or HD, = x− d is = [...].

Therefore having added the □ □ PH and PN, there will come out □ HN [...], = □ HA i. i. bb+ dd; and taking away from both sides [...]; and multiplying every where by L (powerof2) and diving by [...].

3. And now comparing this Equation with another form, which shall be like an Equation arising from the Solution of some Problem, e. g. with this [...]; to this you'l have = this other, [...].

4. Wherefore, because in these equal forms, first, 2 a is = p, a will be = [...] i. e. the line BA. Secondly, Because aa (or [...]) [...];

Therefore [...], and dividing by [...] = AD.

Thirdly, Because [...] i. e. [...]; therefore dividing by 2L (powerof2), d will be = [...] i. e. [Page 96] [...] i. e. resolving [...] into equivalent Terms ex­pressed by p and q, [...]; which is the other mem­ber of the Central Rule to be found. viz. a is = [...]

Therefore ab will be = [...]

Therefore [...].

THE INDEX OF THE FIRST BOOK.

SECTION I.

CHAP. I.
COntaining the Definitions or Explications of the Terms which relate to the Object of the Mathematicks. Page 1
CHAP. II.
Containing the Explication of those Terms which relate to the Affections of the Objects of the Mathematicks. p. 35

SECTION II.

CHAP. I.
Of the Composition and Division of Quantities. p. 55
CHAP. II.
Of the Powers of Quantities. p. 59
CHAP. III.
Of Progression or Arithmetical Proportionals. p. 66
[Page] CHAP. IV.
Of Geometrical Proportion in general. p. 69
CHAP▪ V.
Of the Proportions or Reasons of Magnitudes of the same kind in particular. p. 90
CHAP. VI.
Of the Proportions of Magnitudes of diverse sorts compared together. p. 114
CHAP. VII.
Of the Powers of the sides of Triangles and regular Fi­gures, &c. p. 12

BOOK II.

SECTION I.

  • COntaining Definitions. p. 154

SECTION II.

CHAP. I.
Of the Chief Properties of the Conick Sections. p. 166
CHAP. II.
Of Parabolical, and Hyperbolical, and Elliptical Spaces. p. 198
CHAP. III.
Of Conoids and Spheroids. p. 204
CHAP. IV.
Of Spiral Lines and Spaces. p. 208
CHAP. V.
Of the Conchoid, Cissoid, Cycloid, and Quadratrix, &c. p. 218
CHAP. VI.
The Epilogue of the whole Work. p: 231
  • An Introduction to Specious Analysis; or, The New Geome­try of Des Cartes, &c. pag. 1
  • Some Examples of Specious Analysis in each kind of Equati­ons. p. 16
    • 1. In Simple Equations. ibid.
    • 2. Some Examples of pure Quadratick Equations. p. 28
    • 3. Some Examples of Quadratick affected Equations. p. 41
    • 4. Some Examples of affected Biquadratick Equations, but like affected Quadratick ones. p. 53
    • 5. Some Examples of Cubick and Biquadratick Equations, both simple and affected, whether reducible or not. p. 61
  • Appendix. p. 93
FINIS.

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