THE ELEMENTS OF EUCLID Explained and Demonstrated in a New and most easie Method With the USES of each PROPOSITION In all the Parts of the MATHEMATICKS By Claude Francois Milliet D'Chales a Jesuite Done out of FRENCH Corrected and Augmented and Illustrated with Nine Copper Plates and the Effigies of EVCLID By Reeve Williams Philomath LONDON Printed for Philip Lea Globemaker at the Atlas and Hercules in the Poultrey near Cheapside 1685. To the Honourable Samuel Pepys Esq Principal Officer OF THE NAVY SECRETARY TO THE ADMIRALTY AND PRESIDENT OF THE ROYAL SOCIETY Honoured Sir THE Countenance and Encouragment you have always given to Mathematical Learning especially as it hath a tendency to Promote the Publique Good has emboldened me humbly to Present your Honour with this little Peice which hath the Admirable Euclid for its Author and the Learned D'Chales for the Commentator the excellency of the Subject with the Apt and Profitable Application thereof in its Vses did first induce me to Translate it for my own use the benefit and quickening in those Mathematical Studies that some professed to have received did prevail with me to make it Publique and the great Obligations I lie under from the many undeserved Favours of your Honour toward me I thought did engage me on this occasion to make some Publique Testimony and Acknowledgment thereof I therefore Humbly beg your Honours Patronage of this little Book and your Pardon for this Address intreating you will be pleased to look upon it with that Benign Aspect as you have been pleased always to vouchsafe to him who is Your Humble and most Obliged Servant REEVE WILLIAMS THE Authors Preface TO THE READER HAving long since observed that the greatest part of those that learn Euclid's Elements are very often dissatisfied therewith because they know not the use of Propositions so inconsiderable in appearance and yet so difficult I thought it might be to good purpose not only to make them as easie as possible but also to add some Vses after each Proposition to shew how they are applicable to Practice And this hath obliged me to change some of the Demnostrations which I looked upon to be too troublesome and above the usual reach of beginners and to substitute others more intelligible For this cause I have Demonstrated the Fifth Book after a method much more clear than that that makes use of equimultiples I have not given all the Vses of the Propositions for that would have made it necessary to recite all the Mathematicks and would have made the Book too big and too hard Wherefore I have only made choice of some of the plainest and easiest to conceive I would not have you to stand too much upon them nor make it your study to understand them perfectly because they depend upon other principles besides for which Reason I have distinguished them with the Italick Letter This is the design of this small Treatise which I willingly publish in a time when the Mathematicks are more than ever studied Milliet D'chales THE TRANSLATORS PREFACE THE Reader having perused the Authors Preface with this farther intimation that he will find the Subject and Scope of this work Succinctly and pertinently presented to him in the Argument before each particular book may I presume expect the loss from me and I shall not at all endeavor to bespeak the Readers acceptance of Euclid's Elements or persuade him to beleive the Necessary and Excellent Vsefulness thereof because every mans Experience so far as he understands them is an abundant Testimony thereunto Neither shall I need to commend the Method with the uses of our Author D'Chales who is well known to the learned of this Age by his Several excellent Mathematical Tracts * Cursus Mathe his Navigation his Local Motion c. for whosoever shall be a while conversant with this book may I presume fiind that instruction and incourgment in the learning of Euclid's Eliments as he hath not before met with in our English Tongue And this as it hath been my own experience since I Translated it from the French for the use of my English Schollars so it is one great cause of its coming abroad into the World for such as had Learned by it found it difficult to attain unless Transcribed which they thought tedious being a subject so Voluminous in Manuscript and full of Schemes I did therefore at their request and the importunity of some Friends condescend to the Printing hereof though not without much averseness to my own mind being unwilling to Expose my self in any Publick thing in this nice Critical Age But that difficulty being now overcome I shall only give the Reader to understand that I have faithfully rendred this Piece into English according to the sense of the Author but here and there omitting some small matters which I judged not so properly related to the subject of this Work and therein will make no want to the Reader nor I hope be any offence to the Ingenious Author himself I have only one thing more to add and that upon the account of an Objection I have met with that here is not all the Books of Euclid and it is true here is not all here are only the first six Books and the Eleventh and Twelfth the other being purposely omitted by our learned Author who judged the understanding of these to be sufficient for all the parts of the Mathematicks * See Argument before the Eleventh Book Page 304. and I could also give Instances of other excellent Authors that are of his Opinion and have taken the like course nay the truth is some very learned in the Mathematicks have reduced the Propositions of these Books to a much lesser Number and yet have thought them a compleat foundation to all the Sciences Mathematical but I shall not trouble my Reader farther on this account not doubting but when he hath perused and well considered our Euclid he will have a better Opinion thereof then any thing I can now say may Justly hope to beget in him and so I shall submit my whole Concernment herein to the Impartial Reader and remain ready to Serve him Reeve Williams Form my School at the Virginia Coff-house in St. Michaels Alley in Cornhill ADVERTISEMENT VVHereas in the French all the Definitions which needed all the Propositions as also all those uses our Author thought fit to Illustrate by Schemes were done in the Book in wood here they are in Copper plates to be placed at the end of the Book The Definitions and uses are in the Plates marked with Arithmetical Charactars or Ziphers but the Propsitions in Alphabetical Ziphers at the begining of the Book or at least when you come to the first Definition you are refâ—rred to the number of the Plate in Which you shall find the Scheme proper thereto as also the Def Prop. and uses belonging unto the Book unless the Plate could not contain them and then you are referred to the next Plate
inscribed by the little Rectangles through which the Circumference of the Circle passeth and all those Rectangles taken together are equal to the Rectangle AL. Imagine that the Semi-Circle is made to roul about the Diameter EB the Semi-Circle shall describe a hemi-sphere and the inscribed Rectangles will describe inscribed Cylinders in the semi-sphere and the Circumscribed will describe other Cylinders Demonstration The Circumscribed Cylinders surpass more the inscribed than doth the hemi-sphere surpass the same inscribed Cylinders seeing that they are comprehended within the Circumscribed Cylinders Now the Circumscribed surpass the inscribed by so much as is the Cylinder AL therefore the hemi-sphere shall surpass by less the inscribed Cylinders than doth the Cylinder made by the Rectangle AL. The Cylinder AL is less than the Cylinder MP for there is the same Ratio of a great Circle of the Sphere which serveth for Base to the Cylinder AL as of MP to R so then by the foregoing a Cylinder which hath for Base a great Circle of the Sphere and the Altitude R would be equal to the Cylinder MP Consequently the hemi-sphere which surpasseth the quantity D by the Cylinder MP and the inscribed Cylinders by a quantity less than AL surpasseth the inscribed Cylinders by less than the quantity D. Therefore the quantity D is less than the inscribed Cylinders What I have said of a hemi sphere may be said of a whole Sphere LEMMA II. LIke Cylinders inscribed in Two Spheres are in Triple Ratio of the Diameters of the Spheres Lemma Fig. II. If the two like Cylinders CD EF be inscribed in the Spheres A B they shall be in Triple Ratio of the Diameters LM NO Draw the Lines GD IF Demonstration The like Right Cylinders CD EF are like so then there is the same Ratio of HD to DR as of QF to FS as also the same Ratio of KD to KG as of PF to PI. And consequently the Triangles GDK IFP are like by the 6th of the 6th so there shall be the same Ratio of KD to PF as of GD to IF or of LM to ON Now the like Cylinders CD EF are in Triple Ratio of KD and PF the Diameters of their Bases by the 12th therefore the like Cylinders CD EF inscribed in the Spheres A and B are in triple Ratio of the Diameters of the Spheres PROPOSITION XVIII THEOREM SPheres are in triple Ratio of their Diameters The Spheres A and B are in Triple Ratio of their Diameters CD EF. For if they be not in Triple Ratio one of the Spheres as A shall be in a greater Ratio than Triple of that of CD to EF therefore a quantity G less than the Sphere A shall be in Triple Ratio of that of CD to EF and so one might according to the first Lemma inscribe in the Sphere A Cylinders of the same height greater than the quantity G. Let there be inscribed in the Sphere B as many like Cylinders as those of the Cylinder A. Demonstration The Cylinders of the Sphere A to those of the Sphere B shall be in Triple Ratio of that of CD to EF by the preceding Now the quantity G in respect of the Sphere B is in Triple Ratio of CD to EF there is then the same Ratio of the Cylinders of the Sphere A to the like Cylinders of the Sphere B. So then the Cylinders of A being greater than the quantity G the Cylinder B that is to say inscribed in the Sphere B would be greater than the Sphere B which is impossible Therefore the Spheres A and B are in Triple Ratio of that of their Diameters Coroll Spheres are in the same Ratio as are the Cubes of their Diameters seeing that the Cubes being like solids are in Triple Ratio of that of their Sides FINIS ERRATA PAge 14. Line 21. read AFD p. 23. l. 17. for DF r. EF. l. 16. r. DFE l. 27. r. FD. p. 24. l. 1 and 9. for FB r. FD. p. 26. l. 8. r. HI p. 52. l. 14. r. ACB p. 55. l. 22. r. ACF p. 71. l. 3. r. ACB p. 82. l. 19. r. ABC p. 117. l. 13. r. GFE p. 125. l. 9. r. AD. p. 178. l. 16. r. CFD p. 194. l. ult r. AF. p. 197. l. 2. r. CDA p. 218. l. 7. r. ⅓ and ⅓ p. 268. l. 24. r. CE. p. 281. l. 3. r. ECD p. 289. l. 4. r. are simular p. 294. l. 13. r. eight ¾ p. 309. l. 15. r. 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The Elements of Euclide explained and demonstrated after a new and most easie method with the uses of each Proposition in all the parts of the Mathematicks by Claude Francois Millet Dechales a Jesuite 17. New Maps of the World four quarters and of all the Countries and of all sizes made according to the latest discoveries extant may be had pasted upon Cloath and Coloured also Sea Plats Mathematical Projection Books and Instruments whatsoever are made and sold by Philip Lea. FINIS EUCLIDS ELEMENTS with the uses EUCLID London Printed for Phillip Lea Globe Maker at the Atlas and Hercules in Cheapside Near Friday Street There all Sorts of Globes Spheres Maps Sea-plats Mathematical Books and Instruments are Made and Sold Plate 2. Propositions and Uses of the first Book See Plate 3. Plate 1. Definitions of the first Book Propositions and Uses of the first Book See Plate 2. Plate 3. Propositions Uses of the first Book Definitions Propositions Uses of the Second Book Plate 4 Definitions of the third Book Propositions Propositions and Uses Plate 5. Definitions Propsitions of the fouth Book Plate 6. Definitions of the Sixth Book Propositions Uses Plate 7. Definitions of the Eleventh Book Propositions Plate 8. Proposi of the Twelfth Book
the double Area of the Triangle I make use thereof in several other Propositions as in the seventh PROPOSITION XIV PROBLEM TO describe a Square equal to a right lined Figure given To make a Square equal to a Right lined Figure A make by the 45th of the 1st a Rectangle BC DE equal to the Right lined Figure A if its sides CD DC were equal we should have already our desire if they be unequal continue the Line BC until CF be equal to CD and dividing the Line BF in the middle in the Point G describe the Semicircle FHB then continue DC to H the Square of the Line CH is equal to the Right lined Figure A draw the Line GH Demonstration The Line BF is equally divided in G and unequally in C thence by the 5th the Rectangle comprehended under BC CF or CD that is to say the Rectangle BD with the Square CG is equal to the Square of GB or to its Equal GH Now by the 47th of the 1st the Square of GH is equal to the Square of CH CG therefore the Rectangle BD and the Square of CG is equal to the Squares of CG and of CH and taking away the Square of CG which is common to both the Rectangle BD or the Right lined Figure A is equal to the Square of CH. USE THis Proposition serveth in the first place to reduce into a Square any Right lined Figure whatever and whereas a Square is the first Measure of all Superficies because its Length and Breadth is equal we measure by this means all right lined Figures In the second place this Proposition teacheth us to find a mean Proportion between two given Lines as we shall see in the Thirteenth Proposition of the Sixth Book This Proposition may also serve to square curve lined Figures and even Circles themselves for any crooked or curve lined Figure may to sence be reduced to a Right lined Figure as if we inscribe in a Circle a Polygon having a thousand sides it shall not be sensibly different from a Circle and reducing the Polygone into a Square we square nearly the Circle THE THIRD BOOK OF Euclid's Elements THis Third Book explaineth the Propriety of a Circle and compareth the divers Lines which may be drawn within and without its Circumference It farther considereth the Circumstances of Circles which cut each other or which touch a streight Line and the different Angles which are made as well those in their Centers as in their Circumferences In fine it giveth the first Principles for establishing the Practice of Geometry by the which we make use and that very commodiously of a Circle in almost all Treatises in the Mathematicks DEFINITIONS 1. Def. of the 3 Book Those Circles are equal whose Diameters or Semidiameters are equal 2. Fig. 1. A Line toucheth a Circle when meeting with the Circumference thereof it cutteth not the same as the Line