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Elements of geometry Euclides UB TObingen " o - o N O cO o N~ . s g Vv N « = e — L S S, s D ey e, e P — SR BTGy e i s s b s ca e b S s sty . —————— S o s Lo ez D s e = oo D — By s ot SRR, s s eeas smers R — ‘ Nt ' ' ) ﬁkﬁi&*’ ;’i{e& 'Y ;ka’“.\\ ::’!/‘v:;{gf‘i v&:ﬂ‘:} ."- G r.; ,y ‘J %{' ¢{¢ "“N ﬁv“kvb J‘\")M,L&Vv;’:/(/ Lp ‘f{“‘, ‘*\P“V }{in/g {;hm/ A ,r ’ | f /m EU*W»“ w\ ulr 'Z‘»‘av%ﬂ«t /“"’!(«fdmvzir’- v ""‘/ *f‘\h«c vd Wa’»—(/»« 5‘,).* ,n&f? @Vﬂw i'vhfu fﬁk“&/fW$s b . ‘?{"‘ ‘ '~ l’fm*k{u ff‘%*w‘:t /"’H,w fg?u\*' L- H-;\i s ‘k»‘ 4 " { ; % oM ETDT R Y CONTAINING THE PRINCIPAL'PROPOSITIONS. N Yok . s FIRST 81X, AND THE ELEVENTH AN | TWELFTH BOOKS WITH NOTES CRITICAL AND EXPLANATORY. By JOHN BONNYCASTLE, OF THE ROYAL MILITARY ACADEMY, WOOLWICH, 0O N B O X FPRINTED FOR J. JOHNSON, N©92, 8ST:. PAUL'S CHURCH-~YARD: MDCCLXXXIX, P RAE PFOANGLE O T all the works of antiquity which have been tranfmitted to the prefent times, none are more univerfally and defervedly efteemed than the Elements of Geometry” which go under the name of Evcrip, In many other branches of {cience the moderns have far furpafied their mafters ; but, after a lapfe of more than two thoufand years this per- formance ftill maintains 1ts original pre- eminence, and has even acquired additional celebrity from the fruitlefs attempts whiclk have been made to eftablith a different {yftem. It 1s, however, generahy allowed, that the Elements, as they now ftand, are attended with many difhiculties, which greatly retard the progrefs of learners, on their firft en- trance upon this ftudy, and prevent them ' from applying to other branches of know- ledge, which, in the prefent advanced ftate of the fciences, are equally ufeful and im- portant, Among other obftacles of this kind - A 2 | may % § i AP R ETF ACE may be mentioned the theory of parallel lines, the doGrine of proportion, and many things in the eleventh and twelfth books, relatino: to, {olids, which are ufually found extremely | embamaﬁ}n ; and notwith{tanding the num- | beriefs eﬁ'orts which have been made to eluci- date and explain them, are ftill liable to man y objections. On this account, 1t has been found necef= iluy, n moft of our academlcal lﬁﬁlﬁhl()ﬂs to have recourfe to fome of the more com- pendlous rud1ments of later writers, who, by means of a different arrangement, have endeavoured to new-model the 1Ubj€&.», and to render it Iefs complex and elaborate. But the greater part of them are fo ill dweﬁed ' that they ferve rather to miflead the }eamm than to afford him any afliftance. For, be- fides bemg deﬁment in order and method, fome of thefe authors have treated: the fub- ]e& algebzazcally ; and others, by mtlom ducmg a number of excepuonab}e principles, and a vag ue unfatlsfa&ozy mode of demon- firation, have de cgraded the fcience, and de- ’pmvcd it of feme of its moft ﬁnm g ad- vantages. i ﬁ'ﬁ’f & P ¥ PREBBACEE ¢ It 1s, therefore, thedefign of ti;e fdﬁowing performance, to obviate thefe obje@ions, and to render the fubjet more familiar and per-- fpicuous, without weakening its evidence, or deftroying 1ts elegance and fimplicity. For this purpofe, many propofitions in Evcrip, which are of little or no ufe in their appli- - Gation, and were only introduced into the Elements as neceflary links in the chain of reafoning, are here omitted ; and others {ub- ftituted in their place, which are equally con- ducive to that end, and at the fame time more ufeful and concife. By this means all the moft eflential principles of the fcience have been brought into a fhorter compafs, and the demonftrations, which lead to its fublimer truths, fo continued, as to render their connection as obvious and comprehen- five as poffible. Great care has alfo been taken to preferve that methodical precifion and rigour of proof, which, in treating of this fubject, are requi-. fites of nearly equal importance with the fcience itfelf. For independently of its other advantages, Geometry has always been con- fidered as an excellent logic, which in form- 4rot ng W P;REFACE. ing the mind, and eftablithing a habit of clofe thinking and juft reafoning, in every en- quiry after truth, is far fuperior to all the dialetical precepts that have yet been in- vented ; the fimplicity of its firk principles s the clearnefs and certainty of its demonftra- tions ; the regular concatenation of its parts ; and the Tniverfality of its application being fuch as no other fubje& can boaft. For thefe reafons, it was judged neceffary to adhere as clofely as poflible to the plan of the original Elements; this being, in many refpets, much more natural and Judmous than any of thofe which have fince been pro- poled by other writers. But as the work was rather defigned'as a regular Inftitution of the moit ufeful principles of the fcience, than a ftri&¢ abridgment of Evcrip, fome al- terations have been made, both in the ar=- rangement of the propoﬁuons and the mode of demonftration ; the latter of which, in - particular, 1t 1s ‘prefumed, will be found confiderably improved, being here delivered In a more conventent form, and rendered as clear and explicit as the nature of the fub- jeét would admit. | In PR'EFACE vii ~ 1In the firft fix books, every thing has been demonftrated with a fcrupulous accuracy ; and it was at firt defigned that the fame method fhould have been obferved through- t; but this, in treating of the folids, was found incompatible with the plan of the ‘work, it being here fcarcely pofiible to fol- low the fri&k principles of Evcrip without becoming prolix and obfcure. It was there- fore thought proper, in this part of the per- formance, to adopt a mode of pmof which though not geometrically exact, is far more perfpicuous than the former, and eqwl}y fatisfatory and convincing to the mind; efpecially in the way it is here given, which is fomething lefs exceptionable than that of CAvALERIUS, by whom it was firft intro- duced. Many other particulars might be mention- ed, in which this performance will be found to differ from moft others of the like nature; but as they confit chiefly of improvements and emendations which are too obvicus to efcape the notice of the reader, any further account of them would be unneceflary. It is fufficient to -obferve that much time and | | attention Wi P REFALH - attention have been beftowed upon the work ; and that nothing which was judged effential to the {cience, or ufeful in facilitating its at- - tamment, has been omitted. The acknow- ledged intricacy of fome propofitions in the fifth and fixth books, made it neceffary to abridge that part of the fubje@ more confider- ably than the former ; but it 1s conceived that what 1s here given will be fully {ufficient to anfwer all the purpofes of the learner. - To avoid critical objections were a vain endeavour : they may be made againft every {yftem of Geometry now extant; and to Evcwrip as well as to other writers. Of this abundant proofs are given by the Commen- tators; and in the Notes at the end of the prefent work, where many things of this kind are pointed out which have hitherto efcaped notice. Thele were added chiefly for the information of young ftudents, and ought to be carefully confulted by thofe who with to obtain a juft idea of the fcience, and the | principles upon which it 1s founded. £ B T HE E LM BN ]S O F G =Rl MR R A Y Bk Y O] I.» DEFINITIONS: 1. A Solid is that which has length, breadth and thicknefs, , 2. A Superficies is one of the bounds of 2 folid, and has length and breadth without thicknefs. 3. A Line is one of the bounds of a fuperficies, and ~has length without breadth or thicknefs. \\ ‘,,// . o ST AE ‘ 4. A Point is one of the extremities of a line, and has neither length, breadth, nor thicknefs. 5. A right line is that which has all its parts lymg in the fame direGtion. 2 ~ ELEMENTS OF GEOMETRY. 6. A plane fuperficies is that which is everywhere perfectly flat and even. \ - 7. A plane rectilineal angle is the inclination or open- ing of two right lines which meet in a point. 2 8. One right line is faid to be perpendicular to ano- ther, when it makes the angles on both fides of it equal to each other. | T A __.\_._—-- / 9. A right angle is that which is made by two right lines that are perpendicular to each other. | 10. An obtufe angle is that which is greater than 2 right angle. 11. An acute angle is that which is lefs than a right angle, s 12. A of i SOORSTNES prRSFINS e 12. A figure is that which is inclofed by one or more boundaries. 13. A circle is a plane figure, contained by one line, called the circumference, which is -every where equally diftant from a point within the figure, called the centre. PN 14. Re&tilineal figures are thofe which are contained by right lines. 15 All plane ﬁgures, bounded by three right lines, are called triangles. 16. An equilateral triangle, is that thch haq all its fides equal to each other, 17. An ifofceles triangle, is that which has only two of its fides equal to each other, £\ 18. A right-angled triangle, is that which has one right angle ; the fide which is oppofite to the right angle being called the hypotenufe, B o 1g. An 4 ELEMENTS OF GEOMETRY. 19- An obtufe-angled triangle, is that which has one ebtufe angle. , | i 20. Parallel right lines are fuch as are in the fame plane, and which, being produced ever fo far both ways, will never meet, 21. Every plane figure, bounded by four right lines, is called a quadrangle, er quadrilateral. 22. A parallelogram, is a quadranglé whofe oppofite fides are parallel. | / " 23. Thediagonal of a quadrangle, is a right line join- ing any two of its oppofite angles. 24. The bafe of any figure is that fide upon which it 1s fuppofed to ftand ; and the vertical angle is that which is oppofite to the bafe. e i ihiaiasd NoTEe, When an angle is exprefled by means of three letters, the one which ftands at the angular point, mufk always be placed in the middle. POSTU- o e S centr O:';f;l BooE PTHE IINST, ‘s POSTULATES, 1. Let it be granted that a right line may be drawn from any one given point to another. 2. That a terminated rlght line, may be produced to any length in a right line. - 3. That a circle may be defcribed from any pomt as a centre, at any diftance from that centre. 4. And that a right line, which meets one of two parallel right lines, may be produced till it meets the other. AXIOMS. 1. Things which are equal to the fame thing are equal to each other, 2. If equals be added to equals the wholes will be equal, 3. If equals be taken from equéls the remainders will be equal, . If equals be added to unequals the wholes WIH be unequal 6 ELEMENTS OF GEOMETRY. g, If equals be taken from unequals the remainders will be unequa]. 6. Things which are double of the fame thmg are cqual to each other. | 7. T hings which are halves of the fame thing are equal to each other. 8. The whole is equal to all its parts taken together. 9. Magnitudes which coincide, or fill the fame fpace, are equal to each other. REMARK S A ProrosiTiON, is fomething which is either pro- pofed to be done, or to be demonftrated. A ProBLEM, isfomething' whick is propofed to be done. ' A THEOREM, is fomething which is propofed to be - demonftrated. | ' A LeEmMMA, is fomething which is previoudly demon- ftrated, in order to render what follows more eafy. ACOROLLARY, is a confequent truth, gamed from fome preceding truth, or demonftration. A ScuoriuMm, is a remark or-obfervation made upon - fomething going before it. fc 1 1rl 2 /] ACE) 10- o= rom BOOK THE FIRST. 7 PROPOSETION L Wedsiin Uron a given finite right line to de- fcribe an equilateral triangle. S Let AB be the given right line ; it is required to de- fcribe an equilateral triangle upon it. ~ From the point A, at the diftance aB, defcribe the circle scp (Pof. 3.) ' ' And from the point B, at the diftance Ba, defcribe the circle Ace (Pof. 3.) Then, becaufe the two circles pafs through each other’s centres, they will cut each other. And, if the right lines ca, cB be drawn from the point of interfection ¢, asc will be the equilateral triangle re- quired. For, fince A is the centre of the circle Bcp, ac is equal to AB (Def. 13.) And, becaufe B is the centre of the circle ack, Bc is alfo equal to AB (Def. 13.) But things which are equal to the fame thing are equal to each other (A4x. 1); therefore Ac is equal to cB. And, fince Ac, cB are equal to each other, as well as to AB, the triangle, ABc is equilateral ; and it is defcribed upon the right line A3, as was to be done, B 4 CPERO. 3 FLEMENTS OF GEOMETRY. PRO P 1I. Paoetis. From a given point to draw a right line equal to a given finite right line. | Let A be the given point, and Bc the given right line; it is required to draw a right line from the point A, that fhall be equal to Bc. / Join the points a, 2, (Pef. 1.) ; and upon BEA defcrxbe the equilateral triangle BaAD (Prop. 1.) From the point B, at the diftance Bc, defcribe the cir- cle cer (Pof. 3.) cutting pB produced in F. And from the point p, at the diftance pF, defcribe the circle ruc (Pgf. 3.), cutting DA produced in G, and AG will be equal to Bc, as was required. | For, fince B is the centre of the circle ceF, BC is equal to BF (Def. 13.) | ‘ And, becaufe b is the centre of the circle FHG, DG is equal te DF (Def. 13.) But the part pa is alfo equal to the part 18, (Def-16.), whence the remainder AG will be equal to the remainder B i(pdri3.) ' And ﬁnce AG, BC have been each proved to be equal to BF, AG will alfo be equal to Bc (Ax. 1.) A right line AG, has, therefore, been drawn from the point A, equal to the right line Bc, as was to be done. SCHO= el BOOK THE FIRST. 9 b3 ScuorruMm. When the point A is at one of the ex- tremities B, of the given line Bc, any right line, drawn from that point to the circumference of the circle cz¥F, will be the one required. | PROD I Prosrewm. From the greater of two given right lines, to cut off a part equal to the lefs. o Let A and ¢ be the two given right lines; it is re- quired to cut off a part from AB, the greater, equal to c the lefs. From the point A draw the right line AD equal to ¢ (Prop. 2.) ; and from the centre A, at the diftance AD, defcribe the circle DEF (Pgf. 3.) cutting AB in E, and aE will be equal to c as was required. | For, fince a is the centre of the circle EpF, AE will be equal to ap (Def. 13.) But ¢ is equal to Ap, by conftruion ; therefore AE will alfo be equal to ¢ (Ax. 1.) Whence, from AB, the greater of the two given lines, there has been taken a part equal to c the lefs, Wthh was to be done, . ScuoriuMm. When the two given lines are {o ﬁtw- ated, that one of the extremities of c falls in the point A, the former part of the conftruction becomes un- neceflary. . FRor, 10 - ELEMENTS OF GEOMETRY. P\R {) P. IV. T uarosremMm. If two fides and the included angle of one triangle, be equal to two fides and the in- cluded angle of another, each to each, the tmangles will be equal in all refpe&s. F Lo Ll B Ry E ‘Let Arc, DEF be two triangles, having ca equal to FD, CB to FE, and the angle c to the angle r ; then will the two triangles be equal in all refpects.. v For concecive the triangle ABc to be apphed to the triangle DEF, fo that the point ¢ may coincide with the point F, and the fide ca with the fide Fp. | “Then, becaufe ca coincides with FD, and the angle c is equal to the angle ¥ (&y Hyp.), the fide ce will alfo coincide with the fide rE. And, fince ca is equal to ¥p, and cB to FE (2 Hyp. ), the point A will fall upon the point D, and the point B upon the point E. - But right lines, which have the {fame extremities, muft coincide, or otherwife their parts would not lie in the fame dire&ion, which is abfurd (Def. 5.); therefore As falls upon, and is equal to DE, " And, becaufe the triangle aBc is coincident with the triangle DEF, the angle A will be equal to the angle b, the angle B to the angle E, and the two triangles will be equal in all refpe&s (4x. g.) Q. E. D. PROP, D €O ROk THE FIRST. ' gr PROE V¥V, THEOR\EM. The angles at the bafe of an ifofceles tri- angle are equal to each other. Let amc be an ifofceles triangle, having the fide ca equal to the fide c¢B ; then will the angle cAB be equal to the angle cBA. For, in ca and CB produced take any two equal parts cp, ck (Prep.3), and join the points AE, BD: Then, becaufe the two fides ca, cE of the trmvl CAE, are equal to the two fides ¢B, cD of the triangle ¢BD, and the angle ¢ is common, the fide AE will alfo be equal to the fide BD, the angle caE to the angle cBD, and the angle D to the angle E (Prop. 4.) And fince the whole ¢b is equal to the whole ce (4 Conft.), and the parc cA to the part cB (by Hyp.), the remaining part AD will alfo be equal to the remaining part BE (Ax. 3.) L b The two fides pA, DB, of the triangle pas, being, therefore, equal to the two fides EB, EA of the triangle EBA, and the angle D to the angle E, the angle ABp will alfo be equal to the angle BAE (Prop. 4.) ‘ And if from the equal angles caE, cBD, there be taken the equal angles BAE, ABD, the remaining angle CAB 12 . ELEMENTS OF GEOMETRY. cas will be equal to the remaining angle cea (4x. 3.) 0L D CororLrarY. Every equilateral triangle is alfo equi- angular. | PROP. VI. THzoREM. If two angles of a triangle be equal to cach other, the fides which are oppofite to them will alfo be equal. C D / . A _ B Let anc be a triangle, having the angle cAB equal to the angle cBa ; then will the fide ca be equal to the fide cB. | For if ca be not equal to cB, one of them muft be oreater than the other ; let ca be the greater, and make aD equal to Bc (Prop. 3.), and join BD. Then, becaufe the two fides ap, AB, of the triangle ADB, are equal to the two fides Bc, BA, of the triangle AcB, and the angle DAB is equal to the angle cBA (4y Hyp.), the triangle Ape will be equal to the triangle ACB (Prop. 4.), the lefs to the greater, which is abfurd. The fide ca, therefore, cannot be greater than the ‘fide cB; and, in the fame manner, it may be fhewn that it cannot be lefs; confequently they are equal to each other. Q. E. D. Coror, Every equiangular triangle is alfo equi- lateral. PR OP, ¢Qua SBOOK T-HI FIRST. 18 PR OP VII. Trneckin If the three fides of one triangle be equal to the three fides of another, each to each, the angles which are oppofite to the equal fides will alfo be equal. - Let aBc, DEF be two triangles, having the fide as equal to the fide pE, Ac to pF, and BC to EF ; then will the angle Ace be equal to the angle DrE, BAC to EDF, and ABC to DEF. For, let the triangle DFE be applied to the triangle ACB, fo that their longeft fides, pE, ap, may coincide, and the point r fall at ¢ ; and join ca. ‘Then, fince the fide ac is equal to the fide r, or AG (by Hyp.), the angle acc will be equal to the angle AGc (Prop. 5.) And, becaufe the fide Bc is equal to the fide EF, or BG (&y Hyp.), the angle Bcc will be equal to the angle BGC (Prop. 5.) But fince the angles acc, BcG are equal to the angles AGC, BGC, each to each, the whole angle ace will be equal to the whole angle ace (4x. 8.) ‘ And, becaufe Ac is equal to AG, BC to G, and the angle AcB to the angle AGE, the angle car will, alfo, ' be 14 ELEMENTS OF GEOMETRY. be equal to the angle GAB, and the angle ABC to the angle aBG (Prop. 3.) ' But the triangles AGB, DFE, are identical ; confe- quently the angles of the triangle pre will, alfo, be equal to the correfponding angles of the triangle AcE. Q. E. D. Pt Vi THEOREM. All right angles are equal to eachi other. " ‘ G ¥ A R i e, Let asc, DEF be each of them right angles ; then will ABc be equal to DEF. ' For conceive the angle DEF to be applied to the angle ‘arc, fo that the point E may coincide Wlth the point 8, and the line Ep with the line BA. And if F does not comcxde with Bc, let it fall, if | poffible, without the angle ABC, in the direftion BG; and produce AB to H. " Then, becaufe the angles ABc, ABG dre fight angles (ly Hyp.), the lines cB, GB will be each perpendlculal to.au (Def. 8. g.) : And, fince 2 right line which is perpendlcul’ar to ano= ther right line, makes the angles on each fide of it equal (D¢ 8. ), the angle cBA will be equal to the ang,le CBH, and the angle GBA to the angle GBH. But the angle GBA 18 greater than the angle cBa, or its cqual CBH ; confequently the angle GBH Is alfo greatet than o 0 t0 d BOOK THE FIRST. 15 than the angle csm ; that is, a part is greater than the whole, which is abfurd. "The line £F, therefore, does not fall without the angle A®C; and in the fame manner it may be fhewn that it does not fall within it ; confequently EF and Bc will co- mcxde, and the angle DEF be equal to the angle Azsc, as was to be ﬂlewn. PROP 1IX. Proéosrewm. To bifect a given re¢tilineal angle, that is, to divide it into two equal parts. Let Bac be the given re&ilineal angle ; it is required to divide 1t into two equal parts. Take any point D in AB, and from ac cut off ar equal to Ap (Prop. 3.), and join DE. Upon bk defcribe the equilateral triangle pFE (Prop. 1.), and join AF ; then will AF bife& the angle BAC, as was required. For ap is equal to AE, by conftruétion; »oF is alfo equal to FE (Def. 16.), and AF is common to each of the triangles AFD, AFE. But when the three fides of one triangle are equal to the three fides of another, each to each, the angles which are oppofite to the equal fides are, alfo, equal ( Prop. 7.) o) ¢ The 16 ~ ELEMENTS OF GEOMETRY. The fide pF, therefore, being equal to the fide FE, the angle DAF will be equal to the angle FAE; and con- fequently the angle BAC is bifected by the right line AF, as was to be done. | P ROP KA. PrROBLE M. To bife& a given finite right line, that is, to divide it into two equal parts. | / L c : Let ac be the given right line ; it is required to divide it into two equal parts. ’ | Upon ac defcribe the equilateral triangle aAce (Prop. 1.), and bifect the angle ABC by the right line BD (Prop. 9.) ; then will Ac be divided into two equal parts at the point D, as was required. For aB is equal to B¢ (Def. 16.), BD 1s common to each of the triangles ADB, cDB, and the angle ABD 1is equal to the angle cBD (4y Confl.) ‘ But when two fides and the included angle of one tri- ~angle, are equal to two {ides and the included angle of another, each to each, their bafes will alfo be equal (Pr’bp. 4.) The bafe AD is, therefore, equal to the bafe nc; and, confequently, the right line ac is bifeCted in the point D, as was to be done. | F 4 P RO T {0 vide BOOK THE FIRST. 17 PR OP Xk PROBL“}EM. At a given point, in a given rlght lme, to erect a perpendicular, & /N /' / A»——// | X B K D ¥ Let AB be the given right line, and p the given point in it; it is required to draw a right line, from the pomt D, that fhall be perpendicular to as. ‘T'ake any point E, in AB, and make DF equal to pE (Prep. 3.), and upon EF defcribe the equilateral trian- gle EcF (Prop. 1.) | Join the points b, ¢ ; and the right lme CD lel be. perpendicular to AB, as was required. For cE is equal to cF (Def. 16), ED to DF (by Confi.) and ¢p is common to each of the triangles Ecp, Feb. The three fides of the triangle Ecp being, therefore, equal to the three fides of the triangle Fcp, each to each, the angle Enc will, alfo, be equal to the angle FDC (Prop. 7.) But one right line is perpendicular to another when the angles on both fides of it are equal (Def. 8.); there- fore cp is perpendicular to AB; and it is drawn from the point D as was to be done. ScroLiuM. If the given point be at, or near, the end ~of aB, the line muft be produced. C | B KO P, 18 ELEMENTS OF GEOMETRY, PR OP. XII. ProBLEM - To draw a light linevperpen‘dicular toa given right line, of an unlimited length, from a given point without it. i . Let &z be the given right line, and c the given point ; §t is required to draw a right line from the point C, - that fhall be perpendicular to as. . - Take any point b, in AB, and from that poxnt with the diftance pc, deferibe the cm:le CGE, cuttmg AB in G. Join cc, and from the point G, with the diftance c¢, defcribe the circle # E m, cutting the former in E. Through the points c, E draw the right line CFE, cut- ‘ting AB in F, and cF will be perpendicular to As, as was required. v For, join the points D,c, D,E, ‘and G,E : Then, becaufe pc is equal to DE, GC to GE (Def 13.) ‘and b6 common to each of the triangles pcG, DEG, the angle cpG will be equal to the angle GDE (Prop. 7.) “And, fince' ¢ is equal to DE, DF common to each of the triangles DCF, DEF, and the angle cpG equal 3 te Lett wil two rig‘ For ; they i Buti point B They Dy 8 48D (4 Wl 1 04 BOOK THE FIRST.: 19 to the angle GDE, the angle pre lel alfo be equal to the angle DFE (Prop. 4.) A oagy : But one line is perpendicular'to another when the angles on both fides of it are equal (.Def 8.) 5 therefore cr is perpendxcular to AB;-and it-is drawn from the point c, as was to be done. . PR OP, XIL Tuzorru. The angles which one rxght lme makes with another, on the fame fide of it, are to. gether equal to two rxght angles. s Let the right hne AB fall upon the right line cp; then will the angles ABC, ABD, taken together, be equal to two right angles. For if the angles ABC, ABD be equal to each other, they will be, each of them, right angles (Def. 8 and g.) But if they be unequal, let 8" be drawn, from the point B, perpendicular to cp (Prop. 11.) . Then, fince the angles EBC, EBD are right angles (Def. 8.), and the angle E®D is equal to the angles £ra, ABD (4. 8.), the angles Esc, EBA and aBD will be equal to two right angles. i But 20 ELEMENTS OF GEOMETRY. But the angles EBC, EBA are, together, equal to the angle asc (4x. 8.); confequently the angles ABc, ABD are, alfo, equal to two right angles. Q. E.D. Cororr. All the angles which can be made, at any point B, on thefame fide of the right line CD, are, to- gether, equal to two right angles P RO.D, \XIV;_... THEOREM. If a right line meet two other right lines, in the fame point, and make the angles on each fide of it together equal to two right angles, thofe lines will form one continued - right line. / g C B D Let the right line AB meet the two right lines ¢B, 8D, at the point B, and make the angles ABc, ABD together equal to two right angles, then will 8D be in the fame right line with cB. < , For, if it be not, let fome other line BE be in the fame right line with cB. Thcn, becaufe the rlght line ae falls upon the right line CBE, the angies ABC, ABE, taken together, are equal to two right angles (Prop o But the ‘mgxe% aBc, ABD are alfo equal to two nght angles (by FHyp.); confequently the angles ABC, ABE are equal to the angles ABC, ABD. And, Anc a2y maini abfird Th with ¢ but 3D 1S 0 i 0ppoL Let the Dol angle For, AB, thy two Ifg Ang, line p equal ¢ The tual ¢ nes, Q0 ight ued 8; BD? ;ether fame ime \ ‘}f,“‘r]t 11 » equl | right pf A6 t\ﬁéa ‘BOOK THE FIRST. 2 And, if the angle ABc, which is common, be taken away, the remaining angle ABE will be equal to the re- maining angle ABD ; the lefs to the greater, which.irs | abfurd. : The line BE, therefore, is not in -the fame right line with cB ; and the fame may be proved of any other line but BD ; confequently CBD Is one continued right lme, as was to be fhewn. PROP . XV_. THEOREM. If two right lines mterfe& each other, the eppoﬁte angles will be equal | Let the two right lines AB, ¢ interfect each other in the point E; then will the angle AEC be equal to the angle DEB, and the angle AED to the angle CEB. For, fince the right line cE falls upon the rlght line AB, the angles AEC, CEB, taken together, are equal to two right angles (Prop. 13.) And, becaufe the right line e falls upon the right line cp, the angles BED, CEB, taken together, are alfo equal to two right angles (Prop. 13.) The angles AEc, cEB, taken together, are, therefore, equal to the angles BED, CER taken together (Ax. 1.) C 3 And, 22 ELEMENTS OF GEOMETRY. And, if the angle cEB, which is common, be taken away, the remaining angle’ AEc will be equal to the re- maining angle BED (Ax. 3.) And, in the fame manner, it may be fhewn that the angle AED is equal to the angle ces. Q. E. D: Cororr. All the angles made by any number of right lines, ‘meeting in a pomt are together equal to four right angles. PR OP. XV THEOREM. If one fide of a triangle be produced, the outward angle will be greater than either of the inward oppofite angles. { - Let aBc be a triangle, having the fide AB plroduced ‘to p; then will the outward angle cBD be greater than' mther of the inward oppofite angles BAC or Acs. For, bife¢t Bc in E (Prop. 10.), and join AE; in which, produced, take £F equal to AE (Prop. 3 ), and join BF. Then, fince AE is equal to EF, EC to EB (%y conjt‘ ) and the angle aEC to the angle BEF (Prop. 15.), the angle ACE will, alfo, be equal to the angle EBF (Prop. 4.) But Buf fequer An may & its e =y BOOK THEF‘“IRST; S o But the angle ¢sD is greater than the angle EBF ; con-. fequently it is alfo greater than the angle ack. ‘ And, if cB be produced to G, and AB be bifected, it may be thewn, in like manner, that the angle ABG, or its equal cBD, Is greater than cAB, Q DR PR OP - XVIL TSR R P, The greater fide of every triangle is oppo- fite to the greater angle; and the greater - angle to the greater fide. G A : D B Let asc be a triangle, having the fide Ap greater than the fide ac ; then will the angle acB be greater than the’ angle ABC. ‘ For, fince AB is greater than ac,let Ap be taken equal to ac (Prop. 3.), and join cp. Then, fince cpB is a triangle, the outward angle apc is greater than the inward oppofite angle psc (Prop. 16.) But the angle acp is equal to the angle apc, becaufe Ac is equal to AD ; confequently the angle acp is, alfo, - _ greater than DBC or ABC, And, fince acp is only a part of ace, the whole an- gle Acs muft be much greater than the angle apc. Again, let the angle acs be grcager than the anvle ABC, then will the fide AB be greater than the fide ac. C 4 For 24 ELEMENTS OF GEOMETRY. For, if aB be not greater than ac, it muf’c be either equal or lefs. | But it cannot be equal, for then the angle AcB would be equal to the angle agc (Prop. 5.), which it is not. Neither can it be lefs, for then the angle acs would be lefs than the angle aBc (Prop. 17.), which it is not. The f{ide as, therefore, is neither equal to ac, nor lefs than it; confequently it muft be greaterg Q. E. D. "PROP., XVIII. THEOREM. ~ Any two fides of a triangle, taken 'toge-- ther, are greater than the third fide. v : B Let aBc be a triangle ; then will any two fides of it, taken together, be greater than the third fide. For, in ac produced, take cp equal to ¢ (Prop 2 ), and 3 join BD. ‘Then, becaufe cp is equal to cB (&y confi. ), the angle . cps will be equal to the angle cep (Prop. s.) But the angle ABD is greater than the angle cBp, confequently it muft alfo be greater than the angle aps. And, fince the greater fide of every triangle, is op- poﬁte to the greater angle (Prep. 17.), the fide Ap is greater than the fide as. ‘ But Ap is equal to ac and cB taken ‘togeth'er (by confi.) ; therefore Ac, cB are alfo greater than AB. And, Let which, requir Ay B, | Dra EF ¢Ql Fro Circle: With ¢ DG in The its e of the Ang tque] o the BOOK THE FIRST. 