AB 3. Fig. 2. Circles touch each other when Meeting they cut not each other as the Circles AB and C. 4. Fig. 3. Right Lines in a Circle are equally distant from the Center when Perpendiculars drawn from the Center to those Lines are equal As if the Lines EF EG being Perpendiculars to the Lines AB CD are equal AB CD shall be equally distant from the Center because the Distance ought always to be taken or measured by Perpendicular Lines 5. Fig. 4. A Segment of a Circle is a Figure terminated on the one side by a streight Line and on the other by the Circumference of a Circle as LON LMN 6. The Angle of a Segment is an Angle which the Circumference maketh with a streight Line as the Angle OLN LMN 7. Fig. 5. An Angle is said to be in a Segment of a Circle when the Lines which form the same are therein as the Angle FGH is in the Segment FGH 8. Fig 6. An Angle is upon that Arch to which it is opposite or to which it serveth for a Base as the Angle FGH is upon the Arch FIH which may be said to be its Base 9. Fig. 6. A Sector is a Figure comprehended under two Semidiameters and under the Arch which serveth to them for a Base as the Figure FIGH PROPOSITION I. PROBLEM To find the Center of a Circle IF you would find the Center of the Circle AEBD draw the Line AB and divide the same in the middle in the Point C at which Point erect a Perpendicular ED which you shall divide also equally in the Point F. This Point F shall be the Center of the Circle for if it be not imagine if you please that the Point G is the Center draw the Lines GA GB GC Demonstration If the Point G were the Center the Triangles GAC GBC would have the sides GA GB equal by the definition of a Circle AC CB are equal to the Line AB having been divided in the middle in the Point C. And CG being common the Angles GCB GCA would then be equal by the 8th of the 1st and CG would be then a Perpendicular and not CD which would be contrary to the Hypothesis Therefore the Center cannot be out of the Line CD I further add that it must be in the Point F which divideth the same into two equal Parts otherwise the Lines drawn from the Center to the Circumference would not be equal Corollary The Center of a Circle is in a Line which divideth another Line in the middle and that perpendicularly USE THis first Proposition is necessary to demonstrate those which follow PROPOSITION II. THEOREM A Streight Line drawn from one point of the Circumference of a Circle to another shall fall within the same Let there be drawn from the Point B in the Circumference a Line to the Point C. I say that it shall fall wholly within the Circle To prove that it cannot fall without the Circle as BVC having found the Center thereof which is A draw the Lines AB AC AV. Demonstration The Sides AB AC of the Triangle ABC are equal whence by the 5th of the 1st the Angles ABC ACB are equal And seeing the Angle AVC is exteriour in respect of the Triangle AVB it is greater than ABC by the 16th of the 1st it shall be also greater than the Angle ACB Thence by the 19th of the 1st in the Triangle ACV the side AC opposite to the greatest Angle AVC is greater than AV and by consequence AV cannot reach the circumference of the Circle seeing it is shorter than AC which doth but reach the same wherefore the Point V is within the Circle the same may be proved of any Point in the Line AB and therefore the whole Line AB falls within the Circle USE IT is on this Proposition that are grounded those which demonstrate that a Circle toucheth a streight Line but only in one Point for if the Line should touch two Points of its Circumference it would be then drawn from one
when the Terms between them are taken twice that is to say as antecedent and as consequent As if there be the same Reason of A to B as of B to C and of C to D. 11. Then A to C shall be in duplicate Ratio of A to B and the Ratio of A to D shall be in triplicate Ratio to that of A to B. It is to be taken notice that there is a great deal of difference between double Ratio and duplicate Ratio We say that the Ratio of four to two is double that is to say four is the double of two whence it followeth that the number two is that which giveth the Name to this Ratio or rather to the Antecedent of this Ratio So we we say double triple quadruple quintuple which are Denominations taken from those numbers duo tres quatuor quinque compared with unity for we better conceive a Reason when its terms are small But as I have already taken notice those Denominations fall rather on the Antecedent than on the Reason it self we call that double triple Reason or Ratio when the Antecedent is double or triple to the consequent but when we say the Reason is duplicate we mean a Reason compounded of two like Reasons as if there be the same Reason of two to four as of four to eight the Reason of two and eight being compounded of the Reason of two and four and of that of four and eight which are alike and as equal the Reason or Ratio of two to eight shall be duplicated by each Three to twenty seven is a duplicated Reason of that of three to nine The Reason of two to four is called subduple that is to say two is the half of four but the reason of two to eight is duple of the sub-duple that is to say that two is the half of the half of eight as three is the third of the third of twenty seven where you see there is taken twice the Denominator ½ and ⅓ In like manner eight to two is a duplicate reason of eight to four because eight is double to four but eight is the double of the double of two If there be four terms in the same continued Reason that of the first and last is triple to that of the first and second as if one put these four Numbers two four eight sixteen the reason of two to sixteen is triple of two to four for two is the half of the half of the half of sixteen As the reason of sixteen to two is triple of sixteen to eight for sixteen is the double of eight and it is the double of the double of the double of two 12. Magnitudes are homologous the Antecedents to the Antecedents and the Consequents to the Consequents As if there be the same Reason of A to B as of C to D A and C are homologous or Magnitudes of a like Ratio The following Definitions are ways of arguing by Proportion and it is principally to demonstrate the same that this Book is composed 13. Alternate Reason or by Permutation or Exchange is when we compare the Antecedents one with the other as also the consequents For example if because there is the same reason of A to B as of C to D I conclude there is the same reason of A to C as of B to D this way of reasoning cannot take place but when the four terms are of the same Specie that is to say either all four Lines or Superficies or Solids Proposition 16. 14. Converse or Inverse Reason is a comparison of the Consequents to the Antecedents As if because there is the same reason of A to B as of C to D I conclude there is the same reason of B to A as there is of D to C. Proposition 16. 15. Composition of Reason is a comparison of the Antecedent and Consequent taken together to the Consequent alone As if there be the same Reason of A to B as of C to D I conclude also that there is the same reason of AB to B as of CD to D. Prop. 18. 16. Division of Reason is a comparison of the excess of the Antecedent above the Consequent to the same Consequent As if there be the same reason of AB to B as of CD to D I conclude that there is the same reason of A to B as of CD Prop. 17. 17. Conversion of Reason is the comparison of the Antecedent to the difference of the Terms As if there be the same reason of AB to B as of CD to D I conclude that there is the same reason of AB to A as of CD to C. Proposition 18. 18. Proportion of Equality is a comparison of the extream Quantities in leaving out those in the middle A B C D E F G H. As if there were the same reason of A to B as of E to F and of B to C as of F to G and of C to D as of G to H. I draw this Consequence that there is therefore the same reason of A to D as of E to H. 19. Proportion of Equality well ranked is that in which one compareth the Terms in the same manner of Order as in the preceding Example Prop. 22. 20. Proportion of Equality ill ranked is that in which one compareth the Terms with a different Order As if there were the same reason of A to B as of G to H and of B to C as of F to G and of C to D as of E to F. I draw this Conclusion that there is the same reason of A to D as of E to H. Prop. 28. Here is all the ways of arguing by Proportion There is the same reason of A to B as of C to D therefore by alternate reason there is the same reason of A to C as of B to D and by inversed reason there is the same Reason of B to A as of D to C and by composition there is the same reason of AB to B as of CD to D. By Division of Reason if there be the same reason of AB to B as of CD to D there is the same reason of A to B as of C to D and by Conversion there is the same reason of AB to A as of CD to C. By reason of Equality well ranked if there be the same reason of A to B as of C to D and also the same reason of B to E as of D to F there will be the same reason of A to E as of C to F. By reason of Equality ill ranked if there be the same reason of A to B as of D to F and also the same reason of B to E as of C to D there will be the same reason of A to E as of C to F. This Book contains twenty five Propositions of Euclid to which there has been added ten which are received The first six of this Book are useful only to prove the following Propositions by the method
whereon the head thereof you shall find the Book it belongeth to and the Propositions or uses continued in their order EIGHT BOOKS OF Euclid's Elements With the Vses of each PROPOSITION The FIRST BOOK EUCLID's Design in this Book is to give the first Principles of Geometry and to do the same Methodically he begins with the Definitions and Explication of the most ordinary Terms then he exhibits certain Suppositions and having proposed those Maxims which natural Reason teacheth he pretends to put forward nothing without Demonstration and to convince any one which will consent to nothing but what he shall be obliged to acknowledge in his first Propositions he treateth of Lines and of the several Angles made by their intersecting each other and having occasion to Demonstrate their Proprieties and compare certain Triangles he doth the same in the First Eight Propositions then teacheth the Practical way of dividing an Angle and a Line into two equal parts and to draw a Perpendicular he pursues to the propriety of a Triangle and having shewn those of Parallel Lines he makes an end of the Explication of this First Figure and passeth forwards to Parallelograms giving a way to reduce all sorts of Polygons into a more Regular Figure He endeth this Book with that Celebrated Proposition of Pythagoras and Demonstrates that in a Rectangular Triangle the Square of the Base is equal to the sum of the Squares of the Sides including the Right Angle DEFINITIONS 1. A Point is that which hath no part This Definition is to be understood in this sence The quantity which we conceive without distinguishing its parts or without thinking that it hath any is a Mathematical Point far differing from those of Zeno which were alltogether indivisible since one may doubt with a great deal of Reason if those last be possible which yet we cannot of the first if we conceive them as we ought 2. A Line is a length without breadth The sense of this Definition is the same with that of the foregoing the quantity which we consider having length without making any reflection on its breadth is that we understand by the word Line although one cannot draw a real Line which hath not a determinate breadth it is generally said that a Line is produced by the motion of a Point which we ought well to take notice of seeing that by a motion after that manner may be produced all sorts of quantities imagine then that a Point moveth and that it leaveth a trace in the middle of the way which it passeth the Trace is a Line 3. The Two ends of a Line are Points 4. A streight Line is that whose Points are placed exactly in the midst or if you would rather have it a streight Line is the shortest of all the Lines which may be drawn from one Point to another 5. A Superficies is a quantity to which is given length and breadth without considering the thickness 6. The extremities of a Superficies are Lines 7. A plain or straight Superficies is that whose Lines are placed equally between the extremities or that to which a streight Line may be applyed any manner of way Plate I. Fig. 1. I have already taken notice that motion is capable of producing all sorts of quantity whence we say that when a Line passeth over another it produces a superficies or a Plain and that that motion hath a likeness to Arithmetical Multiplication imagine that the Line AB moveth along the Line BC keeping the same situation without inclining one way or the other the Point A shall describe the Line AD the Point B the Line BC and the other Points between other Parallel Lines which shall compose the Superficies ABCD. I add that this motion corresponds with Arithmetical Multiplication for if I know the number of Points which are in the Lines AB BC Multiplying of them one by the other I shall have the number of Points which Composeth the Superficies ABCD as if AB contains four points and BC six saying Four times Six are Twenty Four the Superficies AB CD should be Composed of Twenty Four Points Now I may take for a Mathematical Point any quantity whatsoever for Example a Foot provided I do not subdivide the same into Parts 8. A plain Angle is the opening of Two Lines which intersect each other and which Compose not one single Line Fig. 2. As the opening D of the Lines AB CB which are not parts of the same Line A Right Lined Angle is the opening of two streight Lines It is principally of this sort of Angles which I intend to treat of at present because experience doth make me perceive that the most part of those who begin do mistake the measuring the quantity of an Angle by the length of the Lines which Composeth the same Fig. 3 4. The most open Angle is the greatest that is to say when the Lines including an Angle are farther asunder than those of another Angle taking them at the same distance from the Points of intersection of their Lines the first is greater than the Second so the Angle A is greater than E because if we take the Points B and D as far distant from the Point A as the Points G and L are from the Points E the Points B and D are farther asunder than the Points G and L from whence I conclude that if EG EL were continued the Angle E would be of the same Measure and less than the Angle A. We make use of Three Letters to express an Angle and the Second Letter denotes the Angular Point as the Angle BAD is the Angle which the Lines BA AD doth form at the Point A the Angle BAC is that which is formed by the Lines BA AC the Angle CAD is comprehended under the Line CA AD. Fig. 3. The Arch of a Circle is the measure of an Angle thus designing to measure the quantity of the Angle BAD I put one Foot of the Compasses on the Point A and with the other I describe an Arch of a Circle BCD the Angle shall be the greater by how much the Arch BCD which is the measure thereof shall contain a greater portion of a Circle and because that commonly an Arch of a Circle is divided into Three Hundred and Sixty equal Parts called Degrees It is said that an Angle containeth Twenty Thirty Forty Degrees when the Arch included betwixt its Lines contains Twenty Thirty Forty Degrees so the Angle is greatest which containeth the greatest number of Degrees As the Angle BAD is greater than GEL the Line CA divideth the Angle BAD in the middle because the Arches BC CD are equal and the Angle BAC is a part of BAD because the Arch BC is part of the Arch BD. 10. When a Line falling on another Line maketh the Angle on each side thereof equal Those Angles are Right Angles and the Line so falling is a Perpendicular Fig. 5. As if the Line AB falling on CD