23 And, in the fame manner, it may be ﬂiewn, that any other two fides, taken together, are greater than the third fide. Q. E. D. PROP. XIX. ProzLEwM. To defcribe a triangle, whofe fides fhall be equal to three given right lines, provided any two of them, taken together, be greater Ithan the third. Let A, B, c be the three given right lines, any two of which, taken together, are greater than the third ; it is required to make a triangle whofe fides fhall be equal to A, B, C refpectively. | Draw any right line bG ; on which take DE equal to a, EF equal to B, and FG equal to ¢ (Prop. 3.) \ From the point g, with the diftance Ep, defcribe the circle KHD, cutting DG in kK ; and from the point F, with the diftance rc, defcribe the circle GHL, cutting DG in L. Then, becaufe EG is greater than D (4y Hyp.), or its equal EK, the point G, which is in the circumference of the circle guL, will fall without the circle kD, And, becaufe Fp is greater than FG (4y Hyp.), or its equal rFr, the point b, which is in the circumference of the circle KHD, will fall without the circle GHL. But 26 ELEMENTS OF GEOMETRY. But fince a part of the circle curL falls without the circle kup, and a part of the c'irclé kup falls without the circle cuL, nelth«*r of the circles can beincluded with- in the other. : \ Again, becaufe bE, ¥G, or their equals Ex, rL are, together, greatér than Er (4y Hyp.), the two circles can sieither touch nor fall wholly without eich ether. They muft, therefore, cut one another, in fome point # 3 and if the right lines EH, FH be drawn, Eur will .~ be the triangle required. For, fince E is the centre of the e o KHD, EH is equal to ED (Def. 13.) ; but ED is‘equal to A (&y Conft.) 3 therefore EH is alfo equal to A. - And; becaufe ¥ is the centre of the circle cur, rm is equal to FG (Dey" 13.) ; but FG is €qual to ¢ (& Con/t.) ; therefore FH is alfo equal to c. And fince EF is, likewife, equal to 5 (4 Conft.), the three fides of the triangle EuF are refpectively equal te the thyee given lines A, B, ¢, which was to be thewn. PR OB 4 0&75 0 % BOOK THE TIRSE. 27y A PR O XX Pros L E M. At a given point, in a given right line, to make a rectilineal angle equal to a given reClilineal angle. 7 Let pE be the given right line, » the given point, and BAC the given rectilineal angle ; it is required to make an angle at the point D that fhall be equal to sac. T'ake any point ¥ in AB, and from the point A, at the diftance AF, defcribe t‘ue circle FGs, cutting Ac in G and join FG. Make bk equal to AF, and KE equal to F6 (Prop. 3) 3 and from the points D, K, at the diftances bk, k&, de- {cribe the circles xLr and nrm, cutting each other in 1. Through the points b, L draw the right line pn, and the angle DN will be equal to BAC, as was required. For,join KL then fincé AG is equal to AF (Def. 13.), and AF is equal to DK (& Confl.); AG Wln alfo be equal to px (Ax.1.) But px is equal to pL (Def 13.); con‘fequently AG is alfo equal to ‘DL (./Yx I.) ; and FG is equal to KE of KL (by Conft.) | The three fides of the triangle px1 are, therefore, equal to the three fides of the triangle il each to each whencc 28 ELEMENTS OF GEOMETRY. whence the angle kDL is equal to the angle FAG, or BAC (Prop. 7.)s and it is made at the point b, as was to be done. - | PROP. XXI. THEOREM. If two triangles be mutually equiangular, and have two correfponding fides equal to each other, the other correfponding fides will alfo be equal, and the two trlangles will be equal in all rcfpc&s. A B Let the triangles ABc, DEF be mutually equiangular, and have the fide A8 equal to the ﬁde DE ; then will the fide ac be alfo equal to the fide DF, the fide BC to the fide £F, and the two triangles will be equal in all refpects. For, if Ac be not equal to DF, one of them muft be greater than the other ; let Ac be the greater, and make AG equal to DF (Prop. 3.); and join BG. Then, fince the two fides AB, AG, are equal to the two fides DE, DF, each to each, and the angle GaB is equal to the angle ¥DE {(by Hyp ), the angle ABG Wlll, “alfo, be equal to the angle DEF (Prop. 4.) But the angle DEF is equal to the angle aABc (by Hyp.) ; confequently the angle aBc will, alfo, be equal to the angle ABC, the lefs to the greater, which is abfurd. The 9 Th DF;. ¢anne If lingy eich BAC b 10 "BOOK THE FIRST. 29 The fide ac, therefore, cannot be greater than the fide DF ; and, in the fame manner, it may be fhewn that it cannot be lefs ; confequently it muft be equal to it. And, fince the twofides aAc, AB, are equal to the two fides DF, BE, each to each, and the angle caB is equal to the angle FDE, the fide Bc will alfo be equal to the fide £F, and the two triangles Wln be equal in all re- fpe@s (Prop. 4.) Q.E. D, PROP, XXII. THEOREM. If a right line interfe& two other right lines, and make the alternate angles equal to each other, thofe lines will be parallel, S o Lol Ve Let the right line EF interfe& the two right lines as, ¢p, and make the alternate angles AEF, EFD equal to each other ; then will aB be parallel to cp. | For, if they be not parallel, let them be produced, and they will meet each other, either on the fide ac, or on the fide BD (Def. 20.) ' Suppofe them to meet in the point G, on the fide BD. Then, fince FGE is a triangle, the outward angle AEF is greater than the inward oppofite angle D (Prop. 16.) But 3@ ~ ELEMENTS OF GEOMETRY. But the angles, AEF, EFD, are equal to each other (4 .I[yp ) ;. whence they are equal and unequal at ithe famc~ tlme, which is abfurd. c . The lines AB, cD, therefore, cannot meet on the ﬁdc BD 5 and, in the fame manner, it may be fhewn that they cannot meet on the fide Ac; confequendy they muft be parallel to each other. (.Def gy (LB D Cororr. Right lines which are perpendxcular to the fame riglit line are parallel to each other. PR O P XA “THEORE M. If a right line" interfect two-other: right lines, and make'the outward angle equal to the inward oppofite-one, on the fame fide ; or the two inward angles, on the fame fide, together equal to two right angles, thofe Lines will be parallel. Let the right line EF interfeé the two right lines A, ~ ¢p, and make the outward angle EGB equal to the in- ward angle uD ; or the two inward angles 8GH, GHD together equal to two right angles; then will AB be pa- rallel to cp. | | Yo, . each oth the ang But and mak Lines wil t0 (D, Again totwo I £qual fo B3GR W A/’)C) TemainIn gle oh) But 48 wil, Let ngs ot & BOOK THE FIRST.. 31 For, fince the angles EGB, GHD are equal to each other (&y Hyp.), and the angleo AGH, EGB are alfo equal to each other (Prap 15.), the angle acu will be equal to the angle Hp, (dx. 1.) o - But when a right hne interfeCs two Other rlght Ime and makes the alternate angles equal to each other, thofe lines will be parahel (Prap. 22.) ; therefore AB is parallel to cD. | by i Again, fince the angles BGH, GHD b are, together2 equal to two. right angles (y Hjyp.), and AGH, BGH are, alfo, equal to two right angles (P;'op 12:Js. the angles AGH, BGH will be equal to the angles BGH, GHD (4x.1.) - And,, if the common angle BcHu be taken away, the Temaining angle AGH will be equal to the remaining an- gle cuD (ﬂx &5, ‘ | But thefe are alternate angles ; therefore; inthis cafe, a8 will, alfoy be parallel to cp (Prop, 22.) Q. E. D. PROP. XXIV. THEOREM. If a right line interf{ect two parallel right lines, it will make the alternate angles equal to each other. Let the right line EF interfect the two parallel right lines AB, ¢p; then will the angle AEF be equal to the alternate angle EFD. For 32 ELEMENTS OF GEOMETRY, For if they be not equal, one of them muft be greater than the other ; let EFD be the greater; and make the angle E¥B equal to AEF (Prep. 20.) ' Then, fince AB, cp are parallel, the right line Fs, which interfe&ts cp, being produced, will meet aAB in fome point B (Pof. 4.) ~ And, fince EFB is a triangle, the outward angle AEF will be greater than the inward oppolfite angle EFB (Prop. 16.) But the angles AEF, EFB are equal to each other (Jy Confl.) whence they are equal and unequa] at the fame time, which is abfurd. | . The angle EFp, therefore, is not greater than the angle _AEF ; and, in the fame manner, it may be thewn that it is not lefs; confequently they muft be equal to each other;: LY B s | CoroLL. Right lines which are perpendicular to one of two parallel right lines, are alfo perpendicular to the mher. ‘ PROP. BOOK THE FIRST. 33 PR OP. XXV. Turonr Er‘vi’.' If a right line interfeét two parallel riglit lines, the outward angle will be equal to the nward oppolite one, on the fame fide ; and the two inward angles, on the fame fide, will be equal to two right angles. Let the right line &F interfe@ the two parallel right lines AB, cD ; then will the outward angle EGB be equil to the inward oppofite angle GuD; and the two inward angles BGH, cuD will be equal to two right angles. For, fince the right line EF interfe@s the two parallel = 3 right lines Ar, cp; the angle aAcu will be equal to the alternate angle cup (Prop. 24.) But the angle acH is equal to the oppofite angle rgn (Prop. 18.) ; therefore the angle EcB will, alfo, be equal to the angle cup. s Again, fince the right line g6 falls upon the right line EF, the angles EGB, BGH, taken together, are equal to two right angles (Prop. 13.) | - But the dngle EGE has been fhewn to be equal to the angle GHD ; therefore, the angles BGH, GHD, taken to- gether, will, alfo, be equal to two rightangles. Q. E. D, D CoroLL , A 34 ELEMENTS OF GEOMETRY. CorowLL. If aright line interfect two other right lines, - and make the two inward angles, on the fame fide, toge- ther lefs than two right angles, thofe lines, being produced, will meet each other. | PROP XXVI THEOREM. Right lines which are parallel to the fame right line, are parallel to each other. . w/d A G AN 5 . // \ G i a ) Let the right lines AB, cb be each’of them parallel to £F, then will AB be parallel to cb. For, draw any right line GK, cutting the lines AB, EF, ¢D, in the points G, H and K. ' Then, becaufe AB is parallel to EF (&y Hyp.), and GH interfeCls them, the angle AGH is equal to the alternate angle cuF (Prop. 24.) . ‘ ; And becaufe cp is parallel 0 EF (& Hyp.), and HK interfe&ls them, the outward angle GHF is equal to the inward angle HKD (Prap, 25-) : But the angle AGH has been fhewn to be equal to the angle GHF § therefore the angle AGK 1S alfo équal to the angle GKD. S And, fince the right line GK interfe€ts the two right " lines AR, cD, and makes the angle AGK equal to the alternate angle ckp, AB will be parallel to CD, as was to be fhewn. PROBR Let It Te {nall be T (Pray (D, £D The gether, wil in Let draw Tequire For, CKhﬂ G 10 the But to A3 & Was r 2 19N A v 1GLilw . @s was to be done. BOOK THE FIRSTe 38 PROP. XXVIL. ProBLEM. Through a giver‘i’ point, to draw a right line parallel to a given right line. | ¢ 7 . e R S i) B Let aB be the given right line, and c the given point 3 it is required to draw a right line through the point c that fhall be parallel to AB. Take any point D in AB, and make DE equal to DC (Prop. 3.); and from the points ¢, E, with the diftances cD, ED, defcribe the arcs rs, nn. Then, fince any two fides of the triangle EcD are, to~ .,%etﬁer, greater than the third fide (Prgp. 18.), thofe arcs will interfe& each other (Prop. 19.) _ . Let them interfect at ¥ 3 and through the points r, c, draw the line cH, and it will be parallel to aB, as was required. S For, fince the fides cF, FE of the triangle EFC are each equal to the fide cp, or DE, of the triangle cDE, (4y Conjt.) and Ec is-common, the angle EcF will be equal | to the angle cep (Prop. 7.) | But thefe are alternate angles ; therefore G is parallel to AB (Prop. 24.) ; and it is drawn through the point c, \ \ D a PROBP: ‘36 ELEMENTS OF GEOMETRY. { PROYP XXV ThHebwsln If one fide of a triangle be produced, the “outward angle will be equal to the two in- ward oppofite angles, taken together; and the three angles of every triangle, taken to- ~gether, are equal to two right angles. C /\/E 4 A B D Let arc be a triangle, having one of its fides AB pro- duced to » ; then will the outward angle cBD be equal to the two inward oppofite angles Bca, caB, taken toge- ther ; and the three angles BCA, CAB and ABc, taken to- gether, are equal to two right angles. - For through the point g, draw the right line BE parallel to Ac (Prop. 28.) : Then, becaufe BE is parallel to ac, and cr mterfe@cs them, the angle cBE will be equal to the alternate an- gle Bca (Prop.24.) And becaufe BE is parallel to ac, and AD m(exfe&s ¢hem, the outward angle EBD will be equal to the inward angle caz (Prop. 25.) ~ But the angles cBE, EBD are equal to the whole ang]e - ¢BD ; therefore the outward angle cBp is equal to the two inward oppofite angles ECA, CAB taken toorethel. : & ‘ And two tri foure EXtrems For The thEm’ 1 A (} BOOK 'THE FIRST. 37 And if; to thefe equals, there be added the angle ABC, the angles cBp, aBc, taken together, will be equal to the three angles Bca, caB and aBc, taken together. But the angles cBD, ABC, taken together, are equal to two right angles (Prop. 13.); confequently the three an- gles Bca, cAB and ABc, taken together, are alfo equal to two right angles, | | CoroLL. 1. If two angles of one triangle, be equal to two angles of another, each to each, the remaining angles will alfo be equal. | CoroLL. 2. Any quadrilateral may be diyided into two triangles; therefore all the four angles of fuch a figure, taken together, are equal to four right angles. £ PR OP XXIX. Turesiis Right lines joining the correfponding ex- tremes of two equal and parallel right lines are themfelves equal and parallel. / /._C | , /’/ Let aB, pc be two equal and parallel right lines ; then will the right lines Ap, BC, which join the correfponding extremes of thofe lines, be alfo equal and parallel. For draw the diagonal, or right line ac : 'Then, becaufe AB is parallel to pc, and Ac interfelts them, the angle nca will be equal to the alternate angle ¢AB (Prop. 24.) Ba : And 38 ELEMENTS OF ‘GEOMETRY. And, becaufe ar is equal to pc (& Hyp.), Ac com- ~mon to each of the triangles ABc, ADC, and the angle pca - equal to the angle cas, the fide aAp will alfo be equal to the fide Bc, and the a‘;]gle DAC to the angle AcB (Prop. 4.) - Since, therefore, the right line ac interfects the two right lines ap, Bc, and makes the alternate angles equal to each other, thofe lines will be parallel (Prop.23.) But the line Ap has been proved to be equal to the line ‘BC; confequently theyare both equal and parallel.. Q.E.D. i PR OP, XXX, ’T_HEQRE'M. The oppofite fides and angles of any paral- lelogram are equal to each other, and the diagonal divides it into two equal parts, A Let ArcD be 2 parallelégram, whofe mdiagonal is Ac then will its oppofite fides -and angles be equal to each other, and the diagonal ac will divide it into two equal parts. iy For, fince the fide aAD is parallel to the fide Bc (Ddf. 22.), and the right line Ac interfeéts them, the angle pac will be equal to the alternate angle aAcs (Prop. 24.) And, becaufe the fide pc is parallel to the fide aB (Def. 22.), and Ac interfe@s them, the angle pca will be equal to the alternate angle caB (Prop. 24.) - Since, therefore, the two angles pac, pca, are egual " to the two angles ACB, CAB, €ach to each, the remain- mng ing ang asc (! whole @ But, gular, equi o the two Par: the fir 2 &y Let fime by then will gram gp For, | terfedks inward ¢ And, | Interfof the inyy; Jince, fDA, al | S8 AP0 L 15 ALy t ¢ach 0 equﬁl BOOK THE FIRST: 29 ing angle Apc will alfo be equal to the remaining angle ABC (Prop. 29. Cor.) and the whole angle paB to the whole angle pcs. \ But, the triangles cpaA, ABC, bemg mutually equian- gular, and having Ac common, the fide pc will alfo be equal to the fide AB, and the fide Ap to the fide Bc, and the two triangles will be equal in all refpe&s (Prop. 21.) - Q. E. D. PROP. XXXI., THEOREM. Parallelograms, and triangles, ftanding upon the fame bafe, and between the fame parallels, are equal to each other. pEERT - B | // \ l/ i | Let AE, BD be two parallelograms ftanding upon the fame bafe AB, and between the fame parallels A, »E ; then will the parallelogram AE be equal to the parallelo- gram BD. For, fince Ap is parallel to Bc (Def. 22.), and DE in- terfects them, the outward angle Ecp will be equal to the inward oppofite angle FpA (Prop. 25.) And, becaufe AF is parallel to BE (Defi 22.), and pE interfets them, the outward angle AFp will be equal to the inward oppofite angle Bec (Prop. 25.) * Since, thesefore, the angle EcB is equal to the angle FDA, and the angle AFD to the angle BEC, the remaining D4 angle A0 ' ELEMENTS OF GEOMETRY., angle cBe will be equal to the remaining angle par (Prop. 29. Cor. 1.) But the fide ap is alfo equal to the fide Bc (Prop. 31.); confequently, fince the triangles ADF, BCE are mutually equiangular, and have two correfponding fides equal to each other, they will be equal in all refpeéts (Prop. 21.) If, theréfore, from the whole figure ABED, there be taken the triangle BcE, there will remain the pa- rallelogram Bp ; and if, from the fame figure, there be taken the triangle ADF, there will remain the parallelo- pram A, | But if equal things be taken from the fame thing, the remainders will be equal ; confequently, the parallelogram AE is equal to the parallelogram sDp. Again, let ABc, ABF be two triangles, ftanding upon the fame bafe AB, and between the {ame parallels,. a5, cr 3 then will the triangle aBc be equal to the triangle ABF. -+ For produce c¥, both ways, to » and E, and draw AD parallel to Bc, and BE to AF (Prop. 28.) Then, fince 8D, AE, are two parallelograms, ftand- ing upon the fame bafe AB, and between the fame parallels AB, DE, they are equal to each other (Prop. 32.) And, becaufe the diagonals ac, BF bifect them (Prop. 31.), the triangle Apc will alfo be equal to the tnangle - asF. Q. E. D.. RO P M'Y' le ' 44 AW AD , ftande he fame 14 ) { {\Prﬁ?o riangle ROP| BOOK THE FIRST. 41 PROP., XXXII. THEOREM, Ifa parallelogram and a triangle ftand upon the fame bafe, and between the fame parallels, the parallelogram will be double the triangle. Let the parallelogram ac and the triangle ars fland upon the fame bafe AB, and between the fame parallels AB, DE; then will the parallelogram ac be double the triangle AEB. For join the points B, D ; then will the parallelogram . Ac be double the triangle ApB, becaufe the dlagonal DB divides it inte two equal parts (Prop. 31.) | - But the triangle ADB is equal to the triangle AEB, be- caufe they ftand upon the fame bafe AB, and between the fame parallels AB, DE (Prop. 32.); whence the paral- lelogram Ac is alfo double the triangle aEs. Q_E. D. CoroLL. Ifthe bafe of the parallelogram be half that of the triangle, or the bafe of the triangle be double “that of the parallelogram, the two figures will be equal to each other, P RO 42 ELEMENTS OF GEOMETRY, PROP. XXXIIL Prosrewm. To make a parallelogram that fhall have its oppofite fides equal to two given right lines, and one of its angles equal to a given retilineal an gle. D : ol Bb o R d s Let aB and ¢ be two given right lines, and D a given reQilineal angle ; it is required to make a parallelogéam that fhall have its oppofite fides equal to AB and C, and one of its angles equal to p. At the point A, in the line AB, make the angle BAF ~ equal to the angle 0 (Prop. 20) and the fide AF equal o (i a0) Alfo, make FE parallel and equal to aB (Prop 28 and 3.), and join BE ; then will AE be the parallelo.crram o quired. . For, fince FE is para]lel and equal to AB (by Conft.), s& will be parallel and equal to A (Prop. 30.); whence the figure AE is a parallelogram. And, becaufe aF is equal to ¢ (&y Conft.) BE will allo ' be equal to ¢ ; and the angle BAF was made equal to the angle p. The oppofite fides of the parallelogram AE are, there- fore, equal to the two given lines AB and c; and one of its angles is equal to the given angle b, as was to be done. - Ar lines termal (Pry Tl paral] alfoy By AB Is grim "BOOK THE FIRST. PROP. XXXIV. THEOREM. If two fides of a triangle be biféé’ced,h he right line joining the points of bifection, will be parallel to the bafe and equal to one half of i C Ve v Let aBc be a triangle, whofe fides ca, cB are bifeted in the points D, E ; then will the right line DE, joining thofe points, be parallel to AB, and equal to one half of it. For, in DE produced, take EF equal to ED (Prop 3)s and 1 Jom BF : Then, fince Ec is equal to EB (&y Hyp.), ED to EF (by Conjt.) and the angle DEC to the angle BEF (Prop. 15. ), the fide BF will alfo be equal to the fide bc, or its equal pA, and the angle EFB to the angle Epc (Prop. 4.) And, bécaufe the right line DF interfeéts the two rilght‘ lines cp, ¥B, and makes the angle pc equal to the al- ternate angle £¥B, BF will be parallel to pC or DA (Prop. 24.) The right lines BF, AD, therefore, bemg equal and parallel, the lines DF, AB, joining their extremes, will alfo be equal and parallel (Prop. 30.) But pr is the double of DE (by Confl.); confequently AB is alfo the double of DE ; that is DE is the half of AB. 9 Q. E. D. PROP. L 44 ELEMENTS OF GEOMETRY. # B P T Pucei e w ., BB . o v To divide a given finite right line into any propofed number of equal parts. Let as be the given right line ; it is required to divide it into a certain propofed number of equal parts. From the point A, draw any right line ac, in which take the equal parts AD, DE, EC, at pleafure, (Prop. 1.) to the number propofed. Join Bc ; and parallel thereto draw the right lines EF, DG, (Prop. 28.) cutting AB, in F and G ; then will Az be divided into the {ame number of equal parts with Ac, ‘as was required. For take EH, CK, each equal to BG (Prop. 3 .), and join p, H and E, K. * Then, fince DG is parallel to EF (by Confl.), and AE interfe&@s them, the outward angle apc will be equal to to inward oppofite angle pEH (Prop. 25.) And, becaufe the fides Ap, DG of the triangle AcD, are equal to the fides DE, EH of the trlangle DHE (by Con/?. ), and the angle ADG is equal to the angle DEH, the bafe AG will alfo be equal to the bafe pH, and the angle pAG to the angle EpH (Prop. 4.) But, fince the right line AE mterfe&s the two right lines DG, EF, and makes the outward angle EDH equal ' to to the | But confequ the line with A odny 4 le AGDy hale AG DAG (0 o ight g equal {0 BGOK THERIEST. ¥ 45 to the inward oppofite angle pac, pu will be parallel to AG or GF (Prop. 23.) | And, in the fame manner it may be thown that £k is alfo equal to AG, and parallel to AG or Fs. The figures GH, FK, therefore, being parallelograms, the fide pu will be equal to the fide GF, and the ﬁde EK, to the fide F8 (Prop. 31.) | ‘ But pH, EK have been each proved to be equa] to AG ; confequently GF, FB are, alfo, each equal to aG ; whence the line aB is divided into the fame number of eq_ual parts with Ac, as was to be done. BOOK 46 ELEMENTS OF GEOMETRY, B O.0 K DEFINITIONS. 1. A reftangle is a parallelogram whofe angles are all right angles. 2. A fquare is a ré&angle, whofe fides are all equal to each other. 3. Every reGtangle is faid to be contained by any two: of the right lines which contain one of the right angles. 4. If two right lines be drawn through any point in the diagonal of a parallelogram, parallel to its oppofite fides, the figures which are interfected by the diagonal are called parallelograms about the diagonal. 5. And the other two parailelograms, which are not interfe@ed by the diagonal, are called complements to the parallelograms which are about the diagonal, e N " o S0, are all ul fo oint 1 aopollte : ,lagoml gre ot f$ 10 i BOOK THE SECOND. 47 6. In every parallelogram, either of the two parallelo- srams about the diagonal, together with the two comple«- ments, is called a gnomon, N l Bl 40 Aoty i S oM e -1 e altxtude of any figure is a perpendxcular drawn from the vertical angle to the bafe, PROP, I ProsLEnm Upon a given right line to defcribe a fquars, B G A ; Tet AB be the given right lme, it is required to de- {cribe a fquare upon it. Make ap, BC, each perpendicular and equal to AR (I 11 and 3.), and join Dc ; then will Ac be the fquare required. : For, fince the angles DAB, ABC are right angles (&y Confl.), ap will be parallel to sc (1. 22 Cor.) ( And becaufe ap, Bc are equal and parallel, AB, DC will, alfo, be equal and parallel (I. 30.) | But Ap, BC are each equal to aB (4y Conft.) ; whence AD, AB, EC and cp are all equal to each other, The a8 ELEMENTS OF GEOMETRY. +The figure a¢, therefore, is an equilateral parallc]od’ gram ; and it has, likewife, all its angles right angles. For the angle paB is equal to the angle pcs, and the angle ABc to the angle apc (I. 30.) But the angles pAB, Anc are right angles (2y Confd.) ; confequently the angles bce, ADc are, alfo, right angles. The figure ac, therefore, being both equilateral and - rectangular, is a fquare; and it is defcribed upon the line AB, as was to be done, PROP 1L THEOREM. Reé’cangles and Squares contained under equal lines are equal to each other. D V c B G e : ’ / i e et f A (i - E X Let BD, FH be two reftangles, having the fides ag, Be equal to the fides £F, FG, each to each; then will the rectangle 8D be equal to the reGtangle rH. For draw the diagonals ac, G : . Then, fince the two fides AB, BC are equal to thetwo .ﬁdes EF, FG, each to each (Jy Hyp.), and the angle Bis equal to the angle F (I 8.), the triangle aBc will he equal to the triangle EFc (I. 4.) But the diagonal of every parallelogram divides it into two equal args (1. 30.) ; whence the halves bemcr equal, the wholes will alfo be equal. . The reftangle 5D is, therefore, equal to the re&angle TH; and in the fame manner it may be proved when the ﬁgures are fquares. | Q. E. D. | 5 PROP. The e € Letp AB e & dizgona) Fur i greater Bl BK Then, angle 13 ingles, ¢ FEG (], But th ing each b 3); triangle* The BF ang Cannot b And angle ap b qual Aot Naet . 7 e two BT {488 yiil 06 # 1nfo N BOOK THE SECOND, 49 PR OP III. THEOREM: The fides and diagonals of equal fquares are equal to each other. Let 8D, rH, be two equal fquares ; then will the fide - AB be equal to the fide ¥, and the diagonal ac to the diagonal EG. ' ol For if A, EF be not equal, one of them muft be greater than the other ; let AB be the greater, and make BL, BK each equal to EF or FG (I. 3.); and join LK. ‘Then, becaufe BL is equal to FE, Bk to FG, and the angle LBK to the angle EFG, being each of them ‘right angles, the triangle Brk will be equal to the triangle FEG (l. 4.) But the triangle FEG is equal to the triangle Bac, be- ing each of them the halves of the equal f'quares FH, BD (I. 30.) ;5 whence the triangle BLK is alfo equal to the triangle BAc, the lefs to the greater, which is abfurd. The fide asn, therefore, is not greater than the fide EF ; and in the fame manner it may be proved that it cannot be lefs ; confequently they arc equal to each other. And becaufe AB is equal to Er, BC to Fo, and the angle ABC to the angle erc (I. 8.), the fide ac will alfo be equal to the fide ec (1. 4.) Q. K, Dy E PROP - 50 ELEMENTS OF GEOMETRY. PROP: IV..THEOREM, -~ The {quare of a greater line is greater than the {quare of a lefs; and the greater {quare has the greater fide. T W Let the right line Az be greater than the right line E¥ 3 then will Bp, the fquare of AB, be greater than FH, the fquare of EF. : For fince AB is greater than EF, and BC than FG (& Hyp.), take BK, a part of BA, equal to EF, and BL, 2 part of Bc, equal to FG (I. 3.); and jein KL. Then, becaufe B is equal to FE; BL to FG, and the angle kBL to the angle EFG (L. 8.), the triangle BLK will be equal to the triangle FGE (L. 4+) But the triangle Bca is greater than the triangle BLK, whence it is alfo greater than the triangle FGE. And fince the fquare BD is double the triangle Bca, and the fquare ¥H is double the triangle FGE (I. 30.), the fquare Bp will alfo be greater than the {quare FH. Again, let the {quare BD be greater than the fquar::j FH ; then will the fide AB be greater than the {ide EF. Forif AB be not greater than EF, it muft be either equal gq it, or lefs ; but it cannot be equal to it, for then the 3 fquare\ But becau(e e fm BOOK THE SECOND. 3 {quare BD would be equal to the fquare ru (IL 2. ), which it is not. Neither can it be lefs, for then the fquare zp would be lefs than the fquare ru (II. 4.), which it is not; confe- quently AB is greater than EF¥, 2s was to be thewn. PROP. Vo Tneab s Parallelograms and triangles, having equal. bafes and altitudes, are equal to each other. n A b3 asie bl x g / e | N 7 dit / Gl auBa 5 B i ol bt N Lol S P e evige A A R B roM oot E Let ac, Ec be two parallelograms, having the bafe B equal to the bafe Er, and the altitude pK to the altitude HL ; then will the parallelogram 'ac be equal to the pa-- rallelogram G, For upon AB, EF, produced if neceﬁ"ary let fall the perpendiculars cm, N (1. 12.) | Then, fince mp, NH aré re&angular parallelograms, the fide pc is equal to the fide KM, and the fide G to the fide L~ (L. 30.) But pc is alfo equal to A, and ne to EF (I 30.) g therefore K M is equal to AR, and LN to EF. And, fince AB is equal to EF (by Hyp.), kM will be equal to LN ; and confequently the re&:anéle MD is equal to the retangle na (Il 2.) i But the rectangle mMp is equal to the parallelogram ac, becaufe they ftand upon the {ame bafe DC, and between the fame parallels pc, Am. E 2 ! Anﬁ, 52 ELEMENTS OF GEOMETRY. And, for the fame reafon, the reftangle NH is equal «0 _the parallelogram EG ; whence the parallelogram Ac is equal to the parallelogram EG. ~ Again, let ABD, EFH be two tnancrles, havmg the bafe AB equal to the bafe EF, and the altitude DK to the altitude B ; then will the triangle ABD be equal to the triangle EFH. | For, if the parallelograms Ac, EG be compleated, they will be equal to each other, by the former part of the propofition. And fince the diagonals DB, HF divide them into two equal parts (1. 