side 13. The streight Lines have not the same common Segment Fig. 20. I would say that of two streight Lines AB CD which meeteth each other in the Point B is not made one single Line BD but that they cut each other and separate after their so meeting For if a Circle be described on the Center B AFB shall be a Semicircle seeing the Line ABD passing through the Center B divideth the Circle into two equally the Segment CFD should be also a Semicircle if CBD were a streight Line because it passeth through the Center B therefore the Segment CFD should be equal to the Segment AFD a part as great as the whole which would be contrary to the Ninth Axiom ADVERTISEMENT WE have two sorts of Propositions some whereof considereth only a truth without descending to the practice thereof and we call those Theorms The other proposeth something to be done or made and are called Problems The first numbers of Citations is that of the Proposition the second that of the Book As by the 2 of the 3 is signified the second proposition of the third Book but if one meet with one number thereby is meant the Proposition of the Book whose Explication is in hand PROPOSITION I. PROBLEM UPon a finite Right Line AB to describe an Equilateral Triangle ACB From the Centers A and B at the distance of AB describe the Circles cutting each other in the Point C from whence draw two Right Lines CA CB thence are AC AB BC AC equal wherefore the Triangle ACB is equilateral which was to be done USE EUclid has not applyed this Proposition to any other use but to demonstrate the two following Propositions but we may apply it to the measuring of an inaccessible Line Use 1. As for Example let AB be an inaccessable Line which is so by reason of a River or some other Impediment make an Equaliteral Triangle as BDE on Wood or Brass or on some other convenient thing which having placed Horizontally at a station at B look to the Point A along the Side BD and to some other Point C along the side BE then carry your Triangle along the Line BC so far that is untill such time as you can see the Point B your first station by the side CG and the Point A by the Side CE I say that then the Lines CB and CA are equal wherefore if you measure the Line BC you will likewise know the length of the Line AB PROPOSITION II. PROBLEM AT a Point given A to make a Right Line AC equal to a Right Line given BC. From the Center C at the distance CB describe the Circle CBE joyn AC upon which raise the Equilateral Triangle ADC produce DC to E from the Center D and the distance DE describe the Circle DEH Let DA be produced to the Point G in the Circumference thereof then AG is equal to CB for DG is equal to DE and DA to DC wherefore AG CE BC AG are equal which was to be done SCHOL THe Line AG might be taken with a pair of Compasses but the so doing answers no Postulate as Proclus well intimates PROPOSITION III. PROBLEM TWo Right Lines A and BC being given from the greater BC to take away the Right Line BE equal to the lesser A. At the Point B draw the Right Line BD equal to A the Circle described from the Center B at the distance BD shall cut off BE equal to BD equal to A equal to BE which was to be done The use of the two preceding Propositions are very evident since we are obliged very often in our Geometrical Practices to draw a Line equal to a Line given or to take away from a greater Line given a part equal to a lesser PROPOSITION IV. THEOREM IF Two Triangles ABC EDF have two Sides of the one BA AC equal to two Sides of the other ED DF each to his correspondent side that is BA to ED and AC to DF and have the Angle A equal to the Angle D contained under the equal Right Lines they shall have the Base BC equal to the Base EF and the Triangle BAC shall be equal to the Triangle EDF and the remaining Angles B C shall be equal to the remaining Angles E F each to each which are subtended by the equal Sides If the Point D be applied to the Point A and the Right Line DE placed on the Right Line AB the Point E shall fall upon B because DE is equal to AB also the Right Line DF shall fall upon AC because the Angle A is equal to D Moreover the Point F shall fall on the Point C because AC is equal to DF Therefore the Right Lines EF BC shall agree because they have the same terms and so consequently are equal wherefore the Triangle BAC is equal to DEF and the Angles B E as also the Angles C F do agree and are equal which was to be demonstrated USE Use 4. SVppose I was to measure the inaccessible Line AB I look from the Point C to the Point A and B then I measure the Angle C thus I place a Board or Table Horizontally and looking successively with a Ruler towards the Points A and B I draw two Lines making the Angle ACB then I measure the Lines AC and BC which I suppose to be accessible I turn about my Board or Table towards some other place in the Field placing it again Horizontally at the Point F and looking along those Lines I have drawn on my Table I make the Angle DEF equal to the Angle C I make also FD FE equal to CA CB Now according to this Proposition the Lines AB DE are equal wherefore whatsoever the Length of the Line DE is found to be the same is the measure of the inaccessible Line AB Another USE may be this Use 4. SVppose you were at a Billiard Table and you would strike a Ball B by reflection with another Ball A Admit CD be one Side of the Table now imagine a perpendicular Line BDE I take the Line DE equal to DB I say that if you aim and strike your Ball A directly towards the Point E the Ball A meeting the Side of the Table at F shall reflect from thence to B for in the Triangle BFD EFD the Side FD is common and the Sides BD DE equal the Angles BFD EFD equal by the Proposition the Angles AFC DFE being opposite are also equal as I shall demonstrate hereafter therefore the Angle of incidence AFC is equal to the Angle of reflection BFD and by consequence the reflection will be from AF to FB PROPOSITION V. THEOREM IN every Isosceles Triangle the Angles which are above the Base are equal as also those which are underneath Let ABC be an Isosceles Triangle viz that the side AB AC be equal I say that the Angle ABC ACB are equal as also the Angles GBC HCB which ly under
under AB and AC shall be Three times 8 or 24 the square of AC 3 is 9 the Rectangle comprehended under AC 3 and CB 5 is 3 times 5 or 15. It is evident that 15 and 9 are 24. USE A Â 43 C 40. 3 B Â 3 120. Â 9. 129 Â Â THis Proposition serveth likewise to Demonstrate the ordinary practice of Multiplication For Example if one would Multiply the Number 43 by 3 having separated the Number of 43 into two parts in 40 and 3 three times 43 shall be as much as three times 3 which is Nine the Square of Three and Three times Forty which is 120 for 129 is Three Times 43. Those which are young beginners ought not to be discouraged if they do not conceive immediately these Propositions for they are not difficult but because they do imagine they contain some great Mystery PROPOSITION IV. THEOREM IF a Line be Divided into Two Parts the Square of the whole Line shall be equal to the Two Squares made of its parts and to Two Rectangles comprehended under the same parts Let the Line AB be Divided in C and let the Square thereof ABDE be made let the Diagonal EB be drawn and the Perpendicular CF cutting the same and through that Point let there be drawn GL Parallel to AB It is evident that the Square ABDE is equal to the Four Rectangles GF CL CG LF The Two first are the Square of AC and of CB the Two Complements are comprehended under AC CB. Demonstration The Sides AE AB are equal thence the Angles AEB ABE are half Right and because of the Parallels GL AB the Angles of the Triangles of the Square GE by the 29th shall be equal as also the Sides by the 6th of the 1. Thence GF is the Square of AC In like manner the Rectangle CL is the Square of CB the Rectangle GC is comprehended under AC and AG equal to BL or BC the Rectangle LF is comprehended under LD equal to AC and under FD equal to BC. Coroll If a Diagonal be drawn in a Square the Rectangles through which it passeth are Squares USE A 144 B 22 C 12 THis Proposition giveth us the practical way of finding or extracting the Square Root of a Number propounded Let the same be the number A 144 represented by the Square AD and its Root by the Line AB Moreover I know that the Line required AB must have Two Figures I therefore imagine that the Line AB is Divided in C and that AC representeth the first Figure and BC the Second I seek the Root of the First Figure of the Number 144 which is 100 and I find that it is 10 and making its Square 100 represented by the Square GF I Subtract the same from 144 and there remains 44 for the Rectangles GC FL and the Square CL. But because this gnomonicall Figure is not proper I transport the Rectangle FL in KG and so I have the Rectangle KL containing 44. I know also almost all the Length of the Side KB for AC is 10 therefore KC is 20 I must then Divide 44 by 20 that is to say to find the Divisor I double the Root found and I say how many times 20 in 44 I find it 2 times for the Side BL but because 20 was not the whole Side KB but only KC this 2 which cometh in the Quotient is to be added to the Divisor which then will be 22. So I find the same 2 times precisely in 44 the Square Root then shall be 12. You see that the Square of 144 is equal to the Square of 10 to the Square of 2 which is 4 and to twice 20 which are Two Rectangles comprehended under 2 and under 10. PROPOSITION V. THEOREM IF a Right Line be cut into equal parts and into unequal parts the Rectangle comprehended under the unequal parts together with the Square which is of the middle part or difference of the parts is equal to the Square of half the Line If the Line AB is Divided equally in C and unequally in D the Rectangle AH comprehended under the Segments AD DB together with the Square of CD shall be equal to the Square CF that is of half of AB viz. CB. Make an end of the Figure as you see it the Rectangles LG DI shall be Squares by the Coroll of the 4th I prove that the Rectangle AH comprehended under AD and DH equal to DB with the Square LG is equal to the Square CF. Demonstration The Rectangle AL is equal to the Rectangle DF the one and the other being comprehended under half the Line AB and under BD or DH equal thereto Add to both the Rectangle CH the Rectangle AH shall be equal to the Gnomon LBG Again to both add the Square LG the Rectangle AH with the Square LG shall be equal to the Square CF. ARITHMETICALLY LEt AB be 10 AC is 5 as also CB. Let CD be 2 and DB 3 the Rectangle comprehended under AD 7 and DB 3 that is to say 21 with the Square of CD 2 which is 4 shall be equal to the Square of CB 5 which is 25. USE THis Proposition is very useful in the Third Book we make use thereof in Algebra to Demonstrate the way of finding the Root of an affected Square or Equation PROPOSITION VI. THEOREM IF one add a Line to another which is Divided into Two equal parts the Rectangle comprehended under the Line compounded of both and under the Line added together with the Square of half the Divided Line is equal to the Square of a Line compounded of half the Divided Line and the Line added If one add the Line BD to the Line AB which is equally Divided in C the Rectangle AN comprehended under AD and under DN or DB with the Square of CB is equal to the Square of CD Make the Square of CD and having drawn the Diagonal FD draw BG Parallel to FC which cuts FD in the Point H through which passeth HN Parallel to AB KG shall be the Square of BC and BN that of BD. Demonstration The Rectangles AK CH on equal Bases AC BC are equal by the 38th of the 1st The Complements CH HE are equal by the 43d of the 1st Therefore the Rectangles AK HE are equal Add to both the Rectangle CN and the Square KG the Rectangles AK CN that is to say the Rectangle AN with the Square KG shall be equal to the Rectangles CN HE and to the Square KG that is to say to the Square CE. Arithmetically or by Numbers LEt AB be 8 AC 4 CB 4 BD 3 then AD shall be 11. It is evident that the Rectangle AN three times 11 that is to say 33 with the Square of KG 16 which together are 49 is equal to the Square of CD 7 which is 49 for 7 times 7 is 49. USE Fig. 6. MAurolycus measured the whole Earth by one single
Square of the Line added are double to the Square of half the Line and to the Square which is Composed of the half Line and the Line added If one supposeth AB to be Divided in the middle at the Point C and if thereto be added the Line BD the Squares of AB and BD shall be double to the Squares of AC and CD added together Draw the Perpendiculars CE DF equal to AC Then draw the Lines AE EF AG EBG Demonstration The Lines AC CE CB beng equal and the Angles at the Point C being Right The Angles AEC CEB CBE DBG DGB shall be half Right and the Lines DB DG and EF FG CD shall be equal The Square of AE is double to the Square of AC the Square of EG is double to the Square of EF or CD by the 47th of the 1st Now the Square of AG is equal to the Squares of AE EG by the 47th of the 1st Therefore the Square of AG is double to the Squares of AC CD The same AG by the 47th of the 1st is equal to the Squares of AD BD or GD Therefore the Squares of AD BD are double the Squares of AC CD ARITHMETICALLY LEt AB be 6 parts AC 3 CB 3 BD 4 the Square of AD 10 is 100. the Square of BD 4 is 16 which are 116. The Square of AC 3 is 9 the Square of CD 7 is 49. Now 49. and 9 is 58 the half of 116. PROPOSITION XI PROBLEM TO Divide a Line so that the Rectangle comprehended under the whole Line and under one of its parts shall be equal to the Square of the other part It is proposed to Divide the Line AB so that the Rectangle comprehended under the whole Line AB and under HB be equal to the Square of AH Make a Square of AB by the 46th of the 1st Divide AD in the middle in E then Draw EB and make EF equal to EB Make the Square AF that is to say that AF AH be equal I say that the Square of AH shall be equal to the Rectangle HC comprehended under HB and the Line BC equal to AB Demonstration The Line AD is equally divided in E and there is added thereto the Line FA thence by the 6th the Rectangle DG comprehended under DF and FG equal to AF with the Square of AE is equal to the Square of EF equal to EB Now the Square of EB is equal to the Squares of AB AE by the 47th of the 1st therefore the Squares of AB AE are equal to the Rectangle DG and to the Square of AE and taking away from both the Square of AE the Square of AB which is AC shall be equal to the Rectangle DG taking also away the Rectangle DH which is in both the Rectangle HC shall be equal to the Square of AG. USE THis Proposition serveth to cut a Line in extream and mean Proportion as shall be shewn in the Sixth Book It is used often in the 14th of Euclid's Elements to find the Sides of Regular Bodies It serveth for the 10th of the Fourth Book to inscribe a Pentagone in a Circle as also a Pentadecagone You shall see other uses of a Line thus divided in the 30th of the Sixth Book PROPOSITION XII THEOREM IN an obtuse angled Triangle the Square of the side opposite to the obtuse Angle is equal to the Squares of the other two sides and to two Rectangles comprehended under the side on which one draweth a Perpendicular and under the Line which is between the Triangle and that perpendicular Let the Angle ACB of the Triangle ABC be obtuse and let AD be drawn perpendicular to BC. The Square of the side AB is equal to the Squares of the sides AC CB and to two Rectangles comprehended under the side BC and under DC Demonstration The Square of AB is equal to the Squares of AD DB. by the 47th of the 1st the Square of DB is equal to the Squares of DC and CB and to two Rectangles comprehended under DC CB by the 4th therefore the Square of AB is equal to the Squares of AD DC CB and to two Rectangles comprehended under DC CB in the place of the two last Squares AD DC Put the Square of AC which is equal to them by the 47th of the 1st the Square of AB shall be equal to the Square of AC and CB. and to two Rectangles comprehended under DC CB. USE THis Proposition is useful to measure the Area of a Triangle it s three sides being known for Example If the side AB was twenty Foot AC 13 BC 11 the Square of AB would be four hundred the Square of AC one hundred sixty nine and the Square of BC one hundred twenty one the Sum of the two last is Two hundred and ninety which being subtracted from four hundred leaves one hundred and ten for the two Rectangles under BC CD the one half-fifty five shall be one of those Rectangles which divided by BC 11 we shall have five for the Line CD whose Square is twenty five which being subtracted from the Square of AC one hundred sixty nine there remains the Square of AD one hundred forty four and its Root shall be the side AD which being multiplied by 5½ the half of BC you have the Area of the Triangle ABC containing 66 square Feet PROPOSITION XIII THEOREM IN any Triangle whatever the Square of the side opposite to an acute Angle together with two Rectangles comprehended under the side on which the Perpendicular falleth and under the Line which is betwixt the Perpendicular and that Angle is equal to the Square of the other sides Let the proposed Triangle be ABC which hath the Angle C acute and if one draw AD perpendicular to BC the Square of the side AB which is opposite to the acute Angle C together with two Rectangles comprehended under BC DC shall be equal to the Squares of AC BC. Demonstration The Line BC is divided in D whence by the 7th the Square of BC DC are equal to two Rectangles under BC DC and to the Square of BD add to both the Square of AD the Square of BD DC AD shall be equal to two Rectangles under BC DC and to the Squares of BD AD in the place of the Squares of CD AD put the Square of AC which is equal to them by the 47th of the 1st and instead of the Squares of BD AB substitute the Square of AB which is equal to them the Squares of BC AC shall be equal to the Square of AB and to two Rectangles comprehended under BC DC USE THese Propositions are very necessary in Trigonometry I make use thereof in the eighth Proposition of the third Book to prove That in a Triangle there is the same Reason between the whole Sine and the Sine of an Angle as are between the Rectangle of the sides comprehending that Angle and
the 1st because the Sides BC BD are equal the Angle ABC shall be double of each The Second case is when an Angle encloseth the other and the Lines making the same Angles not meeting each other as you see in the second figure the Angle BID is in the Center and the Angle BAD is at the Circumference Draw the Line AIC through the Center Demonstration The Angle BIC is double to the Angle BAC and CID double to the Angle CAD by the preceding case Therefore the Angle BID shall be double to the Angle BAD USE THere is given ordinarily a practical way to describe a Horizontal Dial by a single opening of the Compass which is grounded in part on this Proposition Secondly when we would determine the Apogaeon of the Sun and the excentricity of his Circle by Three observations we suppose that the Angle at the Center is double to the Angle at the Circumference Ptolomy makes often use of this Proposition to determine as well the excentricity of the Sun as the Moon 's epicycle The first Proposition of the Third Book of Trigonometry is grounded on this PROPOSITION XXI THEOREM THe Angles which are in the same Segment of a Circle or that have the same Arch for Base are equal If the Angles BAC BDC are in the same Segment of a Circle greater than a Semicircle they shall be equal Draw the Lines BI CI. Demonstration The Angles A and D are each of them half of the Angle BIC by the preceding Proposition therefore they are equal They have also the same Arch BC for Base Secondly let the Angles A and D be in a Segment BAC less than a semi-circle they shall notwithstanding be equal Demonstration The Angles of the Triangle ABE are equal to all the Angles of the Triangle DEC The Angles ECD ABE are equal by the preceding case since they are in the same Segment ABCD greater than a Semi-circle the Angles in E are likewise equal by the 15th of the 1st therefore the Angles A and D shall be equal which Angles have also the same Arch BFC for Base USE Prop. XXI IT is proved in Opticks that the Line BC shall appear equal being seen from A and D since it always appeareth under equal Angles We make use of this Proposition to describe a great Circle without having its Center For Example when we would give a Spherical figure to Brass Cauldrons to the end we may work thereon and to pollish Prospective or Telescope Glasses For having made in Iron an Angle BAC equal to that which the Segment ABC contains and having put in the Points B and C two small pins of Iron if the Triangle BAC be made to move after such a manner that the Side AB may always touch the Pin B and the Side AC the Pin C the Point A shall be always in the Circumference of the Circle ABCD. This way of describing a Circle may also serve to make large Astrolabes PROPOSITION XXII THEOREM QVadrilateral figures described in a Circle have their opposite Angles equal to Two Right Let a Quadrilateral or four sided figure be described in a Circle in such manner that all its Angles touch the Circumference of the Circle ABCD I say that its opposite Angles BAD BCD are equal to two Right Draw the Diagonals AC BD. Demonstration All the Angles of the Triangle BAD are equal to Two Right In the place of its Angle ABD put the Angle ACD which is equal thereto by the 21st as being in the same Segment ABCD and in the place of its Angle ADB put the Angle ACB which is in the same Segment of the Circle BCDA So then the Angle BAD and the Angles ACD ACB that is to say the whole Angle BCD is equal to Two Right USE PTolomy maketh use of this Proposition to make the Tables of Chords or Subtendents I have also made use thereof in Trigonometry in the Third Book to prove that the sides of an Obtuse Angled Triangle hath the same reason amongst themselves as the Sines of their opposite Angles PROPOSITION XXIII THEOREM TWo like Segments of a Circle described on the same Line are equal I call like Segments of a Circle those which contain equal Angles and I say that if they be described on the same Line AB they shall fall one on the other and shall not surpass each other in any part For if they did surpass each other as doth the Segment ACB the Segment ADB they would not be like And to demonstrate it draw the Lines ADC DB and BC. Demonstration The Angle ADB is exteriour in respect of the Triangle BDC Thence by the 21th of the 1st it is greater than the Angle ACB and by consequence the Segments ADB ACB containeth unequal Angles which I call unlike PROPOSITION XXIV THEOREM TWo like Segments of Circles described on equal Lines are equal If the Segments of Circles AEB CFD are like and if the Lines AB CD are equal they shall be equal Demonstration Let it be imagined that the Line CD be placed on AB they shall not surpass each other seeing they are supposed equal and then the Segments AEB CED shall be described on the same Line and they shall then be equal by the preceding Proposition USE Use 24. CVrved Lined Figures are often reduced to Right Lined by this Proposition As if one should describe Two like Segments of Circles AEC ADB on the equal sides AB AC of the Triangle ABC It is evident that Transposing the Segment AEC on ADB the Triangle ABC is equal to the figure ADBCEA PROPOSITION XXV PROBLEM TO compleat a Circle whereof we have but a part There is given the Arch ABC and we would compleat the Circle There needeth but to find its Center Draw the Lines AB BC and having Divided them in the middle in D and E draw their Perpendiculars DI EI which shall meet each other in the Point I the Center of the Circle Demonstration The Center is in the Line DI by the Coroll of the 1st It is also in EI it is then in the Point I. USE Use 25. THis Proposition cometh very frequently in use it might be propounded another way as to inscribe a Triangle in a Circle or to make a Circle pass through three given points provided they be not placed in a streight Line Let be proposed the Points A B C put the Point of the Compass in C and at what opening soever describe Two Arks F and E. Transport the foot of the Compass to B and with the same opening describe Two Arcks to cut the former in E and F. Describe on B as Center at what opening soever the Arches H and G and at the same opening of the Compass describe on the Center A Two Arks to cut the same in G and H. Draw the Lines FE and GH which will cut each other in the Point D the Center of the Circle The Demonstration is evident enough for if you had drawn
ABC Draw the touch Line FED by the 17th of the 3d. and make at the Point of touching E the Angle DEH equal to the Angle B and the Angle FEG equal to the Angle C by the 23d of the 1st Draw the Line GH the Triangle EGH shall be equi Angled to ABC Demonstration The Angle DEH is equal to the Angle EGH of the Alternate Segment by the 32d of the 3d. now the Angle DEH was made equal to the Angle B and by consequence the Angles B and G are equal The Angles C and H are also equal for the same reason and by the Coroll 2. of the 32d of the 1st the Angles A and GEH shall be equal Therefore the Triangles EGH ABC are equiangled PROPOSITION III. PROBLEM TO describe about a Circle a Triangle equiangled to another If one would describe about a Circle GKH a Triangle equiangled to ABC one of the Sides BC must be continued to D and E and make the Angle GIH equal to the Angle ABD and HIK equal to the Angle ACE then draw the Tangents LGM LKN NHM through the Points G K H. The Tangents shall meet each other for the Angles IKL IGL being Right if one should draw the Line KG which is not drawn the Angles KGL GKL would be less than two Right therefore by the 11th Axiom the Lines GL KL ought to concur Demonstration All the Angles of the Quadrilateral GIHM are equal to four Right seeing it may be reduced into Two Triangles the Angles IGM IHM which are made by the Tangents are Right thence the Angles M and I are equivalent to Two Right as well as the Angles ABC ABD Now the Angle GIH is equal to the Angle ABD by construction therefore the Angle M shall be equal to the Angle ABC For the same reason the Angles N and ACB are equal and so the Triangles LMN ABC are equiangled PROPOSITION IV. PROBLEM To inscribe a Circle in a Triangle IF you would inscribe a Circle in a Triangle ABC divide into Two equally the Angles ABC ACB by the 9th of the 1st drawing the Lines CD BD which concurr in the Point D. Then draw from the Point D the Perpendiculars DE DF DG which shall be equal so that the Circle described on the point D at the opening DE shall pass through F and G. Demonstration The Triangles DEB DBF have the Angles DEB DFB equal seeing they are Right the Angles DBE DBF are also equal the Angle ABC having been divided into Two equally the Side DB is common therefore by the 26th of the 1st these Triangles shall be equal in every respect and the Sides DE DF shall be equal One might demonstrate after the same manner that the Sides DF DG are equal One may therefore describe a Circle which shall pass through the Points E F G and seeing the Angles E F G are Right the Sides AB AC BC touch the Circle which shall by consequence be inscribed in the Triangle PROPOSITION V. PROBLEM To describe a Circle about a Triangle IF you would describe a Circle about a Triangle ABC divide the Sides AB BC into Two equally in D and E drawing the Perpendiculars DF EF which concurr in the Point F. If you describe a Circle on the Center F at the opening FB it shall pass through A and C that is to say that the Lines FA FB FC are equal Demonstration The Triangles ADF BDF have the Side DF common and the Sides AD DB equal seeing the Side AB hath been divided equally and the Angles in D are equal being Right Thence by the 4th of the 1st the Bases AF BF are equal and for the same reason the Bases BF CF. USE WE have often need to inscribe a Triangle in a Circle as in the first Proposition of the Third Book of Trigonometry This practice is necessary for to measure the Area of a Triangle and upon several other occasions PROPOSITION VI. PROBLEM