30.), the trxangle asD will alfo be equal to the triangle EFH. O. D Cororr. Parallelograms and Triangles ftanding upon equal bafes, and between the {ame parallels, are equal to eacp other. : ‘P“RO 1 VI. THEORE M. 4 The complements of the parallelograms which are about the diagonal of any paral- lelogram are equal to each other. « Let ac be a parallelogram, and Ax, kc, complements about the diagonal b ; then will the complement Ak be equal to the complement K. For For, the trian And, diagonz the tni be equal ) Par nil of Let 3y it llzgo fouares, For B, the. And ling (3, 4 (1, 3%:311\3 g paal BOOK THE SECOND, 53 For, fince ac is 2 parallelogram, whofe diagonal is BD, the triangle p AB will be equal to the triangle Bcp (I. 30.) And, becaufe Ec, HF are alfo parallelograms, whofe ~diagonals are DK, KB, thg triangle DGk will be equal to the triangle DEK, and the triangle KF3 to the trlangle kHB (l. 30.) But, fince the triangles pck, KFB are, together, equal to the triangles DEK, KHB, and the whole triangle pag to the whole triangle pca, the remaining part ax will be equal to the remaining part kc. Q: E: D. R OPY. VH. THEGREM. : Paréllelograms which are about the diago- nal of a fquare are themielves {quares. D G 3 ‘(!7““*') | P | - D h: E“ /;Tj‘i ST ! | | ]// | | SRR A H 52 Let 8D be a {quare, and HE, FG parallelograms about its diagonal Ac; then will thofe parallelograms alfo be fquares For fince the fide of the fquare AB is equal to the fide BC, the angle caB will be equal to the angle ace (I. s.) And becaufe the right line cH is parallel to the right ~line cB, the angle aku will alfo be equal to the angle &en (1. 2%.) The angles cAB, AKH are, therefore, equal to each other; and confequently the fide am is equal to the fide nk (L 6.) \ | E 3 But 54 ELEMENTS OF GEOMETRY. _But the fide An is equal to the fide EK, and the fide Hk to the fide AE (I. 30.); whence the figure HE is™ _equilateral. | It has alfo all its angles right angles: _ For eaH is a right angle, being the angle of a fquare ; and HG, EF are each of them parallel to the fides of the fame fquare, whence the remaining angles will alfo be right angles (I.25.) The figure HE, theréfore, being equilateral, and hav- ing all its angles right angles, is a {quare: and the fame ~may be proved of the figure ra. LD, 5 PROP, VI THEOREM. < The reCtangles contained under a given line and the feveral parts of another line, any how divided, are, together, equal to the retangle of the two whole lines, ¥ M X G A O oS O S / B D i c _._.‘._AT__ % Let & and sc be two right lines, one of which, B¢, is divided into feveral parts in the points p, E; then will the re&angle of A and BC, be equal to the fum of the reGtangles of A and BD, A and DE, and A and Ec. - For make BF perpendicular to sc (I. 11.) and equal to a (L 3.), and draw Fc parallel to BC, and DH, EI and and CG € they me¢ Then. r (11, caufe o Let ty te poing With {he i 43, Of [0 4o 0¢ ¢ 1A aVe ma .,Ll’ v ,« ¥ D - ) BOOK!THEYSECOND.. - and cG each parallel to BF (1. 27.), producing them till they meet FG in the points H, I, G. : - Then, fince the re&angle Bu is contained byBIﬁ and Br (II. Dcy" 3.), it is alfo contained by 8D and A, be- caufe BF is equal to A (by Con/t.) | | And, fince the re&angle pr1 is contained by pE and DH, it is alfo contained by pE and a, becaufe DH 1s equal to 8F (L. 30.), or a. | The re&tangle EG, in like manner, is contained by ec and A; and the reCtangle BG by Bc and a. But the whole reétangle BG, is equal to the re&ang]es BH, DI and EG, taken together; whence the reftangle of A and Bc is alfo equal to the reCtangles of A and zp, a and DE and A and Ec, taken together. AN PR O I T dE e If a right line be divided into any two parts, the retangles of - the whole line and each of the parts, are, together, equal to the {quare of the whole line. T by ‘ A2 o f i | | | 3 1 Let the right line AB be divided into any two parts in ‘the point ¢ ; then will the reCtangle of aB, ac, together with the reftangle of AB, BC, be equal to the fquare of A, E e For, 56 ELEMENTS OF GEOMETRY. . For, upon aB defcribe the fquare ap (II. 1.), and through ¢ draw cr parallel to AE or D (I. 27.) ‘ Then, fince the re&angle AF is contained by AE, Ac, it is alfo contained by aB, Ac, becaufe AE is equal to AB (II. Def. 2.) And, fince the re&angle cp is contained by Bp, BC, it is alfo contained by aB, Bc, becaufe BD is equal to AB. But ADp, or the {quare of AB, is equal to the rectangles AF, CD, taken together; whence the reftangle as, ac, together with the reftangle AB, Bc, is alfo equal to the {quare of AB. ke 1), - PREY X CErinin If a right line be divided into any two parts, the rectangle of the whole line and one of the parts, is equal to the reftangle of the two parts, together with the {quare of the aforefaid part. . F D A £ B Let the nght line AB be divided into any two parts in | ‘the point ¢ ; then will the reftangle of As, BC be equal to the I'€C)[ahgu. of ac, cB, together with the {quare of cz. For upon cB defcribe the fquare ce (II. 1.), and | through A draw AF parallel to co (L. 27.), meeting Ep, produced, in F. ' Then, Jy and \Ey AC, equd to BDy BC, 2l to AB, tangles AB AC, al to the 5D, AU ¢ and ngle of e of BOOK THE §ECOND.: §7 Then, fince AE is a reGtangle, contained by ap, ek, it is alfo contained by aB, BC, becaufe BE is equal to BC (II Def. 2.) And, in like manner, AD is a reGtangle contamed by , AC, €D, or by Ac, cB; and CE is the fquare of cs (fy Conf.) But the rectangle AE is equal to the retangle ap, and the fquare cE, taken together; whence the re&angle of AB, Bc is alfo equal to the re&tangle of Ac, cB together with the fquare of cz, | Q. E. D. PR OP. XI TﬁEOkEM. If a right line be divided into any two parts, the {quare of the whole line will be equal to the {quares of the two parts, toge- ther with twice the retangle of thofe parts. E I‘Q F | 5 ) e | \\ | SR S ¢ Ak B Let the right line aB be divided into any two parts in the point ¢ ; then will the fquare of AB be equal to the fquares of Ac, cB together with twice the re@angle of AC, CB. For upon AB make the fquare AD (II 1.), and draw the diagonal EB ; and make ck, FH parallel to AE, ED (L 27.): "Then, 5"8 ELEMENTS OF GEOMETRY. Then, fince the parallelograms about the diagonal of 2 fquare are themfelves fquares (II. 7) FK will be the fquare of FG, or its equal Ac, and cH of cs. And fince the complements of the parallelograms about the diagonal are equal to each other (IL. 6.), the comple- ment AG will be equal to the complement Gp. But AG is equal to the re&tangle of ac, cB, becaufe - cG isequal to ¢B (II. Def. 2.); and ¢ is alfo equal to the reftangle of ac, cB, becaufe Gk is equal to GF (Def. 11, 2.) or ac (L 30.), and 6H to cB (1. 30.) The two rectangles Ac, G are, therefore, equal to twice the reCtangle of Ac, cB; and ¥k, cH have been proved to be equal to the fquares of ac, cB. \ But thefe two rectangles, together with the two fquares, make up the whole fquare AD ; confequently the fquare AD is equal to the {quares of Ac, cB, together with twice the rectangle of Ac, cB. . 2 £ D). Cororr. If a line be divided into two equal parts, the fquare of the whole line wxll be equal to four times the {quare of half the lme. PROP if PET: S , 0 edtar togeti The thefe 6 the why And HBy, to But ( fe&angh mon yp) thice th B 1\ llllll 0R BOOK THE SECOND. 59 PR OD X THEORT & If a right line be divided into any two parts, the {quares of the whole line, and one of the parts, are equal to twice the reCtangle of the whole line and that part, together with the fquare of the other part. l\j | J 24 . K sl o S Let the right line AB be divided into any two parts ift the point c ; then will the fquares of AB, BC, be equal to twice the rectangle aAB, BC together with the fquare of Ac. ; For, upon AB make the fquare AD (II 1.), and draw the diagonal BE; and make Fc, HK parallel to BD, BA (I 27, ) . | Then becaufe AG is eqml to'ep (IL. 6.), to each of thefe equals add ck, and the whole Ak will be equal to the whole c¢p. And, fince the doubles of equals are equal, the gnomon HBF, together with cx, will be the double of AK. But ck is a fquare upon cB (II. 7.), and twice the retangle AB, BC is the double of Ak, whence the gno- mon HBF, together with the fquare ck, is, alfo, equal to twice the reftangle AB, BC. And, 60 . ELEMENTS OF GEOMETRY. And, becaufe HF is 2 fquare upon HG or ac (II. 7.J, if this be added to each of thefe equals, the gnomon uzF, together with the fquares ck, HF, will be equal to twice the reangle AB, BC, together with the fquare of ac. But the gnomon uBF, together with the fquares cxk, HF, are equal to the whole {quare Ap, together with the fquare ck; confequently, the fquares of am, BC, are equal to twice the reCtangle aB, Bc together with the fquare of ac. - Q. E, D. PR OP XN Tutoein The difference of the fquares of any two unequal lines, is equal to a reGtangle unden theu {fum and dxﬁ'erence e v A . 0 S > a Let aB, ac be any two unequal lines ; then will the difference of the fquares of thofe lines be equal to a re&- ~ angle under their fum and difference. For, upon AB, Ac make the fquares AE, a1 (II.1.); and in HE, produced, take EG equal to Ac (L. 3.); and make GF parallel to EB (I. 27.); and produce ci, ik till they meet HG, GF in D and F. ‘Then, fince nEisequal to AB (Def.I1. 2.) and EG to ac (by Comfl.), ne will be equal to the fum of am and AC. And | aid £6 be equs But differer I angle equal Fo L u'»} Me th the LD, N } ROORI THESSSECOND,: Ga And becaufe aH is equal to AB, and Ak to ac (II.- Def. 2.), xu will be equal to cs, or the dxﬁ'erence of aB and Ac. : But the retangle kG is contained by G, and HK, whence it is, alfo, contained by the fum and dlﬁ'erence. of AB and Ac. And, fince LE is equal to 1K (1. 30.)or cB (& Cm_z/i 1 and EG to Ac (&y Conf.) c1, or LB, thereCtangle LG will be equal to the re@angle Lc (II. 2.) But the rectangles HL, Lc are, together, equal to the difference of the {quares AE, A1; confequently the ret- angles HL, 1G, or the whole retangle KG, is alfo equal to the difference of thofe fquares. | Q. E.D, PROP. XIV. THEoREM. In any right angled triangle, the fquarc'of the hypotenufe is equal to the fum of the {quares of the other two fides. & ¥ XC e D BB jf . /,,41& E j Let aBc be a right angled triangle, having the right angle acB; then will the {quare of the hypotenufe aB he equal to the fum of the fquares of Ac and cBs For, on AB, defcribe the fquare ae (Il. 1.), and on AC, cB the fquares AG, BH; and, through the point c, draw 62 ELEMENTS OF GEOMETRY. draw cL parallel to AD Or BE (L 27.)3 and join BF, €D, AK and CE. | Then, fince the right lme AC meets the two right lines GC;s CB in the point ¢, and makes each of the angles ~ AcG; AcB a right angle (by Hyp. and Def. 2.), cc will be in the fame right line with cs (I. 14.) ; * And, becaufe the angle FAc is equal to the angle nae (I. 8.), if the angle cAB be added to each of them, the whole angle FaB will be equal to the whole angle nac. The fides FA, aB, are, alfo, equal to the fides ca, AD, each to each, (Def. 2.), and their included angles have, likewife, been fhewn to be equal ; whence the tri« angle ABF is equal to the triangle acp (L. 4.) But the fquare AG is double the triangle asr (1. 32.) and the parallelogram AL is double the triangle Acp (I 32.); confequently the parallelogram AL is equal to the fquare AG (4. 6.) And, in the fame manner, it may be demonftrated, that the parallelogram BL is equal to the fquare BH; therefore the whole fquare AE is equal to the fquares ac and BH taken together. | o Ao, Cororr. The difference of the fquares of the hypo- - tenufe and either of the other fides is equal to the fquars of the remaining fide. PROP \ If angle theo thofe (] A (fy {Q‘Jll‘: ( And 8 equy ingles b <5 L atling ! n ’ & (4 ua.gleb 1 ™ 6¢ wil A AG LED. | Q ""pu % iquare 207 BOOK. THE SECOND. 63 ) O P XV THEOREM If the fquarc of one of t‘he fides of a tris angle be equal to the fum of the fquares of fthe other two fides, the angle contamed by thofe fides will be a right angle. Let ABc be a triangle ; then if the fquare of the fide AR be equal to the fum of the {quares of Ac, cB, the an- gle ace will be a right angle. For, at the point ¢, make cp at right angles to ce (I. 11.), and equal to ac (I. 3.); and join Ds. Then, fince the fquares of equal lines are equal (II. 2.), the {quare of pc will be equal to the fquare of Ac. And, if, to each of thefe equals, there be added the iquare of cB, the fquares of nc, cB will be equal to the {gpares of Ac, cB. But the fquares of pc, ce are equal to the fquare of D (II. 14.), and the fquares of Ac, cB to the {fquare of AR (by Hyp.); whence the fquare of BD is equal to the {quare of AB. And fince equal fquares have equal fides (1I; 3.}, As i3 cqual to 2D ; BC is alfo common to each of the tris angles ABC, DBC, and Ac is equal to cp (&y Conf.); ' con- 64 ELEMENTS OF GEOMETRY. confequently the angle AcB is equal to the angle scop (1 7.) - But the angle 5cp is 2 rxght angle (&y Conjt.), whence the angle Acs is alfo a right angle. Q. E. D. PROP. XVI. THEOREM. “The difference of the fquares of the two fides of any triangle, is equal to the diffe- rence of the fquares of the two lines, or diftances, included between the extremes of the bafe and the perpendicular. - D Let aBc be a triangle, having cp perpendicular to AB; then will the difference of the fquares of Ac, cB be equal . to the difference of the fquares of Ap, DB. ~ For the fum of the {quares of AD, pc is equal to the fquare of ac (IL 14.); 2nd the fum of the fquares of ‘BD, DC is equal to the fquare of Bc (IL. 14.) The difference, therefore, between the. fum of the fquares of ap, pc and the fum of the fquares of BD, D, ~ is equal to the difference of the {quares of acy CB. And, fince pc is common, the difference between the fum of the {quares of AD, pc, and the fum of the fquares of D, DC is equal to the difference of the fquares of ° AD, DB. : 3 But BOOK' THE SECOND. 65 ¢ BCD But things which are equal to the fame thing are equal to each other ; confequently the difference of the fquares ’i’em’ of Ac, ¢5 is equal to the difference of the fquares of ' D, AD, DB. | Q. EaBy Cororrs The reGtangle under the fum and difference- of the two fides of any triangle, is equal to the reGangle under the bafe and the difference of the fegments of the o bafe (II. 13.) e 5y Of PROP. XVI. Turoaem, In any obtufe-angled triangle, the fquare oi the fide {fubtending the obtufe angle, is greater than the fum of the fquares of the other two fides, by twice the re¢angle of the bafe and the diftance of the perpendmum lar from the obtufe angle. 0 AB; . . » equdl | e | 7] ‘ ‘ ‘V\V to the / | f e } ‘ o | , A B D T _______ Let asc be a triangle, of which asc is an obtufe an- | E gle, and cp perpendicular to AB; then will the {quare of B Ac be greater than the {quares of ap, Bc, by t‘mce the l i % retangle of AB, BD. — ":s. o8 For, fince the right line ap is divided into two parts, in the point B, the fquare of aD is equal to the {quares of 66 ELEMENTS OF GEOMETRY. AB, BD, together with twice the re@angle of Ap, BD (il 1) And if, to each of thefe equals, there be added the fquare of pc, the fquares of Ap, pcC will be equal to the {quares of AR, BD and Dc, together with twice the reét-- angle of AB, BD. ' But the fquares of Ap, DC are equal to the {quare of Ac, and the fquares of BD, Dc to the {quare of Bc (II. 14.) 