To inscribe a Square in a Circle TO inscribe a Square in the Circle ABCD draw to the Diameter AB the Perpendicular DC which may pass through the Center E. Draw also the Lines AC CB BD AD and you will have inscribed in the Circle the Square ACBD Demonstration The Triangles AEC CEB have their Sides equal and the Angles AEC CEB equal seeing they are Right therefore the Bases AC CB are equal by the 4th of the 1st Moreover seeing the Sides AE CE are equal and the Angle E being Right they shall each of them be semi-right by the 32d of the 1st So then the Angle ECB is semi-right And by consequence the Angle ACB shall be Right It is the same of all the other Angles therefore the Figure ACDB is a Square PROPOSITION VII PROBLEM To describe a Square about a Circle HAving drawn the Two Diameters AB CD which cut each other perpendicularly in the Center E draw the touch Lines FG GH HI FI through the Points A D B C and you will have described a Square FGHI about the Circle ACBD Demonstration The Angle E and A are Right thence by the 29th of the 1st the Lines FG CD are parallels I prove after the same manner that CD HI FI AB AB GH are parallels thence the Figure FCDG is a parallelogram and by the 34th of the 1st the Lines FG CD are equal as also CD IH FI AB AB GH and by consequence the Sides of the Figure FG GH HI IF are equal Moreover seeing the Lines FG CD are parallels and that the Angle FCE is Right the Angle G shall be also Right by the 29th of the 1st I demonstrate after the same manner that the Angles F H and I are Right Therefore the Figure FGHI is a Square and its Sides touch the Circle PROPOSITION VIII PROBLEM To inscribe a Circle in a Square IF you will inscribe a Circle in the Square FGHI divide the Sides FG GH HI FI in the middle in A D B C and draw the Lines AB CD which cutteth each other in the Point E. I demonstrate that the Lines EA ED EC EB are equal and that the Angles in A B C D are Right and that so you may describe a Circle on the Center E which shall pass through A D B C and which toucheth the Sides of the Square Demonstration Seeing the Lines AB GH conjoyns the Lines AG BH which are parallel and equal they shall be also parallel and equal therefore the Figure AGHB is a parallelogram and the Lines AE GD AG ED being parallel and AG GD being equal AE ED shall be also equal It is the same with the others AE EC EB Moreover AG ED being parallels and the Angle G being Right the Angle D shall be also a Right Angle One may then on the Center E describe the Circle ADBC which shall pass through the Points A D B C and which shall touch the Sides of the Square PROPOSITION IX PROBLEM To describe a Circle about a Square
for Pentagons are the most ordinary You must also take notice that these ways of describing a Pentagon about a Circle may be applyed to the other Polygons I have given another way to inscribe a Regular Pentagon in a Circle in Military Architecture PROPOSITION XV. PROBLEM TO inscribe a Regular Hexagon in a Circle To inscribe a Regular Hexagon in the Circle ABCDEF draw the Diameter AD and putting the Foot of the Compass in the Point D describe a Circle at the opening DG which shall intersect the Circle in the Points EC then draw the Diameters EGB CGF and the Lines AB AF and the others Demonstration It is evident that the Triangles CDG DGE are equilateral wherefore the Angles CGD DGE and their opposites BGA AGF are each of them the third part of two Right and that is 60 degrees Now all the Angles which can be made about one Point is equal to four Right that is to say 360. So taking away four times 60 that is 240 from 360 there remains 120 degrees for BGC and FGE whence they shall each be 60 degrees So all the Angles at the Center being equal all the Arks and all the Sides shall be equal and each Angle A B C c. shall be composed of two Angles of Sixty that is to say One Hundred and Twenty degrees They shall therefore be equal Coroll The Side of a Hexagon is equal to the Semi diameter USE BEcause that the Side of an Hexagon is the Base of an Ark of Sixty degrees and that is equal to the Semi-Diameter its half is the Sine of Thirty and it is with this Sine we begin the Tables of Sines Euclid treateth of Hexagons in the last Book of his Elements PROPOSITION XVI PROBLEM TO inscribe a Regular Pentadecagon in a Circle Inscribe in a Circle an equilateral Triangle ABC by the 2d and a Regular Pentagon by the 11th in such sort that the Angles meet in the Point A. The Lines BF BI IE shall be the Sides of the Pentadecagon and by inscribing in the other Arks Lines equal to BF BI you may compleat this Polygon Demonstration Seeing the Line AB is the Side of the Equilateral Triangle the Ark AEB shall be the third of the whole Circle or 5 fifteenths And the Ark AE being the fifth part it shall contain 3 15 thence EB contains two and if you divide it in the middle in I each part shall be a fifteenth USE THis Proposition serveth only to open the way for other Polygons We have in the compass of Proportion very easie methods to inscribe all the ordinary Polygons but they are grounded on this For one could not put Polygons on that Instrument if one did not find their sides by this Proposition or such like The end of the Fourth Book THE FIFTH BOOK OF Euclid's Elements THis Fifth Book is absolutely necessary to demonstrate the Propopositions of the Sixth Book It containeth a most universal Doctrine and a way of arguing by Propâ—rtion which is most subtile solid and Brief So that all Treatises which are founded on Proportions cannot be without this Mathematical Logick Geometry Arithmetick Musick Astronomy Staticks and to say in one word all the Treatises of the Sciences are demonstrated by the Propositions of this Book The greatest part of Measuring is done by Proportions and in practisal Geometry And one may demonstrate all the Rules of Arithmetick by the Theorems hereof wherefore it is not necessary to have recourse to the Seventh Eighth or Ninth Books The Musick of the Ancients is scarce any thing else but the Doctrine of Proportion applyed to the Senses It is the same in Staticks which considers the Proportion of Weights In fine one may affirm that if one should take away from Mathematicians the knowledge of the Propositions that this Book giveth us the remainder would be of little use DEFINITIONS The whole corresponds to its part and this shall be the greater quantity compared with the lesser whether it contains the same in effect or that it doth not contain the same Parts or quantities taken in general are divided ordinarily into Aliquot parts and Aliquant parts 1. An Aliquot part which Euclid defines in this Book is a Magnitude of a Magnitude the lesser of the greater when it measureth it exactly That is to say that it is a lesser quantity compared with a greater which it measureth precisely As the Line of Two Foot taken Three times is equal to a Line of Six Foot 2. Multiplex is a Magnitude of a Magnitude the greater of the lesser when the lesser measureth the greater That is to say that Multiplex is a great quantity compared with a lesser which it contains precisely some number of times For Example the Line of Six Foot is treble to a Line of Two Foot Aliquant parts is a lesser quantity compared with a greater which it measureth not exactly So a Line of 4 Foot is an Aliquant part of a Line of 10 Foot Equimultiplexes are Magnitudes which contain equally their Aliquot parts that is to say the same number of times 12. 4. 6. 2. A B C D For example if A contains as many times B as C contains D A and C shall be equal Multiplexes of B and D. 3. Reason or Ratio is a mutual habitude or respect of one Magnitude to another of the same Species I have added of the same Species 4. For Euclid saith that Magnitudes have the same reason when being multiplied they may surpass each other To do which they must be of the same Species In effect a Line hath no manner of Reason with a Surface because a Line taken Mathematically is considered without any Breadth so that if it be multiplyed as many times as you please it giveth no Breadth and notwithstanding a Surface hath Breadth Seeing that Reason is a mutal habitude or respect of a Magnitude to another it ought to have two terms That which the Philosophers would call foundation is named by the Mathematicians Antecedent and the term is called Consequent As if we compare the Magnitude A to the Magnitude I this habitude or Reason shall have for Antecedent the quantity A and for consequent the quantity B. As on the contrary if we compare the Magnitude B with A this Reason of B to A shall have for Antecedent the Magnitude B and for consequent the Magnitude A. The Reason or habitude of one Magnitude to another is divided into rational Reason and irrational Reason Rational Reason is a habitude of one Magnitude to another which is commensurable thereto that is to say that those Magnitudes have a common measure which measureth both exactly As the reason of a Line of 4 Foot to a Line of 6 is rational because a Line of two Foot measureth both exactly and when this happeneth those Magnitudes have the same Reason as one Number hath to another For Example because that the Line of two Foot which is the common measure is found twice in the Line
of 4 Foot and thrice in the Line of 6 the first to the second shall have the same Reason as 2 to 3. Irrational Reason is between Two Magnitudes of the same Species which are incommensurable that is to say that have not a common measure As the Reason of the Side of a Square to its Diagonal For there cannot be found any measure although never so little which will measure both precisely Four Magnitudes shall be in the same Reason or shall be Proportionals when the Reason of the first to the second shall be the same or like to that of the third to the fourth wherefore to speak properly Proportion is a similitude of Reason But one findeth it difficult to understand in what consisteth this similitude of Reason It is only to say that two habitudes or Relations be alike For Euclid hath not given a just Definition and which might have explained its Nature having contented himself to give us a mark by which we may know if Magnitude have the same Reason And the obscurity of this Definition hath made this Book difficult I will endeavour to supply this default 5. Euclide saith that Four Magnitudes have the same Reason when having taken the Equi-multiplices of the first and of the third and other Equimultiplices of the second and of the fourth whatever combination is made when the Multiplex of the first is greater than the Multiplex of the second the Multiplex of the third shall be also greater than the Multiplex of the fourth And when the Multiplex of the first is equal or less than the Multiplex of the second the Multiplex of the third is equal or less than the Multiplex of the fourth That then there is the same Reason between the first and second as there is between the third and fourth A B C D 2. 4. 3. 6. E F G H 10 8. 15 12. K L M N 8. 8. 12. 12. O P Q R 6. 16 9. 24 As if there were proposed four Magnitudes A B C D. Having taken the Equi-multiplexes of A and C which let be E and G quintuplex F and H double to B and D. In like manner taking K and M quadruple to A and C L and N double to B and D. Taking again O and Q triple to A and C P and R quadruple to B and D. Now because E being greater than F G is greater than H and K beng equal to L M is equal to N In fine O being lesser than P Q is lesser than R. Then A shall have the same Reason to B as C to D. I believe that Euclid ought to have Demonstrated this Proposition seeing it is so intangled that it cannot pass for a Maxim To explain well what Proportion is it is to say that four Magnitudes have the same Ratio although one may say in general that to that end the first must be alike part or a like whole in respect of the second as is the third compared to the fourth notwithstanding because this Definition doth not convene with the Reason of equality there must be given a more general and to make it intelligible it must be explained what is meant by a like Aliquot part Like Aliquot parts are those which are as many times in their whole as three in respect of nine two in respect of six are alike Aliquot parts because each are found three times in their whole The first quantity will have the same Reason to the second as the third hath to the fourth if the first contains as many times any Aliquot part of the second whatever as the 3d. contains alike Aliquot part of the 4th A B C D. as if A contains as many times a Hundreth a Thousandth a Millionth part of B as C contains a Hundreth a Thousandth a Millionth part of D and so of any other Aliquot parts imaginable there will be the same Reason of A to B as of C to D. To make this Definition yet clearer I will in the first place prove that if there be the same reason of A to B as there is of C to D A will contain as many times the Aliquot parts of B as C doth of D. And I will afterwards prove that if A contains as many times the Aliquot parts of B as C doth of D there will be the same Reason of A to B as of C to D. The first Point seemeth evident enough provided one doth conceive the terms for if A contains one Hundred and one times the tenth part of B and C only One Hundred times the tenth part of D the Magnitude A compared with B would be a greater whole than C compared with D so that it could not be compared after the same manner that is to say the habitude or Relation would not be the same The second point seemeth more difficult to wit whether if this propriety be so found the Reason shall be the same that is to say if AB contains as many times any Aliquot parts whatever of CD as E contains like Aliquot parts of F there shall be the same Reason of AB to CD as of E to F. For I will prove that if there were not the same Reason A would contain more times any Aliquot part of B than C containeth alike Aliquot parts of D which would be contrary to what we had supposed Demonstration Seeing there is the same Reason of AG to CD as of E to F AG will contain as many times KD an Aliquot part of CD as E would contain a like Aliquot part of F. Now AB contains KD once more than AG thence AB will contain once more KD an Aliquot part of CD than E doth contain a like Aliquot part of F which would be contrary to the supposition 6. There will be a greater Reason of the first quantity to the second than of the third to the fourth if the first contains more times any Aliquot part of the second than the third doth contain a like Aliquot part of the fourth As 101 hath a greater Reason to 10 than 200 to 20 because that 101 contains One Hundred and one times the Tenth part of 10 and 200 contains only One Hundred times the Tenth part of 20 which is 2. 7. The Magnitudes or quantities which are in the same Reason are called Proportionals 8. A Proportion or Analogie is a Similitude of Reason or habitude 9. A Proportion ought to have at least three terms For to the end there be similitude of Reason there must be two Reasons Now each Reason having two terms the antecedent and the consequent it seemeth there ought to be four as when we say that there is the same Reason of A to B as of C to D but because the consequent of the first Reason may be taken for antecedent in the second three terms may suffice as when I say that there is the same Reason of A to B as of B to C. 10. Magnitudes are in continued Proportion
Triangle ABC the Line DE is Parallel to the Base BC the Sides AB AC shall be divided proportionally that is to say that there shall be the same Reason of AD to DB as of AE to EC Draw the Lines DE BE. The Triangles DBE DEC which have the same Base DE and are between the same Parallels DE BC are equal by the 37th of the 1st Demonstration The Triangles ADE DBE have the same point E for their vertical if we take AD DB for their Bases and if one should draw through the point E a Parallel to AB they would be both between the same Parallels they shall have thence the same Reason as their Bases by the 1st that is to say that there is the same Reason of AD to DB as of the Triangle ADE to the Triangle DBE or to its equal CED Now there is the same Reason of the Triangle ADE to the Triangle CED as of the Base AE to EC There is therefore the same Reason of AD to DB as of AE to EC And if there be the same Reason of AE to EC as of AD to DB I say that the Lines DE BC would then be Parallels Demonstration There is the same Reason of AD to DB as of the Triangle ADE to the Triangle DBE by the 1st there is also the same Reason of AE to EC as of the Triangle ADE to the Triangle DEC consequently there is the same Reason of the Triangle ADE to the Triangle BDE as of the same Triangle ADE to the Triangle CED So then by the 7th of the 5th the Triangles BDE CED are equal And by the 39th of the 1st they are between the same Parallels USE THis Proposition is absolutely necessary in the following Propositions one may make use thereof in Measuring as in the following figure If it were required to measure the height BE having the length of the staff DA there is the same Reason of CD to DA as of BC to BE. PROPOSITION III. THEOREM THat Line which divideth the Angle of a Triangle into two equal parts divideth its Base in two parts which are in the same Reason to each other as are their Sides And if that Line divideth the Base into parts proportional to the Sides it shall divide the Angle into Two equally If the Line AD divideth the Angle BAC into Two equal parts there shall be the same Reason of AB to AC as of BD to DC Continue the Side CA and make AE equal to AB then draw the Line EB Demonstration The exterior Angle CAB is equal to the Two interior Angles AEB ABE which being equal by the 5th of the 1st seeing the Sides AE AB are equal the Angle BAD the half of BAC shall be equal to one of them that is to say to the Angle ABE Thence by the 27th of the 1st the Lines AD EB are parallel and by the 2d there is the same Reason of EA or AB to AC as of BD to DC Secondly If there be the same Reason of AB to AC as of BD to DC the Angle BAC shall be divided into Two equally Demonstra There is the same reason of AB or AE to AC as of BD to DC thence the Lines EB AD are parallel and by the 29th of the 1st the Alternate Angles EBA BAD the internal BEA and the external DAC shall be equal and the Angles EBA AEB being equal the Angles BAD DAC shall be so likewise Wherefore the Angle BAC hath been divided equally USE WE make use of this Proposition to attain to the Proportion of the sides PROPOSITION IV. THEOREM EQuiangular Triangles have their Sides Proportional If the Triangles ABC DCE are equiangular that is to say that the Angles ABC DCE BAC CDE be equal There will be the same Reason of BA to BC as of CD to CE. In like manner the reason of BA to AC shall be the same with that of CD to DE. Joyn the Triangles after such a manner that their Bases BC CE be on the same Line and continue the sides ED BA seeing the Angles ACB DEC are equal the Lines AC EF are parallel and so CD BF by the 29th of the 1st and AF DC shall be a parallelogram Demonstration In the Triangle BFE AC is parallel to the Base FE thence by the 2d there shall be the same reason of BA to AF or CD as of BC to CE and by exchange there shall be the same reason of AB to BC as of DC to CE. In like manner in the same Triangle CD being parallel to the Base BF there shall be the same Reason of FD or AC to DE as of BC to GE by the 2d and by exchange there shall be the same reason of AC to BC as of DE to CE. USE THis Proposition is of a great extent and may pass for a universal Principle in all sorts of Measuring For in the first place the ordinary practice in measuring inaccessible Lines by making a little Triangle like unto that which is made or imagined to be made on the ground is founded on this Proposition as also the greatest part of those Instruments on which are made Triangles like unto those that we would measure as the Geometrical Square Sinical Quadrant Jacobs Staff and others Moreover we could not take the plane of a place but by this Proposition wherefore to explain its uses we should be forced to bring in the first Book of practical Geometry PROPOSITION V. THEOREM TRiangles whose sides are proportional are equianguler If the Triangles ABC DEF have their sides proportional that is to say if there be the same reason of AB to BC as of DE to EF as also if there be the same reason of AB to AC as of DE to DF the Angles ABC DEF A and D C and F shall be equal Make the Angle FEG equal to the Angle B and EFG equal to the Angle C. Demonstration The Triangles ABC EFG have two Angles equal they are thence equiangled by the Cor. of the 32d of the 1st and by the 4th there is the same reason of DE to EF as of EG to EF. Now it is supposed that there is the same reason of DE to EF as of EG to EF. Thence by the 7th of the 5th DE EG are equal In like manner DF FG are also equal and by the 8th of the 1st the Triangles DEF GEF are equiangular Now the Angle GEF was made equal to the Angle B thence DEF is equal to the Angle B and the Angle DFE to the Angle C. So that the Triangles ABC DEF are equiangular PROPOSITION VI. THEOREM TRiangles which have their sides proportional which include an equal Angle are equiangular If the Angles B and E of the Triangles ABC DEF being equal there be the same reason of AB to BC as of DE to EF the Triangles ABC DEF shall be equiangular Make the Angle FEG equal to the Angle B and
the Lines being in sub-duplicate Ratio shall be also Proportional USE A B C D. 3. 2. 6. 4. 9. 4. 36. 16 E F G H. THis Proposition may be easily applied to Numbers If the Numbers A B C D are Proportional their Squares EF GH shall be so likewise which we make use of in Arithmetick and yet more use thereof in Algebra PROPOSITION XXIII THEOREM EQuiangular Parallelograms have their Ratio compounded of the Ratio of their sides If the Parallelograms L and M be equiangular the Ratio of L to M shall be compounded of that of AB to DE and that of DB to DF. Joyn the Parallelograms in such a manner that their sides BD DF be on a streight Line as also CD DE which may be if they be equiangular Compleat the Parallelogram BDEH Demonstration The Parallelogram L hath the same Ratio to the Parallelogram BDEH as the Base AB hath to the Base BH or DE by the first the Parallelogram BDEH hath the same Ratio to the Parallelogram DFGE that is to say M as the Base BD hath to the Base DF. Now the Ratio of the Parallelogram L to the Parallelogram M is compounded of that of L to the Parallelogram BDEH and that of BDEH to the Parallelogram M. Thence the Ratio of L to M is compounded of that of AB to DE and of that of BD to EG For example if AB be eight and BH five BD four DF seven say as four is to seven so is five to eight ¾ you shall have these three numbers eight five eight ¾ eight to five shall be the Ratio of the Parallelogram L to BDEH the same as that of AB to DE five to eight ¾ shall be that of the Parallelogram BDEH to M. So then taking away the middle term which is five you shall have the Ratio of eight to eight ◠for the Ratio compounded of both or as 4 times 8 or 32 to 5 times 7 or 35. PROPOSITION XXIV THEOREM IN all sorts of Parallelograms those through which the Diameter passeth are like to the greater Let the Diameter of the Parallelogram AC pass through the Parallelograms EF GH I say they are like unto the Parallelogram AC Demonstration The Parallelograms AC EF have the same Angle B and because in the Triangles BCD IF is Parallel to the Base DC the Triangles BFI BCD are equiangular There is therefore by the 4th the same Ratio of BC to CD as of BF to FI and consequently the sides are in the same Ratio In like manner IH is Parallel to BC there shall then be the same Ratio of DH to HI as of DC to BC and the Angles are also equal all the sides being Parallel thence by the 8th Def. the Parallelograms EF GH are like unto the Parallelogram AC USE I Made use of this Proposition in the Tenth Proposition of the last Book of Perspective to shew that an Image was drawn like unto the Original with the help of a Parallelogram Composed of four Lines PROPOSITION XXV PROBLEM TO describe a Polygon like unto a given Polygon and equal to any other Right Lined figure If you would describe a Polygon equal to the Right Lined figure A and like unto the Polygon B make a Parallelogram CE equal to the Polygon B by the 34th of the first and on DE make a Parallelogram EF equal to the Right Lined figure A by the 45th of the first Then seek a mean Proportional GH between CD and DF by the 13th Lastly make on GH a Polygon O like unto B by the 18th It shall be equal to the Right Lined figure A. Demonstration Seeing that CD GH DF are continually Proportional the Right Lined figure B described on the first shall be to the Right Lined figure O described on the second as CD is to DF by the Coroll of the 20th Now as CD is to DF so is the Parallelogram CE to the Parallelogram EF or as B to A seeing they are equal There is thence the same Ratio of B to O as of B to A. So then by the 7th of the 5th A and O are equal USE THis Proposition contains a change of figures keeping always the same Area which is very useful principally in Practical Geometry to reduce an irregular figure into a Square PROPOSITION XXVI THEOREM IF in one of the Angles of a Parallelogram there be described a lesser Parallelogram like unto the greater the Diameter of the greater shall meet the Angle of the lesser If in the Angle D of the Parallelogram AC be described another lesser DG like thereto The Diameter DB shall pass through the point G. For if it did not but it passed through I as doth the Line BID Draw the Line IE Parallel to HD Demonstration The Parallelogram DI is like unto the Parallelogram AC by the 24th Now it is supposed that the Parallelogram DG is also like thereto thence the Parallelograms DI DG would be like which is impossible otherwise there would be the same Ratio of HI to IE or GF as of HG to GF and by the 7th of the 5th the Lines HI HG would be equal The Twenty Seventh Twenty Eighth and Twenty Ninth Propositions are unnecessary PROPOSITION XXX PROBLEM TO cut a Line given into extream and mean Proportion It is proposed to cut the Line AB in extream and mean Proportion that is to say in such a manner that there may be the same Ratio of AB to AC as of AC to CB. Divide the Line AB by the 11th of the second in such a manner that the Rectangle comprehended under AB CB be equal to the Square of AC Demonstration Seeing the Rectangle of AB BC is equal to the Square of AC there will be the same Ratio of AB to AC as of AC to BC by the 17th USE THis Proposition is necessary in the Thirteenth Book of Euclid to find the length of the Sides of some of the five Regular Bodies Father Lucas of St. Sepulchres hath Composed a Book of the Proprieties of a Line which is cut into extream and mean Proportion PROPOSITION XXXI THEOREM A Polygon which is described on the Base of a Right Angled Triangle is equal to the other Two like Polygon described on the other Sides of the same Triangle If the Triangle ABC hath a Right Angle BAC the Polygon D described on the Base BC is equal to the like Polygons F and E described on the Sides AB AC Demonstration The Polygons D E F are amongst themselves in duplicate Ratio of their homologous Sides BC AC AB by the 20th If there were described a Square on those sides they would be also in duplicate Ratio to their sides Now by the 47th of the first the Square of BC would be equal to the Squares of AC AB thence the Polyligon D described on BC is equal to the like Polygons E and F described on AB AC USE THis Proposition is made use