3 whence the fquare of Ac is greater than the fquares of AB, BC by twice the reftangle of aB, 3p. Q. E. D. PR O O Tiionin In any triangle, the {quare of the fide - ‘tending an acute angle, is lefs than the fum_ of the {quares of the bafe and the other fide, by twice the retangle of the bale and the diftance of the perpendxcular from the acute - angle. | | Let aBc be a triangle, of which ABc is an acute an- A A B gle, and cp perpendicular to AB : then will the fquare of AC, be lefs than the fum of the {quares of AB and BC, by twice the rectangle of AB, BD. i . ~,For, fince Az, and B produced, are divided into two ‘parts in the points D, and A, the fum of the fquares of AB, 3 A BD % BOOGR THE SECOND 67 8D is equal to twice the retangle of AB, BD, together with the {quare of ap (II. 12.) And if, to each of thefe equals, there be added the fquare of bc, the fum of the fquares of AB, 8D and DC will be equal to twice the reftangle Of AB, BD, together with the fum of the {quares of AD, DC. - But the fum of the {1133V€a of BD, DC is .e‘qual%to the fquare of BC, and the fum of the fquares of AD, DC to the fquare of ac (II. 14.) ; whence the fquare of Acis lefs than the fum of the {quares of AB, BC, by twice the rectangle of AB, BD. ; - Q. B b P RO P XIX. I HEORE M. - In any triangle, the double of the fquare of a line drawn from the vertex to the mid- dle of the bafe, together with double the {quare of the femi-bafe, is equal to the fum f?hw {quares of the ozu’*‘ two fides. P / / /' \ £ ; ke FEEh o o / \ s AR AR \ / / \ i \ £ “ "\‘ / } \\ / A { | N / / \ "( [ \\ o / \ ] X \ \ Vit \ ; ; \ \ / | Yo s B SO e e e e . ATt g v B RE Y A A & AR 1 13 #h A Let anc be a triangle, and CE a line drawn from the vertex to the middle of the bafe aB; then will twice the fum of the {quares of cE, EA be equal to the fum of the fquares of Ac, CE. For on AB, produced if neceflary, let fall the peryen« dicular ep (L. 12.) Eoa "Then, 68 ELEMENTS OF GEOMETRY. Then, becaufe AEc is an obtufe angle, the fquare of AcC is equal to the fquares of AE, Ec together with twice the retangle of Ae, ep (II. 17.) ~ And, becaufe BEC is an acute angle, the {quare of cB together with twice the retangle of BE, ED is equal to the fquares of 8k, Ec (Il. 18.) And fince AE is cequal to EB (y Conjl.), the fquare of‘ BC together with twice the re@angle of AE, ED is equal to- the {quares of AE, Ec. But if equals be added to equals, the wholes will be cqual ; whence the fquares of Ac, cB, together with twice ~ the re@tangle of AE, ED, are equal to twice the fquares of - AE, EC, togethef with twice the re&tangle of AE, ED. And, if twice the reGtangle of Ax, rc, which is com- mon, be taken away, the fum of the {quares of ac, cB will be equal to twice the fum of the {quares of AE, EC. Q. E. D. PROP XL Turotew. In an ifofceles triangle, the {quare of a line drawn from the vertex to any point in the bafe, together with the reCtangle of the fegments of ‘the bafe, is equal to the {quare of one of the equal fides of the triangle. M / i / AR D B Let ABc bean ifofceles triangle, and cE a line drawn from the vertex to any point in the bafe Ap; then will the 3 - {quare ) BOOK THE SECON D. 6g iquare of CE, together with the reCtangle of AF., EB be equal to the fquare of Ac or ca. For bife the bafe as in b (I. 10.), and join the points E, D. Then, fince Ac is equal to cB, AP to DB, and cp : i1s common to each of the triangles Acp, BCD, the angle cpa will be equal to the angle cpe (I. 7.); and confequently ¢p will be perpendicular to aB (Def. 8 9.) And, becaufe Ack is a triangle, and cp is the per—- pendicular, the difference of the fquares of ac, cE is equal to the difference of the fquares of Ap, DE A 15 ] But, fince 8Eis the fum of ap and DE, and AE is their difference, the difference of the {quares of AD, DE is equal to the reflangle of AE, EB; confequently, the dif= ference of the fquares of ac, cE is alfo equal to the re@- angle AE, EB. And if, to each of thefe equals, there be added the - fquare of cE, the fquare of ac will be equal to the fquare of cE, together w1th the reCtangle of AE, EB. Q. E. D, F 3 S PRO P, v .~ ELEMENTS OF GEOMEIRY. PR Y XXI; TurOoREM. The diagonals of any parallelogram bifect each other, and the fum of their {fquares is equal to the fum of the [quares of the four fides of the parallelogram. D c Bl 7 v / \\\\; b / / P P ,,i // \\ Nl o A B Let aAsch be a parallelogram, whofe diagonals Ac, ED snterfect each other in E; then will AE be equal to £c, and BE to £ and the fum of the {quares of Ac, 8D will be equal to the fum of the {quares of AB, BC, CD and DA. G ‘For fince AB, pc are parallel; and Ac, BD interfe&t them, the angle pce will be equal to the angle EAB (1. 24.), and the angle cDE to the angle esa (L 24.) " The angle pec is likewife equal to the angle AEE (1. 15.), and the fide pc to the fide az (I. 30.); con- _fequently pE is alfo equal to EB, and ck to Ea (I, 21.) Again, fince DB is bifected in E, the fum of the {quares of pc, ce will be equal to twice the fum of the {quares of g, £c {1}.19.) | : And, becaufe pc is equal to AB, and ce to pa (1. 30.) the fum of the fquares of AB, CB, DC and DA are equal to four times the fum of the fquares of DE, EC. “But four times the fquare of DE is equal to the fquare of ep (II. 11. Cor.), and four times the fquare of Ec is ' equal B Z 24 3% SrRTI - of e, EA (IL 13.) BOOK THE SECOND. 71 equal to the fquare of Ac; whence the fum of the {quares of Ac, BD are equal to the fum of the fquares of AB, Bc, cD and DA. 2 k..D. PROP. XXU.. Prosrzum, To divide a given mght hne mto two parts, fo that the rectangle contained bj the whole line and one of the parts, fhall be equal to the {quare of the other part D Let ar be the given right line ; it is required to divide it into two parts, fo that the reGangle of the whole line and one of the parts fhall be equal to the fquare of the other part. Upon aB defcribe the fquare ac (IL 1.), and b1fe&: the fide of it Ap in £ (1. 10.) Join the points B, £ and, in £A produced, take EF equal to EB (I. 3.); and upon AF defcribe the fquare FH (lI. 1.) Then will AB be divided in u fo, that the reé’cangle AB, BH, will be equal to the fquare of an, For, fince DF is equal to the fum of &8 and ED, or its equal E4, and AF is equal to their difference, the re- angle of DF, FA is equal to the difference of the fquares Fa But 2 ELEMENTS OF GEOMETRY. But the re@angle of DF, FA is equal to DG, oecaufc ra is equal to FG (II. Def. 2.); and the difference of the fquares of EB, EA is equal to the {quare of AB (1. 14. Cor.) ; whence DG is equal to Ac. And, if from each of thefe equals, the part DH, which 1s common to both, be taken away, the remainder ac wili be equal to the remainder HC. | But uc is the re&angle of AB, BH; for AB is equal to pc; and AG is the fquare of AH; therefore the right line aB is divided in 1 fo, that the reCtangle of AB, BH is equal to the fquare of AH, which was to be done. BOOK n Lon A , €l d 0K BOOK THE THIRD, 73 \ B O . O-K 1l DEFINITIONS: 1. A radius of a circle, is a right line drawn from the " centre to the circumference. 2 2. A diameter of a circle, is a right line drawn through the centre and terminated b_o\th ways by the circums- ference, § 3. Anarc of a cxrcle, is any part of its periphery, or circumference., | '../'\ % ’ 4. The chord, or fubtenfe, of an arc, 15 2 nght line joining the two extremities of that arc. . . I s o - - - b 3 . % o . L 03 ° 0 ® 5. A femicircle, is a figure contained under any diame- ter and the part of the circumference cut oft by that diamCterv 6. A 74 ELEMENTS OF GEOMETRY. 6. A fegment of a circle, is a figure contained under any arc and the chord of that arc. . ....... 4. A tangent to a circle, is a right line which paffes through a point in the circumference without cutting it. 8. Right lines, or chords, are faid to be equally diftant from the centre of a circle, when perpendiculars drawn to them from the centre are equal. 9. And the right line on which the greater perpendi. cular falls, is faid to be farther from the centre. 10. Anangle in a fegment, is that which is contained by two right lines,.drawn from any point in the arc of the {egment, to the two extremities of the chord of that arc, <) 11. One circle is faid to touch another, when it paflas through a point in its circumference without cutting it. e) PROP. £ . 14185 T BO0OK THE THIRD, 75 P RIEE L PROBI’,EM; o find the centre of a given circle, Let asc be the given circle ; it is requircd to find its centre. Draw any chord AB, and bife& it in D (L IO ), and through the point D draw cE at right angles to aB (L. 11.), and bifect it in F: then’ will the point ¥ be the centre of the circle. | For if it be not, fome other point mu{’c be the centre, either in the line Ec, or out of it. ‘ But it cannot be any other point in the line EC, for if it were, two lines drawn from the centre of the circle to ltS circumference would be unequal, which is abfurd. Neither can it be any point out of that line; for if it can, let G be that point; and join GA, GD and GB. Then, becaufe ¢A is equal to GB (I. Def. 13.), AD tO pB (by Conjfi.), and GD common to each of the trian- gles AGD, BGD, the angle ADG will be equal to the angle 8p6 (1. 7.) ‘ : ~ But when one line falls upon another, and makes the adjacent angles equal, thofe angles are, each of them, right angles (1. qu 8 and 9 o) ' - b 'ELEMENTS OF GEOMETRY. The angle apg, therefore, is equal to the angle apc (1. 8.), the whole to the part, which is abfurd ; confe- quently no point but F can be the centre of the circle. Q.E. D. Cororr. If any chord of a circle be bifeéted, a right line drawn through that point, perpendicular to the chord, will pafs through the centre of the circle. PROP 1. Tuzorsm. If any two points be taken in the circum- ference of a circle, the chord, or right line ~which joins them, will fall wholly within the circle, Let ABE be a circle, and A, B any two points in the circumference ; then will the right line AB, which joins thefe points, fall wholly within the circle. | For find c, the centre of the circle age (IIl.1.), and join ¢, A, ¢, B; and through any point b, in aB, draw the rightline cE, cutting the circumference in E. Then, becaufe ca is equal to cB (I. Def. 13.), the ‘angle cas will be equal to the angle cea (L. 5.) And, fince the outward angle cps of the triangle ACD, is greater than the inward oppofite angle cas (I. 16.), it will alfo be greater than the angle cza, | But The the fan fequen If : centr perpe inpe 4 il {C ) — i‘hmu;‘n then wil For ‘( T}}@n, OBy (b gles Any {e Dgp But g When It 0 Ottr ], Chyg AB, BOOGK "THE: BEBIRD; i But the greater fide of every triangle is oppofite to the greater angle (L. 17.); whence cz, or its equal ck, will be greater than cp. ; \ , - The point p, therefore, falls within the circle ; and ‘the fame may be thewn of any other boint in AB; con- fequently the whole line aAB muft fall within the circle. QG B D PROP. IlI. THEOREM. If a right line, which pafles through the centre of a circle, bife¢t a chord, it will be perpendicular to it ; and if it be perpendiéum lar to the chord, it will bife& it. | C A% E /8 k \J"/ o Let arc be a circle, and ce a right line which pafles through the centre D, and bifects the chord aB in £ then will ce be perpendicular to as. For join the points ADp, DB : Then, becaufe Ap is equal to pe (II. Def. 13.), AE to EB (4y Hyp.), and ED common to each of the trian- gles ADE, BDE, the angle pEA will be equal to the an- gle pze (1. 7.) | But one line is faid to be perpendicular to another, when it makes the angles on both fides of it equal to each other (I. Def.8.); confequently CE is perpendicular tothe chord ag. Again, 28 ELEMENTS OF GEOMETRY. Again, let the right line DE be drawn from the centre ~ p, perpendicular to the chord AB; then will AB be bifect- - ed in the point E. For join the points AD, DB, as before : ' ‘Then, fince the angle DAB is equal to the angle pBA (I. 5.), and the angle AED to the angle DEB, (being each of them right angles) the angle Apz will alfo be equal to the angle EpB (I. 28. Cor. 1.) | And, becaufe the triangles pEA, DEB are mutuaﬂy equiangular, -and have the fide DE common, the fide AE will alfo be equal to the fide EB (I. 21.); whence ABis bifeGted in the point E, 2s was to be fhewn. Corory. If a right line be drawn from the vertex of an ifofceles triangle, to the middle of the bafe, it will be perpendicular to it; and if it be perpendicular to the bafe, it will bife& both it and the vertical angle. PROP. IV. THEOREM. f more than two equal right lines can be drawn from any point in a circle to the cir- cumference, that point will be the centre. 1 a7, \ - f(‘iL/AB i Let annc be a circle, and o a point within it ; then if any three right lines 0A, 08, 0C, drawn from the point o to the circumference, be equal to each other, that point will be the centre. ’ ' ". For | ol i ke | B g | b | i | E | g | ol 1* natd It DR N WIEN A RO oK DHE THERD. 78 For draw the lines aB, ac, and bife& them in the ‘ points ¥, ¢ (L 10. ); and through the centre o, draw FD, GE, cutting the circumference in b and E. Then, fince AF is equal to FB (2y Con/l.), Ao to 0B (&y Hyp.), and oF common to each of the triangles aor, BOF, the angle aro will be equal to the angle Bro (L.7) And becaufe the right b orF falls upon the right line AB, and makes the adjacent angles equal to each other, ‘oF will be perpendicular to az (I. Def. 8.) But when a right line bifects any chord at right angles, it paffes through the centre of the circle (IIL. 1. Cor.) 5 whence the centre muft be fomewhere in the line rp. And, in the fame manner, it may be thewn, that the centre muft be fomewhere in the line GE. But the lines rp, GE have no other point but o which 1s common to them both ; therefore o is the centre of the circle ABD, as was to be thewn. PROP V. TdHrotreu Circles of equal radii are equal to each | other ; and if the circles are equal the radii will be equal Let asc, DEF be two circles,.of which the radii GA, Ge are equal to the radii Hr, HE ; then will the circle ABC be equal to the circle DEF. ; For 86 ELEMENTS OF GEOMETRY. For conceive the circle 'DEF to be applied to the circle #sc, fo that the centre m may coincide with the centr e G " Then, fince the radii HF, HE are equal to the radi GA, GB (by Hyp.), the points F, E will fall in the cir- cumference of the circle asc (I. Def. 13.) ; and the fame may be fhewn of any other point D. But fince any number of points, taken in the cu‘cum- ference of the circle peF, fall in the circumference of the circle aBc, the two circumferences muft coincide, and confequently the circles are equal to each other. Again, let the circle ABc be equal to the circle DEF 5 then will the radii ca, 6B be equal to the radii HF, HE. For if they be not equal, they muft be either greater or lefs: let them be greater; and apply the circles to each other as before. Then, fince the radii GA, GB are greater than the radii HF, HE, the points F, E will fall within the circle ABC ; and the fame may be fthewn of any other point D. But, fince any number of points, taken in the circum- ference of the circle DEF, fall within the circle asc, the whole circle pEF muft, alfo, fall within the circle aBc. The circle DEF is, therefore, lefs than the circle Asc, and equal to it at the fame time (&y Hyp.), which i1s abfurd : whence the radii Ga, GB are not greater than the radii HF, HE. And in the {ame manner it may be thewn that they cannot be lefs ; confequemlv they are equal to each other. QL. 17, CORQLL. Equal circles, or fuch as have equal radii, -or diameters, have equal circumferencess PROP. Fo the diz An¢ ifpoff and jo; g The togethe quil 5 Ang Comm¢ than th But, €3 s & freater The | Crcle point g Circle i the ¢ radi e Clre e fame reum- BOOK THE THIRD. 81 PROP. VI. THEOREM, If two circles touch each other internally, | the centres of the circles and the point of contact will be all in the fame right line, Let the two circles BEG, BDF touch each other inter- nally at the point B ; then will the centres of thofe cir- cles and the point B be in the fame right line. For let A be the centre of the circle BEG, and draw ‘the diameter GB. And if the centre of the circle pF be not in ¢, let, if poflible, fome point ¢, out of that line, be the centre 3 and join A,c, c,B; and produce AC to cut the cxrcles in D and E. - Then, fince Acs is a triangle, the fides ac, ¢z, taken together, are greater than the fide aB (I. 18.), or its equal AE. And if, from thefe equals, the part ac, thlCh is common, be taken away, the remainder c¢B will be greater than the remainder cE. , | But, fince c is the centre of the circle BpF (Ly Hyp.), ¢B is equal to cp (I. Def. 13.) 