of to make all sorts of figures greater or lesser for it is more universal then the 47th of the 1st which notwithstanding is so useful that it seemeth that almost all Geometry is established on this Principle The 32d Proposition is unnecessary PROPOSITION XXXIII THEOREM IN equal Circles the Angles as well those at the Center as those at the Circumference as also the Sectors are in the same Ratio as the Arks which serve to them as Base If the Circles ANC DOF be equal there shall be the same Ratio of the Angle ABC to the Angle DEF as of the Ark AC to the Ark DF. Let the Ark AG GH HC he equal Arks and consequently Aliquot parts of the Ark AC and let be divided the Ark DF into so many equal to AG as may be found therein and let the Lines EI EK and the rest be drawn Demonstration All the Angles ABG GBH HBC DEI IEK and the rest are equal by the 37th of the third so then AG an Aliquot part of AC is found in the Ark DF as many times as the Angle ABG an Aliquot part of the Angle ABC is found in DEF there is therefore the same Ratio of the Ark AC to the Ark DF as of the Angle ABC to the Angle DEF And because N and O are the halfs of the Angles ABC DEF they shall be in the same Ratio as are those Angles there is therefore the same Ratio of the Angle N to the Angle O as of the Ark AC to the Ark DF. It is the same with Sectors for if the Lines AG GH HC DI IK and the rest were drawn they would be equal by the 28th of the Third and each Sector would be divided into a Triangle and a Segment The Triangles would be equal by the 8th of the first and the little Segments would be also equal by the 24th of the third thence all those little Sectors would be equal and so as many as the Ark BF containeth of Aliquot parts of the Ark of AC so many the Sector DKF would contain Aliquot parts of the Sector AGC There is therefore the same Reason of Ark to Ark as of Sector to Sector The end of the Sixth Book THE ELEVENTH BOOK OF Euclid's Elements THis Book comprehendeth the first Principles of Solid Bodies whence it is impossible to establish any thing touching the third species of quantity without knowing what this Book teacheth us which maketh it very necessary in the greatest part of the Mathematicks in the first place Theodosius's Sphericks doth wholly suppose it Spherical Trigonometry the third part of Practical Geometry several Propositions of Staticks and of Geography are established on the principles of Solid Bodies Gnomonicks Conical Sections and the Treatise of Stone-cutting are not difficult but because one is often obliged to represent on paper figures that are raised and which are comprised under several Surfaces I leave out the Seventh the Eighth the Ninth and the Tenth Books of Euclid's Elements because they are unnecessary in almost all the parts of the Mathematicks I have often wondred that they have been put amongst the Elements seeing it is evident that Euclid did Compose them only to establish the Doctrine of incommensurables which being only a curiosity ought not to have been placed with the Books of Elements but ought to make a particular Treatise The same may be said of the Thirteenth and of the rest for I believe that one may learn almost all parts of the Mathematicks provided one underâ—… well these Eight Books of Euclid's Elements DEFINITIONS 1. Def. 1. Plate VII Fig. I. A Solid is a Magnitude which hath length breadth and thickness As the figure LT its Length is NX its Breadth NO its Thickness LN 2. The extremities or terms of a Solid are Superficies 3. A Line is at Right Angles or Perpendicular to a Plane when it is Perpendicular to all the Lines that it meeteth with in the Plane Fig. II. As the Line AB shall be at Right Angles to the Plane CD if it be Perpendicular to the Lines CD FE which Lines being drawn in the Plane CD pass through the Point B in such sort that the Angles ABC ABD ABE ABF be Right 4. A Plane is perpendicular to another when the Line perpendicular to the common section of the Planes and drawn in the one is also perpendicular to the other Plane Fig. III. We call the common Section of the Planes a Line which is in both Planes as the Line AB which is also in the Plane AC as well as in the Plane AD. If then the Line DE drawn in the Plane AD and perpendicular to AB is also perpendicular to the Plane AC the Plane AD shall be Right or perpendicular to the Plane AC 5. Fig. IV. If the Line AB be not perpendicular to the Plane CD and if there be drawn from the Point A the Line AE perpendicular to the Plane CD and then the Line BE the Angle ABE is that of the inclination of the Line AB to the Plane CD 6. Fig. V. The inclination of one Plane to another is the Acute Angle comprehended by the two perpendiculars to the common section drawn in each Plane As the inclination of the Plane AB to the Plane AD is no other than the Angle BCD comprehended by the Lines BCCD drawnin both Planes perpendicular to the common Section AE 7. Plains shall be inclined after the same manner if the Angles of inclination are equal 8. Planes which are parallel being continued as far as you please are notwithstanding equidistant 9. Like solid figures are comprehended or terminated under like Planes equal in number 10. Solid figures equal and like are comprehended or termined under like Plains equal both in Multitude and Magnitude 11. Fig. VI. A solid Angle is the concourse or inclination of many Lines which are in divers Planes As the concourse of the Lines AB AC AD which are in divers Planes 12. Fig. VI. A Pyramid is a solid figure comprehended under divers Planes set upon one Plane and gathered together to one point As the figure ABCD. 13. Fig. VII A Prism is a solid figure contained under Planes whereof the two opposite are equal like and parallel but the others are parallelograms As the figure AB its opposite Plains may be Polygons 14. A Sphere is a solid figure terminated by a single Superficies from which drawing several Lines to a point taken in the middle of the figure they shall be all equal Some define a Sphere by the motion of a semi-circle turned about its Diameter the Diameter remaining immoveable 15. The Axis of a Sphere is that unmoveable Line about which the semia-circle turneth 16. The Center of the Sphere is the same with that of the semi-circle which turneth 17. The Diameter of a Sphere is any Line whatever which passeth through the Center of the Sphere and endeth in its
Superficies 18. A Cone is a figure made when one Side of a Right Angled Triangle viz. one of those that contain the Right Angle remaining fixed the Triangle is turned round about till it return to the place from whence it first moved And if the fixed Right Line be equal to the other which containeth the Right Angle then the Cone is a Rectangled Cone but if it be less it is an Obtuse Angled Cone if greater an Acute Angled Cone 19. The Axis of a Cone is that fixed Line about which the Triangle is moved 20. A Cylinder is a figure made by the moving round of a Right Angled Parallelogram one of the sides thereof namely which contains the Right Angle abiding fixed till the parallelogram be turned about to the same place whence it began to move 21. Like Cones and Cylinders are those whose Axes and Diameters of their Bases are Proportional Cones are right when the Axis is perpendicular to the Plain of the Base and they are said to be Scalene when the Axis is inclined to the Base and the Diameter of their Bases are in the same Ratio We add that inclined Cones to be like their Axes must have the same inclination to the Planes of their Bases PROPOSITION I. THEOREM Plate VII Prop. I. A Strait Line cannot have one of its parts in a Plane and the other without it If the Line AB be in the Plane AD it being continued shall not go without but all its parts shall be in the same Plane For if it could be that BC were a part of AB continued Draw in the Plane CD the Line BD perpendicular to AB draw also in the same Plane BE perpendicular to BD. Demonstration The Angles ABD BDE are both Right Angles thence by the 14th of the first AB BE do make but one Line and consequently BC is not a part of the Line AB continued otherwise two strait Lines CB EB would have the same part AB that is AB would be part of both which we have rejected as false in the Thirteenth Maxim of the first Book USE WE establish on this Proposition a principle in Gnomonicks to prove that the shadow of the stile falleth not without the Plane of a great Circle in which the Sun is Seeing that the end or top of the stile is taken for the Center of the Heavens and consequently for the Center of all the great Circles the shadow being always in a streight Line with the Ray drawn from the Sun to the Opaque Body this Ray being in any great Circle the shadow must also be therein PROPOSITION II. THEOREM LInes which cut one another are in the same Plane as well as all the parts of a Triangle If the Two Lines BE CD cut one another in the Point A and if there be made a Triangle by drawing the Base BC I say that all the parts of the Triangle ABC are in the same plane and that the Lines BE CD are likewise therein Demonstration It cannot be said that any one part of the Triangle ABC is in a Plane and that the other part is without without saying that one part of a Line is in one Plane and that the other part of the same Line is not therein which is contrary to the first Proposition and seeing that the sides of the Triangle are in th same Plane wherein the Triangle is the Lines BE CD shall be in the same Plane USE THis Proposition doth sufficiently determine a Plane by two streight Lines mutually intersecting each other or by a Triangle I have made use thereof in Opticks to prove that the objective parallel Lines which fall on the Tablet ought to be Represented by Lines which concur in a Point PROPOSITION III. THEOREM THe common section of two Places is a streight Line If Two Planes AB CD cut one another their common section EF shall be a streight Line For if it were not take Two Points common to both Planes which let be E and F and draw a strait Line from the point E to the point F in the Plane AB which let be EHF Draw also in the Plane CD a streight Line from E to F if it be not the same with the former let it be EGF Demonstration Those Lines drawn in the Two Planes are two different Lines and they comprehend a space whch is contrary to the Twelfth Maxim Thence they are but one Line which being in both Planes shall be their common section USE THis Proposition is fundamental We do suppose it in Gnomonicks when we represent in a Dial the Circles of the hours marking only the common section of their Planes and that of the Wall PROPOSITION IV. THEOREM IF a Line be perpendicular to two other Lines which cut one another it shall be also perpendicular to the Plane of those Lines If the Line AB be perpendicular to the Lines CD EF which cut one another in the point B in such manner that the Angles ABC ABD ABE ABF be right which a flat figure cannot represent it shall be perpendicular to the Plane CD EF that is to say that it shall be Perpendicular to all the Lines that are drawn in that Plane through the point B as to the Line GBH Let equal Lines be cut BC BD BE BF and let be drawn the Lines EC DF AC AD AE AF AG and AH Demonstration The four Triangles ABC ABD ABE ABF have their Angles Right in the Point B and the Sides BC BD BE BF equal with the side AB common to them all Therefore their Bases AC AD AE AF are equal by the 4th of the 1st 2. The Triangles EBC DBF shall be equal in every respect having the Sides BC BD BE BF equal and the Angles CBE DBF opposite at the vertex being equal so then the Angles BCE BDF BEC BFD shall be equal by the 4th of the first and their Bases EC DF equal 3. The Triangles GBC DBH having their opposite Angles CBG DBH equal as also the Angles BDH BCG and the sides BC BD they shall then have by the 26th of the 1st their Sides BG BH CG DH equal 4. The Triangles ACE AFD having their sides AC AD AE AF equal and the Bases EC DF equal they shall have by the 8th of the 1st the Angles ADF ACE equal 5. The Triangles ACG ADH have the Sides AC AD CG DH equal with the Angles ADH AGC Thence they shall have their Bases AG AH equal Lastly the Triangles ABH ABG have all their sides equal thence by the 27th of the 1st the Angles ABG ABH shall be equal and the Line AB perpendicular to GH So then the Line AB shall be perpendicular to any Line which may be drawn through the point B in the Plane of the Lines CD EF which I call perpendicular to the Plane USE THis Proposition cometh often in use in the first Book of Theodosius for example to Demonstrate that the Axis of the World is
the fourth is tripled of that of the first to the second USE YOu may comprehend by this Proposition that the Celebrated Proposition of the Duplication of the Cube proposed by the Oracle consisteth in this to find two means continually Proportional For if you propose for this first term the side of the first Cube and that the fourth term be the double of this first If you find Two mean proportionals the Cube described on the first Line shall have the same Ratio to that described on the second as the first Line hath to the fourth which shall be as One to Two We also correct by this Proposition the false opinion of those which imagine that like Solids are in the same Ratio as are their sides as if a Cube of one foot long were the half of a Cube of Two Foot long although it be but the eighth part thereof This is the Principle of the Calibres which may be applyed not only to Cannon Shot but also to all like Bodies For example I have seen a person which would make a Naval Architecture and would keep the same Proportions in all sorts of Vessels but argued thus if a Ship of One Hundred Tuns ought to be Fifty Foot long a Ship of Two Hundred ought to be One Hundred Foot by the Keel In which he was much mistaken for instead of making a Ship of twice that Burden he would make one Octuple He ought to have given to the Second Ship to be double in Burden to the First somewhat less than Sixty Three Foot PROPOSITION XXXIV THEOREM EQual Parallelepipedons have their Bases and height Reciprocal and those which have their Bases and height Reciprocal are equal If the Parallepipedons AB CD are equal they shall have their Bases and height reciprocal that is to say there shall be the same Ratio of the Base AE to the Base CF as of the height CH to the height AG. Having made CI equal to AG draw the Plane IK parallel to the Base CF. Demonstration The Parallelepipedon AB hath the same Ratio to CK of the same height as hath the Base AE to CF by the 32d Now as AB to CK so is CD to the same CK seeing that AB and CD are equal and as CD is to CK which hath the same Base so is the height CH to the height CI by the Coroll of the 32d Wherefore as the Base AE is to the Base CF so is the height CH to the height CI or AG. I further add that if there be the same Ratio of AE to CF as of the height CH to the height AG the Solids AB CD shall be equal Demonstration There is the same Ratio of AB to CK of the same height as of the Base AE to the Base CF by the 32d there is also the same Ratio of the height CH to the height CI or AG as of CD to CK we suppose that the Ratio of AE to CF is the same of that of CH to CI or AG so then there shall be the same Ratio of the Solid AB to the Solid CK as of the Solid CD to the same Solid CK Therefore by the 8th of the 5th the Solids AB CD are equal USE THis reciprocation of Bases rendreth these Solids easie to be measured it hath also a kind of Analogy with the Sixteenth Proposition of the Sixth Book which beareth this that parallelograms equiangled and equal have their Sides reciprocal and it Demonstrateth also as well the practice of the Rule of Three PROPOSITION XXXVI THEOREM IF Three Lines are continually proportional the Solid Parallelepipedon made of them is equal to an equiangular parallelepipedon which hath all it sides equal to the middle Line If the Lines A B C be continually proportional the parallelepipedon FE made of them that is to say which hath the side FI equal to the Line A and HE equal to B HI equal to C is equal to the equiangular Parallelepipedon KL which hath its sides LM MN KN equal to the Line B. Let there be drawn from the Points H and N the Lines HP NQ perpendicular to the Planes of the Bases they shall be equal seeing that the Solid Angles E and K are supposed equal in such manner that if they did penetrate each other they would not surpass each other and that the Lines EH KN are supposed equal Therefore the heights HP NQ are equal Demonstration There is the same Ratio of A to B or of FI to LM as of B to C or of LM to HI so then the parallelogram FH comprehended under FI IH is equal to the