5 whence cp will alfo be greater than ce, which is impoffible. The point ¢, therefore, cannot be the centre of the circle BDF; and the fame may be fhewn of any other point out of the line Ag. o AL B & PR OF, Qs ELEMENTS OF GEOMETRY. PROP. VIL. THEOREM. If two circles touch each other externally, the centres of the circles and the point of conta@ will be all in the fame right line. Let the two circles BEG, BDF touch each other ex- ternally at the point B 5 then will the centres of thofe cir- cles and the point B, be in the fame right line, For, let A be the centre of the circle BEG, and draw the diameter GB, which produce tzl it cuts the circle BDF 0.5, And, if the centre of the circle BDF be not in the line - AF, let, if poffible, fome point ¢, out of that line, be the centre ; and join C, A, C, B. Then, fince A is the centre of the circle Lh(;, AE IS equal to a8 (L. Def. 13.) And becaufe ¢ is th centre of the cxrcle BEDF \by Hyp. ), ¢p is equal to B (L. Def. 13.) But aB, BC, togzether, are greater than ac (L. 18.)3 therefore AE, CD, together, are alfo greater than AcC; which is abfurd. : The point c, therefore, cannot be the centre of the circle 8pF 3 and the fame may be fhewn of any other point out of the line AF. | | 2. E..D 1 PROP, 211 ..... ‘BODK MHE THLRD.r 83 ERO R VI Yoedr il Any two chords in a circle, which are equally diftant from the centre, are equal to each other ; and if they be equal to each other, they will be equally diftant from the centre, A D Let ABED be a c,rcle, whofe centre is 0 ; then will any two chords A®, DE, which are equahy dxﬁant from 0, be equal to each other. ( For join the points A0, oD, and let fah the perpendx«” culars oc, OF (L. 12.) Then, fince a right line, drawn from the centre of a circle, at right angles to any chord, bife&ts it (III. Ty ac will be equal to cB, and DF to FE. And, becaufe the angles Aco, pFo are right angles, the fquares of ac, co will be equal to the {quare of a0 (II. 14.), and the fquares of DF, Fo to the fquare of Do. But the fquare of Ao is equal to the {quare of op (II. 2.); confequently the fquares of Ac, co will be equal to the fquares of DF, Fo. And fince oc is equal to oF (IIL Def. 8.), the fquare of oc will be equal to the fquare of oF (II. 2.) ; whence &9 the 84 ELEMENTS OF GEOMETRY. the remaining fquare of ac will alfo be equal to the re- maining fqﬂare of pF; or Ac equal to DF (2.0 and aB to pE (L. £x. 6.) - ; ~ Again, let the chord A be equal to the chord DE; then will oc, oF, or their diftances from the centie, be equal to each other. For the fquares of Ac, co are equal to the {quare of oA (Il 14.)s and the {quares of pr, ¥o to the fquare of oD. | : . ‘ But the fquare of oA is equal to the fquare of oD (1L. 2.) ; therefore the fquafes of ac, co are equal to the fquares of DF, FO. : And fince ac is the half of aB (1II. 3.), and DF is the half of pE (IIL. 3.), the {quare of Ac is equal to the fquare of or (II. 2.) o | The remaining fquare of co is, therefore, equal to the remaining fquare of Fo 3 and confequently €O is equal to “¥o (II. 3.), as was to be thewn. i | Corory. If two right angled triangles, having equal hypotenufes, have two other fides alfo equal, the re- maining fides will likewife be equal, and the triangles will be equal in all refpects. | PROP. Let and the FG, AD! For ¢, ind join Then, 1 o, 4 Buto therefore Again of on ] of op, DF 1§ the e quare 4 { 5 fotne eq\m\ {0 | g, equd the re- tes wil gies W DROP' s $G0R TREPHINBIE - 8y PRO/P. IX. THEOREM. - A dlametcr is the greateft right line that can be drawn in a circle, and, of the reft, that which is nearer the centre is greater than that which is more remote. Let aBcp be a circle, of which the diameter is AD, and the centre o0 ; then if Bc be nearer the centre than FG, AD will be greater than Bc, .and Bc than Fe. ‘ For draw oH, ox- perpendicylar to Bc, FG (L 12.), and j Jo'n 0B, 0€, 0G and OF. Then, becaufe oA is equal to or (I. Def: 13. ), and oD to oc, AD is equal to 0B and oc taken together, But 0B, oc, taken together, are greater than sc (1. 18.); therefore AD is alfo greater than sc. Again, the {quares of oH, HB are equal to the fquare ~of 0B (II. 14.), and the fquares of ok, kF to the fquarc of OF. Butthefquare of oB is equal to thefquare of oF (Il.2.); whence the fquares of oH, HB are equal to the fquares of oK, KF. | And {ince FG is farther from the centre than Bc (by Hjp.), ok will be greater than on (III. Def. g.), andthe fquare of ok than the {quare of on (II, 4.) G 3 The £ - 86 ELEMENTS OF GEOMETRY. The remaining {quare of HB, therefo;e, is greater than the 1emam1ng fquare of K, and HB greater than kKr (. 4.) ~ But BCis the double of BH, and FG is the double of Fk (1. 3.); confequently BC i$ alfo greater than FG. - QE.D, PROP X, THEOREM. A right 1in¢-dréwn perpendicular to the diameter of a circle, at one of its extremi- ties, is a tangent to the circle at that point. \J‘ Let ABcC be a circle whofe centre is E, and diameter AB ; then if DB be drawn perpendlcular to aB, it will be a tangent to the circle at the point B. For in BD take any point ¥, and draw EF, cutting the circumference of the circle in c. L Then, fince the angle EBD is a right angle (6y Hyp.), ~ the angles BEF, EFE will be each of them lefs than a right angle (1. 28. ) And, becaufe the greater fide of every triangle is op- pofite to the greater angle. (1. 17. s the iide EF is greater than the fide £8, or its equal Ec. ~ But fince EF is greater than EC, the point F will fall without the cxrcle ABC and the fame may be thewn of any other point in BD, except B. The o the 1Ml RoOK- - PHE THIRD 87 -The line 8D, therefore, cannot cut the circle, but muft fall wholly without it, and be a tangent to it at the point B, as was to be thewn, , ScuoriuM. A right line cannot touch a circle in more than one point, for if it met it in two points it would fall wholly within the circle (1I1. 2.) | PROTP X Piuoerem. ‘From a given point to draw a tangent toa given circle. | Let A be the given point, and Fpc the given circle; it is required from the point A to draw a tangent to the circle Fpc. | | Find E, the centre of the circle FDC (HI 1.), and join EA; and from the point E, at the diftance EAé defcribe the circle cAB, - Through the point D, draw DB at right angles to Ea (I. 11.), and join EB, Ac; and ac will be the tangeﬁt required. " For, fince E is the centre of the circles FDC, GAB, EA is equal to EB, and ED to EC. And, becaufe the two fides EA, EC, of the tnanOI EAC, are equal to the two fides EB, ED, of the triangle z:gm,’ ~ and the angle E common, the angle Eca will alfo be equal to the angle b8 (L. 4.) G 4 = hHut 88 ELEMENTS OF GEOMETRY. But the angle EpB being a right angle, the angle rca is alfo a right angle; therefore fince Ac is perpen- ~ dicular to the diameter Ec, it will touch the circle Fpc, and be a tangent to it at the point ¢ (IIL 10.) PROP. XII. TuEOREM. If a _rightlinc be a tangent to a circle, and another right line be drawn from the centre to the point of conta®, it will be perpendi- cular to the tangent, - ¥ [ R Let the right line DE be a tangent to the circle ABc at the point B, and, from the centre F, draw the right line FB ; then will F& be perpendicular to DE. For if it be not, let, if poffible, fome other right line ¥G be perpendicular to DE. | N D G K Then, becaufe the angle FGE is a right angle (by Hyp.) the angle G will be lefs than a right angle (I. 28.) And, fince the greater fide of every triangle is oppofite to the greater angle (I. 17.), the fide & will be greater than the fide Fe. ' But F8 is equal to ¥c ; therefore rc will alfo be greater than FG, a part greater than the whole, which is 1m- ~ poflible. | | k The Qi'Ei T o\ 20¢ ) 1?9611&3 , i;ed(\.r : gffatef h 15 Ime The oS S O BB s i BOOK THE THIRD. 89 The line Fo, therefore, cannot be perpendicular to DE; and the fame may be demonftrated of any other line but ¥B ; confequently FB is perpendicular to DE. Q. E. D. PROP. XIII. THEOREM. ~ If a right line be a tangent to a circle, and another right line be drawn at right angles to it, from the point of contad, it will pafs through the centre of the circle. g S, D B Let the right line DE be a tangent to the circle acs at the point B; then if AB be drawn at right angles to b, from the point of contact B, it will pafs through the centre of the circle. : For if it does not, let »,"if poflible, be the centre of the circle; and join FB. Then, fince DE is a tangent to the circle, and FBis a right line drawn from the centre to the point of contaé’c the angle FeE is a right angle (III. 12.) But the angle ABE is alfo a right angle, by conftruc- tion; whence the angle FBE is equal to the anglé ABE ; ihe lefs to the greater, which is impoffible, The point F, therefore, is not the centre; and the fame may be fhewn of any other point which is out of the line AB; confequentiy AB muft pafs through the centre of the circle, as was to be thewn, PROP g0 ELEMENTS OF GEOMETRY. PROP. XIV. THroREM. An angle at the centre of a circle is dou- ble to that at the circumference, when both of them ftand upon the fame. arc, Let the angle BEC be an angle at the centre of the circle ABc, and BAC an angle at the circumference, both ftanding upon the fame arc Bc ; then will the angle BEC be double the angle BAC, " Firft, let E, the centre of the circle, be within the angle Bac, and draw AE, which produce to F. Then, becaufe EA is equal to EB, the angle EAB will be equal to the angle EBA [1e:54) And, becaufe AEB is a triangle; the outward angle BEF will be equal to the two inward oppofite angles EAB, EBA, taken together (I. 28.) But fince the angles EAB, EBA, are equal to each other, they are, together, double the angle EAB; whence the angle BEF is alfo-double the angle EAB. And, in the fame manner it may be fhewn, that the angle FEC is double the angle EAC; confequently the whole angle BEc will alfo be double the whole angle BAC. Again, and t theref angle are ¢ Le in th BDC | For fﬁmicir ?Hd B BOOK THE THIRD. g1 | Again, let E, the centre of the circle Asc, fall without the angle Bac, and join AE. Then, fince the angle BFE,of the triangle EF B, is equal i to t.he angle cra of the triangle car (I. 15.) the re- | maining angles BEF, FBE of the one, are, together, both equal to the remaining angles Fac, Fca of the other But the angle FBE is equal to the angle Ear (L. 5.), and the angle Fca, or Eca, to the angle Eac (I.5.); therefore the angles BEF, EAF, are, together, equal to the angles FAC, EAC. | | | ' And, if the angle EAF, which is common, be taken away, the remaining angle BEF or BEC, will be equal to 1 twice the angle FAc, or BAC, as was to be fhewn. e, botd ‘ | gleBEC 8 ,‘ PROP. XV. THEOREM. hin All angles in the fame fegment of a circle . B are equal. to each other. Ew«,»" & & A 7 s 'ﬂ 5 Fiie S ! ‘,«,_/,___\\D 5 7 QN ole BEF ¥ i # \\\ﬁ'r , EBAy @ Nl OUivly | ence the Let ancp be a circle, and BAC, BDC any two angles in the fame fegment BADC ; then will the angles Bac, hat e BDc be equal to each other. : oy ¢ For, firft, let the fegment BADC be greater than a g femicircle, and having found the centre E, join BE and EC. | Then, 92 ELEMENTS OF GEOMETRY. » Then, fince an angle at the centre of a circle is double - to that at the circumference (I1II. 14. ), the angle Bac will be half the angle BEC. ; o And, for the fame reafon, the angle Bpc will, alfo, be half the angle BEC. But things which are halves of the fame thing are equal to each other ; confequently the angle BAC is equal to the angle BDC. Again, let the fegment BADC be not greater than a femicircle : ' Then, fince the right lines 8D, Ac interfect each other in F, the angle srA will be equal to the oppoﬁte angle pre (L. 135.) s , ‘And becaufe the fegment ABcD is greater than a femi- circle, the angles’ ABp, AcD, which ftand in thas feg~ ment, are equal to each other (IIl. 15.) But fince the two angies BFA, ABF of the triangle FBA, are equal to the two angles prc, rcp of the tri- angle DCF, the remaining angle BAF, or BAC, will alfo be equal to the remaining angle Fpc, or BDC. : Q. E. D, PROP. XVI. THEOREM, “An angle in a femicircle 1s a right angle. ‘Let anc be a femicircle ; then will any ang?e ACB, iR that femicircle, be a right angle. o FN | For, e ) An a femi ¥, AD(] Io§ef $ in wUﬂC CBAC By i For, BOOK THE THIRD. 93 For, find the centre of the circle £ (IIL. 1.) and draw the diameter CED. | Then, becaufe an angle at the centre of a circle is double to that at the circumference (LI, 14), the angle AED will be double the angle aAcp. And, for the {ame reafon, the angle BED W111 be double the angle BCD. : | The angles AED, BED, therefore, taken together, are double the whole angle AcB. 1 But the angles AED, BED, are, together, equal to two right angles (I.13.) confequently the angle acs will be equal to one right angle. Q. E.D. Cororr. The angle Bac, which ftands in a fegment ~greater than a femi-circle, is lefs than a right angle (1. 28.): ‘And the angle BCF, Whmh ftands in a fegment lefs than a femi=circle, is greater than a right angle. | P R’V‘»O P, X Vil THEOREM. “The oppofite angles of any 'quadri*'lateral ﬁ figure, infcribed in a circle, are equal to two right angles. Tet aBcp be a qvadrilateral, infcribed in the circle aDcE; then will the oppofite angles BAD, BCD, taken together, be equal to two right angles. For, 94 ELEMENTS OF GEOMETRY. For, draw the dxagenals Agy HD, and produce the fide BA to E. Then, becaufe the outward ,.angl'e of any triangle, is equal to the two inward oppofite angles taken together (1. 28. ), the angle aD WIH be equal to the angles ABD, ADB. ' | . B i : , And, becaufe all angles in the fame fegment of a circle are equal to each other (II 15.), the angle Asp will be equal to the angle AcD, and the angle ADB to the angle ACB. | - The angle EaD, therefore, which is equal to the angles ABD, ADE, taken together, will alfo be equal ‘to the angles ACD, ACB, taken together, or to the whole an= ~ gle BcD. But the angles EAD, BAD, taken together, are equal to two right angles (1. 13.) ; confequently the angles Bcp; ~ BAD, taken together, will alfo be equal to two right , angles. Q... D, Corovrr. If any fide as, of the quadnlateral ABCDy be produced the outward angle EAD will be equal to the inward oppofite angle Bcp, 0 R, et - fame 1 ference Dray the pery Then gether S ABI J Circle y wil 0 iﬁc BRI T HESTHEEDL . (g% PROP, XVIL PromLEM. Through any three points, not fituated in the fame right line, to defcribe the circum- ference of a circle. n«z,_{__’:\_ S Toa A< s \ Vae \ / / A Let A, B, c, be any three points, not {ituated in the fame right line; it 1s required to defcribe the circums ference of a circle through thofe points. | Draw the right lines AB, BC, and bifect them with the perpendiculars pH, G (1. 10474 11.) ; and join DE. Then, becaufe the angles FED, FDE are lefs than two right angles, the lines pH, EG will meet ‘each other, in fome point ¥ (Cor. 1. 25.) ; and that point will be the centre of the circle required. ‘ For, draw the lines Fa, ¥5 and rc. ‘Then, fince ap is equal to DB, DF common, and the angle aDF equal to the angle ¥pB (I. 8.), the fide ra will alfo be equal to the fide 3 (1. 