parallelogram LN comprehended under LM MN equal to B by the 15th of the 6th the Bases are thence equal Now the heights HP NQ are so likewise therefore by the 31st the Parallelepipedons are equal PROPOSITION XXXVII THEOREM IF four Lines are proportional the Parallepipedons described on those Lines shall be Proportional and if like Parallelepipedons are Proportional their homologous sides shall be so likewise If A to B bein the same Ratio as C to D the like Parallepipedons which shall have for their homologous sides A B C D shall be in the same Ratio Demonstration The Parallelepipedon A to the Parallelepipedon B is in triple Ratio of that of the Line A to the Line B or of that of the Line C to the Line D. Now the Parallelepipedon C to the Parallelepipedon D is also in triple Ratio of that of C to D by the 33d there is therefore the same Ratio of the Parallepipedon A to the Parallepipedon B as of the Parallelepipedon C to the Parallelepipedon D. PROPOSITION XXXVIII THEOREM IF Two Planes be Perpendicular to each other the Perpendicular drawn from one Point of one of the Planes to the other shall fall on the common section If the Planes AB CD being perpendicular to each other there be drawn from the Point E of the Plane AB a Line perpendicular to the Plane CD it shall fall on AG the common section of the Planes Draw EF perpendicular to the common section AG. Demonstration The Line EF perpendicular to AG the common section of the Planes which we supposed perpendicular shall be perpendicular to the Plane CD by the 3d. Def. and seeing there cannot be drawn from the Point E two perpendiculars to the Plane CD by the 13th the perpendicular shall fall on the common Section AG. USE THis Proposition ought to have been after the 17th seeing it hath respect to Solids in general It is of use in the Treatise of Astrolabes to prove that in the Analemma all the Circles which are perpendicular to the Meridian are strait Lines PROPOSITION XXXIX THEOREM IF there be drawn in a parallelepipedon Two Planes dividing the opposite sides equally their common Section and the Diameter doth also cut each other equally Let the opposite sides of the parallelepipedon AB be divided into Two equally by the Planes CD EF their common section GH and
the Diameter BA shall divide each other equally in O. Draw the Lines BG GK AH LH I prove in the first place that they make but one strait Line for the Triangles DBG KMG have their sides BD KM equal seeing they are the halves of equal sides as also GD GM Moreover DB KM being parallel the Alternate Angles BDG GMK shall be equal by the 29th of the first so then by the 4th of the first the Triangles DBG KGM shall be equal in every respect and consequently the Angles BGD KGM Now by the 16th of the 1st BG GK is but one Line as also LH HA thence AL BK is but one Plane in which is found the Diameter AB and GH the common section of the Planes The Plane AL BK cutting the parallel Planes AN CD shall have its common sections GH AK parallel and by the 4th of the 6th there shall be the same Ratio of BK to GK as of BA to BO and of AK to OG by the 4th of the 6th Now BK is double to BG thence BA is double to BO as AK equal to HG is double to GO So then the Lines GH AB cut each other equally in the Point O. Coroll 1. All the Diameters are divided in the Point O. Coroll 2. Here may be put Corollaries which depend on several Propositions for example that Triangular Prisms of equal height are in the same Ratio as are their Bases for the Parallelepipedons of which they are the halves by the 32d are in the same Ratio as their Bases so that the half Bases and the half parallelepipedons that is to say the Prisms shall be in the same Ratio Coroll 3. Polygon Prisms of the same height are in the same Ratio as are their Bases seeing they may be reduced to Triangular Prisms which shall each of them be in the same Ratio as their Bases Coroll 4. There may be applied to the Prism the other Propositions of the Parallelepipedons for example that equal Prisms have their heights and Bases Reciprocal that like Prisms are in triple Ratio of their homologous sides USE THis Proposition may serve to find the Center of gravity of Parallelepipedons and to Demonstrate some Propositions of the Thirteenth and Fourteenth Books of Euclid PROPOSITION XL. THEOREM THat Prism which hath for its Base a Parallelogram double to the Triangular Base of another and of the same height is equal thereto Let there be proposed Two Triangular Prisms ABE CDG of the same height and let be taken for the Base of the first the Parallelogram AE double to the Triangle FGC the Base of the second Prism I say that those Prisms are equal Imagine that the Parallelepipedons AH GI are drawn Demonstration It is supposed that the Base AE is double to the Triangle FGC Now the Parallelogram GK being also double to the same Triangle by the 34th of the 1st the Parallelograms AE GK are equal and consequently the Parallelepipedons AH GI which have their Bases and heights equal are equal and Prisms which are the halves by the 28th shall be also equal The End of the Eleventh Book THE TWELFTH BOOK OF Euclid's Elements EUclid after having given in the preceding Books the general Principles of Solids and explained the way of measuring the most Regular that is to say those which are terminated by plain Superficies treateth in this of Bodies inclosed under curved Surfaces as are the Cylinder the Cone and the Sphere and comparing them with each other he giveth the Rules of their Solidity and the way of measuring of them This Book is very useful seeing we find herein those Principles whereon the most skilful Geometricians have established so many curious Demonstrations on the Cylinder the Cone and of the Sphere PROPOSITION I. THEOREM LIke Polygons inscribed in a Circle are in the samo Ratio as are the Squares of the Diameters of those Circles Plate VIII If the Polygons ABCDE GFHKL inscribed in Circles are like they shall be in the same Ratio as are the Squares of the Diameters AM FN Draw the Lines BM GN FH AC Demonstration It is supposed that the Polygons are like that is to say that the Angles B and G are equal and that there is the same Ratio of AB to BC as of FG to GH whence I conclude by the 6th of the 6th that the Triangles ABC FGH are equiangular and that the Angles ACB FHG are equal so then by the 21st of the 3d. the Angles AMB FNG are equal Now the Angles ABM FGN being in a semi-circle are Right by the 31st of the 3d. and consequently the Triangles ABM FGN are equiangular Therefore by the 5th of the 6th there shall be the same Ratio of AB to FG as of AM to FN and by the 22d of the 11th if one describe Two similar Polygons on AB FG which are those proposed and also Two other like Polygons AM and FN which shall be their Squares there shall be the same Ratio of the Polygon ABCDE to the Polygon FGNKL as of the Square of AM to the Square of FN This Proposition is necessary to Demonstrate the following Proposition LEMMA IF a Quantity be less than a Circle there may be inscribed in the Circle a Regular Polygon greater than that quantity Lemma Fig. II. Let the Figure A be less than the Circle B there may be inscribed in the Circle a Regular Polygon greater than the figure A. Let the figure G be the difference of the figure A and the Circle in such sort that the figures A and G taken together be equal to the Circle B. Inscribe in the Circle B the Squares CDEF by the 6th of the 4th if that Square were greater than the figure A we should have what we pretend to If it be less divide the Quadrants of the Circle CD DE EF FC into Two equally in the Points H I K L in such sort as you may have an Octogon And if that Octogon be less than the figure A subdivide the Arks and you shall have a Polygon of Sixteen Sides then Thirty Two of Sixty Four c. I say in fine you shall have a Polygon greater than the figure A that is to say a Polygon having less difference from the Circle than the figure A in such sort that the difference shall be less than the figure G. Demonstration The Square inscribed is more than the one half of the Circle it being the half of the Square described about the Circle and in describing the Octagon you take away more than the half of the Remainder that is to say the four Segments CHD DIE EKF CLF For the Triangle CHD is the half of the Rectangle CO by the 34th of the 1st it is then more than the half of the Segment CHD it is the same of the other Arks. In like manner in describing a Polygon of Sixteen Sides you take away more than one half of that which remains in the Circle and so of the rest You
perpendicular to the Plane of the Equinoctial In like manner in Gnomonicks we Demonstrate by this Proposition that the Equinoctial is perpendicular to the Meridian in Horizontal Dials it is not less useful in other Treatises as in that of Astrolabes or in the sections of Stone PROPOSITION V. THEOREM IF a Line be Perpendicular to three other Lines which cut one the other in the same Point these three Lines shall be in the same Plane If the Line AB be perpendicular to three Lines BC BD BE which cut one another in the point B the Lines BC BD BE are in the same plane Let the plane AE be the Plane of the Lines AB BE and let CF be the Plane of the Lines BC BD. If BE was the common section of the Two Planes BE would be in the Plane of the Lines BC BD as we pretend it should Now if BE be not the common section let it be BG Demonstration AB is perpendicular to the Lines BC BD it is then perpendicular to the Plane CF by the 4th and 5th Def. AB shall be also perpendicular to BG Now it is supposed that it is perpendicular to BE thence the Angles ABE ABG would be right and equal and notwithstanding the one is a part of the other So then the Two Planes cannot have any other common section besides BE it is therefore in the Plane CF. PROPOSITION VI. THEOREM THe Lines which are perpendicular to the same Plane are parallel If the Lines AB CD be perpendicular to the same plane EF they shall be parallel It is evident that the internal Angles ABD BDC are right but that is not sufficient for we must also prove that the Lines AB CD are in the same Plane Draw DG perpendicular to BD and equal to AB Draw also the Lines BG AG AD. Demonstration The Triangles ABD BDG have the sides AB DG equal BD is common the Angleâ— ABD BDG are right Thence the Bases AD BG are equal by the 4th of the 1st Moreover the Triangles ABG ADG have all their sides equal thence the Angles ABG ADG are equal and ABG being right seeing AB is perpendicular to the Plane the Angle ADG is right Therefore the Line DG is perpendicular to the three Lines CD DA DB which by consequence are in the same Plane by the 5th Now the Line AB is also in the plane of the Lines AD DC by the second thence AB CD are in the same plane Coroll Two Lines which are parallel are in the same Plane USE WE demonstrate by this Proposition that the hour Lines are parallels amongst themselves in all Planes which are parallel to the Axis of the World as in the Polar and Meridian Dials and others PROPOSITION VII THEOREM A Line which is drawn from one parallel to another is in the same Plane which those Lines are The Line CB being drawn from the point B of the Line AB to the point C of its parallel CD I say that the Line CB is in the Plane of the Lines AB CD Demonstration The parallels AB CD are in the same Plane in which if you draw a strait Line from the point C to the point B it will be the same with CB otherwise two strait Lines would inclose a space contrary to the 12th Maxim PROPOSITION VIII THEOREM IF of Two parallel Lines the one be perpendicular to a Plane the other shall be so likewise As if of the Two parallel Lines AB CD the one AB be perpendicular to the Plane EF CD shall be so likewise Draw the Line BD seeing that ABD is a right Angle and that the Lines AB CD are supposed parallel the Angle CDB shall be right by the 30th of the 1st Wherefore if I Demonstrate that the Angle CDG is Right I shall prove by the 4th that CD is perpendicular to the Plane EF. Make the right Angle BDG and make DG equal to AB then draw the Lines PG AG. Demonstration The Triangles ABD BDG have their sides AB DG equal the side BD is common the Angles ABD BDG are right Thence by the 4th of the 1st the Bases AD BG are equal The Triangles ADG ABG have all their sides equal so then by the 6th the Angles ADG ABG are equal This last is right seeing the Line AB is supposed perpendicular to the Plane EF thence the Angle ADG is right and the Line DG being perpendicular to the Lines DB DA it shall be perpendicular to the Plane of the Lines AD DB which is the same in which are the parallels AB CD So then the Angle GDC is a Right Angle by the 5th Def. PROPOSITION IX THEOREM THe Lines which are Parallel to a third are parallel amongst themselves If the Lines AB CD are parallel to the Line EF they shall be parallel to each other although they be not all three in the same Plane Draw in the Plane of the Lines AB EF the Line HG perpendicular to AB in shall be also perpendicular to EF by the 29th of the first In like manner draw in the Plane of the Lines EF CD the Line HI perpendicular to EF CD Demonstration The Line EH being perpendicular to the Lines HG HI is also perpendicular to the Planes of the Lines HG HI by the 4th thence by the 8th the Lines AG CI are so likewise and by the 6th they shall be parallel USE THis Proposition serveth very often in Perspective to determine on the Plane the Image of the parallel Lines and in the sections of stones where we prove that the sides of the Squares are parallel to each other because they are so to some Line in a different Plane In Gnomonicks we prove that vertical Circles ought to be drawn by perpendiculars on the Wall because that the Lines which are their common section with the Wall are parallel to the Line drawn from the Zenith to the Nadir PROPOSITION X. THEOREM IF Two Lines which meet in a point are parallel to Two other Lines in a different Plane they will form an equal Angle If the Lines AB CD AE CF are parallel although they be not all four in the same Plane the Angles BAE DCF shall be equal Let the Lines AB CD AE CF be equal and draw the Lines BE DF AC BD EF. Demonstration The Lines AB CD are supposed parallel and equal thence by the 33d of the first the Lines AC BD are parallel and equal as also AC and EF and by the preceding BD EF shall be also parallel and equal and by the 33d of the first BE DF shall be parallel and equal So then the Triangles BAE DCF have all their sides equal and by the 8th the Angles BAE DCF shall be equal Coroll One might make some Propositions like unto this which would not be unuseful as this If one should draw in a parallel Plane the Line CD parallel to AB and if the Angles BAC DCF are equal the Lines AC CF are parallel