4.) And, in the fame manner, it may be fhewn, that the fide ¥c is alfo equal to the fide FB. | The lines FA, FB and Fc, are, therefore, all equal to each other ; and confequently F 1s the centre of a circle which will pafs through the points A, B and ¢, as was to be fhewn. | S i b | ScHo, ) g6 ELEMENTS OF GEOMETRY, Scuo. If the fegment of a circle be given, and any three points be taken in the circumference, the centre of the circle may be found, as above. PROP. XIX, Turcdarn If the cppoﬁté angles of a quadrilaterai, taken together, be equal to two right angles, a circle may be defcnbed about that quadri= lateral, Let apco be a quadrilateral, whofe oppofite angles DCB, DAB are, together, equal to two right angles : - then may a circle be defcribed about that quadrilateral. For fince the circumference of a circle may be de- {cribed through any three points (III, 18.), let E be the centre of a circle which pafles through the points b, c 8 ; and draw the indefinite right line £ra. And if the circle does not pafs through the fourth point A, let it pafs, if pofiible, through fome other point F, m the line EA, and draw the lines br, rB, and BD. Then, fince the oppofite angles BFD, DCB are, together, - equal to two right angles (III. 17.}, and the angles BAD, pcB are allo equal to two right angles (4y Hyp.), the .éngles BFD, DcB will be equal to the angles BaD, DCE. And And angle angle which be e poﬁﬁ:’e The througt of any whence ABCD, t’q ua 1 equal Ia | Upon then For fo that 1 1)) Upo The, Wil fl neide ¢ and any centre of ateral, amples, ;uadffa nts Dy L th point it F; m » nﬂ(hfr Osw ] o5 BAD) gJ,the ) And RGOK THRE TRHIRD. - O And if from each of thefe equals there be taken the angle pcB, which is common to both, the remaining angle BaD will be equal to the remaining angle 8¥p ; or, which is the fame thing, the two angles prE, EFz will be equal to the two angles pAE, EAB, which is im= poffible (I. 16.) The circumference of the circle, therefore, cannot pafs through the point 7 ; and the fame may be demonftrated of any other point in the line Ea, except the point A whence a circle may be defcribed about the quadrilateral ABCD, as was to be thewn, FEROU P XX THEOREM.‘ ~ Segments of circles, which ftand upon equal chords, and contain equal angles, are equal to each other. (3 | A Let Ace, DFE be two fegments of circles, which ftand upon the equal chords AB, DE, and contain equal angles ; then will thofe fegments be equl to each other. | For let the fegment DFE be applied to the fegment Acg, fo that the point » may fall upon the point 4, and the line 'DE upon the line AB. Then, fince pE is equal to AB (& Hyp.), the point & will fall upon the point B, and the two fegments will coe incide with each uther, Lt | H | For g8 = ELEMENTS OF GEOMETRY. For if they do not, there muft be fome point, in the cir- ‘cumference of one of them, which will fall e1ther Wxthm or without the other.: : Let the pomt F, in the circumference of the circle DFE, be that point, “which fuppofe to fall at ¢ within the cxrcIe ‘Ace ; and draw the lines ace, Bc and G, Then, fince the outward angle acs, of the tnanfrlc BCG, is greater than the inward oppolite angle GCB, it will alfo be greater than the angle DFE, which is eﬂuai t0 GCB, or ACB (by [—]yp i But the angle AGB is allo equal to the angle pFE, be- caufe the fegments in which they fland are identical ; whence they are equal and unequal at the fame time, which is abfurd. The point F, therefore, cannot fall within the circle AcBj and in the fame manner it may be thewn that it cannot fall Wxthout its confequently the fegments muft coincide, and be equal to each other. R 1) CoroLrr. Segments of circles, which ftand upon equal chords, and contain equal angles, have equal circum- ferences. PROP. In €] il 0 Clf the i Let, .&CB, D} the angle the arc 4 For, i cles are Clrcumfer And, { ale equa and the z bafes AB, The o} 2 the , BCA wil , BUI ﬁn; qual to gh the e g, Cqual [0 th ﬂﬂﬂﬂﬂ 00 BOUK T HE "THIRD, 00 P R O P. XXI.{ THEORE M, In equal circles, equal angles ftand upon equal arcs, whether they be at the centres or circumferences ; and if the arcs be equal, the angles will be equal. Let aBc, DEF be two equal circles, having the angles AGB, DHE, at their centres, equal to each other, as alfo the angles AcB, DFE, at their circumferences ; then will the arc AKE be equal to the arc pLE. For, join the points AB, DE: then, fince the cir- cles are equal to each other (4y Hyp.), their radii and circumferences will alfo be equal (III. s5.) And, fince the two fides AG, GB of the triangle aBG, are equal to the two fides DH, HE of the (triangle DEH, and the angle AGB to the angle puz (Jy Hyp.), their bafes aB, DE will likewife be equal (I. 4.) ‘The chord Ap, therefore, being equal to the chord pE, and the angle Acs to the arigl\e DFE (by Hyp.), the arc BcA will alfo be equal to the arc EFD (Cor, 111. 20.) But fince the whole circumference of the circle agc, is equal to the whole circumference of the circle per, and the arc BCA to the arc EFD, the arc aks will alfo be equal to the arc pLE. H 2 Again, 100 ELEMENTS OF GEOMETRY., Again, let the arc Ax8 be equal to the arc pLE; then will the angle Ace be equal to the angle puE, and the angle Acs to the angle DFE. \ For, if acs be not equal to DHE, one of them muft be greater than the other; let AGs be the greater; and make the angle Ack equal to pHE (l. 20.) ‘ Then, fince equal angles ftand upon. equal arcs (III. 21.), the arc Ak will be equal to the arc DLE. But the arc DLE is equal to the arc aAks (&y Hyp.) ; whence the arc Ax is alfo equal to the arc AxB; the lefs to the greater, which is impoffible. The angle acs, therefore, is not greater than the angle DHE ; and in the fame manner it may be proved that it cannot be lefs ; confequently they are equal to each other. ~ And fince angles at the centre are double to thofe at the circumference, the angle ace will alfo be equal to GED. ’ ‘the angle DFE. PROP. XXIL. TrrEoREM. In equal circles, equal chords {ubtend equal arcs, the greater equal to the greater, and the lefs to the lefs ; and if the chords be equal the arcs will be equal. Tet ABc, DEF be two equal circles, in which the chord aB is equal to the chord pE; then will the arc ACB ACB be € ¢ DLE, For, fir and join € 'I‘ben Hy) U5 DA, HE, ingle ACB But equ upon equa to the arc | And fing equal to te arc 4, equal to the Again, | the arc axy AB be equa Forlet ¢ fO'{?' nd' Th\n 111 H’ arre | o v' { ak proved {nat h ath each otler, to thofe at | wklc‘\ thc e 4C | m“ s 8 POGR 1T HE THIRD, 101 ACB be equal to the arc DFF, and the arc AxB to the arc DLE, ’ For, find ¢, 1, the centres of the ecircle and join GA, GB, HD and HE. Then, fince the circles are equal to each other (& Hyp.) their radii and c1rcumferences will alfo be equal (s, | : And, fince the fides AG, GB are equal to the fides DH, HE, and the bale AB to the bafe pE (4y Hyp.), the angle AGB will alfo be equal to the angle puEe (L. 21.) But equal angles, at the centres of equal circles, ftand upon equal arcs (III. 21.) ; therefore the arc AxBis equal to the arc DLE. s (HL 1. ), And fince the whole circumference of the circle ABC is equal to the whole circumference of the circle pEF, and the arc AKB to the arc DLE, the arc acs will alfo be equal to the arc DFE, Again, let ABc, DEF be two equal circles, of whxch the arc AKB is equal to the arc DLE ; then will the chord AB be equal to the chord DE. ; For let G, H be the centres of the circles, found as be- fore; and join AG, GB, DH and HE. | . Then, fince the circles are equal to each other {4y H}p )s the radii AG, 6B will be equal to the radii DH, e (1L 5.) And becaufe the arc AKB is equal to the arc pLE (4y Hyp.), the angles AGB, DHE, at the centres, will be equal (III. 21.) | But, fince the two fides AG, GB are equal to the two fides pH, HE, and the angle AGB to the angle puz, the bafe aB will alfo be equalto thebafepe (I.4.) Q. E.D, H 3 rYROP 102\ ELEMENTS OF GEOMETRY. PROP. XXII. ProsrLEm. To bxﬂ,& a given arc, that is, to divide it into two equal parts. Let ApB be the given arc ; it is required to divide it into two equal parts. Draw the rightline A8, which bife¢t in ¢ (I. 10.) ; and, from the point c, erect the perpendicular co (I. 11.); then will the arc Aps be bifected in the point D, as was required.' . | : , | For, join the points Ap, pB: then, fince the two fides Ac, ¢cp, of the tri\angle ADC, are equal to the two fides Bc, cp of the triangle BDC, and the angle acp to the angle Bco (L. 8. ), the bafe Ap will be equal to the bafe DB (I. 4) And, becaufe pc, or pc produced, pafles through the centre of the circle (IIl. 1 Cor.), the fegments ADE, pBF will be each of them lefs than a femicircle, But equal chords are fubtended by equal arcs, the greater equal to the greater, and the lefs to the lefs (I11. 22.) ; whence the chord Ap being equal to the chord DB, the arc AED will be equal to the arc nraz. ) . ' Scmorrum. An arc of a circle cannot, in gencral be trifected, or divided into three equal parts, by any known method, which is purely geometrical, PFRCOP fegmem For points Ther i 2 I contad And, allo Y DAC, But 4t equ; equa t If o be take CAR unl vaz Wi » |m i wv Hp : « tnn 111 J (e \\‘(O e ACD (0 M t" knOWn R 0%, HO00K THE THIRD. 107 PR OP. XXIV. TrHEOREM. The angle formed by a tangent to a cir- cle and a chord drawn from the point of contaé" 1S eqml to the amﬂe in the alter- nate fpvment, i n Aty : Fr ! \\ s / \ o, Sy A\ ! T V;; 1 \ (// i i ' \ \ /) \ / \ X \ // // \\ \ £ / "'// B A < - “Leét'ec be a tangent to the circle AFDE, and AE a chord drawn from the point of conta& ; then will the angle cAE be equal to the angle AFE in the alternate fegment. | | For draw the d,ameter Ap (III. 1.) and join the points Fy D : Then, becaufe Bc is a tangent to the circle, and Ap is'a line drawn through the centre, from the point of conta&, the angle pac will be a right angle (ILI. 12.) And, becaufe AFD is a femi-circle, the angle pra will alfo be a right angle (III. 16.), and equal to the angle DAC. But fince all angles in the fame fegment of a c1rcle are equal to each other (III. 15.), the angle pre will be equal to the angle DAE. If, therefore, from the equal angles pac, pra, there be taken the equal angles DAE, DFE, the remaining angle caE will be equal to the remaining angle AFE. & ED, H 4 | PR O 104 ELEMENTS OF GEOMETRY. PROP, XXV. Proziewm. Upon a given right line to defcribe a feg- ment of a circle, that fhall contain an angle equal to a given rectilineal angle. Lo S \ Let aB be the given right line, and c the given reéti- lineal angle ; it is required to defcribe a fegment of a cir- cle upon the line AB that fhall contain an angle equal to ¢+ Make the angle BaD equal to ¢ (I. 20.), and, from ‘the point A, draw AE at right angles to ap (I. 11.), and make the angle ABF equal to the angle FAB Loac.) " - Then, fince the angles ABF, FAB, are equal to each other, and lefs than two right angles, the fides aA¥, FB will meet, and be equal to each other (L. 25 Cor. and 1. 6.) From the point ¥, therefore, at the diftance Fa, or #p, defcribe the circle AEB, and AEEA will be the feg- ment required, For let ar be produced to cut the circle in E; and ~ join the points E, B. | | Then, becaufe ap.is perpendicular to the diameter AE, it will be a tangent to the circle at the point A (I1L. 10.) And, becaufe ap touches the circle, and AB is a chord drawn from the point of contact, the angle BAD will o i & i % A b . g ¥ will be (IIL. 2 But frudio the ang ScHo fmi-Git ment rec To that & retll Let 4 ‘mgle; It ARG, the Dray point 4 | ngle D Quired, For, § thord . Wil be e LIS 1 feg- angle BOOK. . THE T.HIRD. I0§ will be equal to the angle AEs in the alternate fegment (II1. 24.) ik But the angle BAD is equal to the angle e, by con- ftru@ion ; confequently the angle Arg is alfo equal to the angle c. LR BT Scrorrum. When the given angle is a right angle, a femi-citcle defcribed upon the given line will be the feg- ment required (1L, 16.) PROP. XXVI PROEBLEM. To cut off a fegment from a given circle, that thall contain an angle equal to a given re@ilineal angle, ANy Let anc be a given circle, and » a given re&ilineal angle ; it is required to cut off a fegment from the circle AxC, that fhall contain an angle equal to b. Draw the light line EF, to touch the circle ABc in the point A (IlI. 10.), and make the angle rAB equal to the angle p (L. 20.); then will ABca be the fegment re- quired, ‘ | | For, fince EF is a tangent to the circle, and AB is a chord drawn from the point of contacl, the angle raB will be equal to the angle Acs in the alternate fegment (IIL. 24.) But 106 FLEMENTS OF GEOMETRY. But the anglc FAB is equal to the angle b, by'c*én# firuCtion ; confequently the angle acs, in the fegment ABCA,- is allo equal’to the angle b. O 2. T ¥ "PROP. XXVII. THEOREM. If two right lines in a circle interfet each other, the retangle contained under the feg- ments of the one, will be equal to the ret- angle contained under the fegments of the other, | ‘ \ /'/ e » ‘ //,« (>'} F// \ A{\»é ND Fo ok Let aB, cp be any two right lines, in the circle Acsp, interfeting each other in the point F; then will the re&- angle contained under the parts AF, FB of the one, be equal to the rep’tan*le"contamed under the parts CF, FD Eiiet «é g o4 of the other. For, through il pomt F, ‘draw ‘the diameter 1 (III, 1.); and, from the cenfre E, draW EG at right angles to Az (I. 12.), and join o Then, fince AEF is a trlangTe, S5 e perpendicular £G divides the chord AB into two equal parts (1L 3.), the line FB will be equal to the difference of the fegments AG, GFE. | | And, becaufe E is the centre of the circle, and AE is qLal to EI or EH, the line 1 will be equal to the fum of L | the RS i ha {an , el AV t\-,._ e} 5 vop© It l- AV § VAV BOOK THE THIRD. 107 the fides AE and EF; and rH will be equal to their dif- ference. But the reGangle under the fum and difference of the two fides of any triangle, .is equal to the reGtangle under the bafe and the difference of the fegments of the bafe (Cor. 11. 16.) ; whence the reCtangle of 1F, FH is equai to the reCtangle of AF, Fa. "And, in the fame nwax’ner;, it may be proved, that the re€tangle of 1F, FH, is equal to the reftangle of pF, Fc: confequenﬂy the reCtangle of. AF, F5 s alfo equal to t‘ie “' reCtangle of DF, L QB D SCHOLIUM. VVhen the two lines interfe&t each other in the centre of the circle, the reCtangles of their fegments will manifeftly be equal, becaufe the fegments themfelves are all equal. PR O P. 108 ELEMENTS OF GEOMETRY. PROP XXVill. T BEOREM. If two right lines be drawn from any point without a circle, to the oppofite part of the circumference, the reGtangle of the: whole and external part of the one, will be equal to the reftangle of the whole and external part of the other. 5 i Let aApr¥B be a circle, and Ac, Bc any two right lines, drawn from the point c, to the oppofite part of the cir- cumference ; then will the rectangle of Ac, cp be equal ~ to the reCtangle of Bc, cF. a For, through the centre E, and the point ¢, draw the right line cu ; and, from the point E, draw EG at right angles to AC (I 12.), and join AE. ‘Then, fince AEc is a triangle, and the perpendncular rc divides the chord AD into two equal parts (III. 3.), the line pc W111 be equal to the difference of the fegments AG, GC. ‘ And, becaufe E is the centre of the c1rcle, and AE is equal to EH, or EI, the line nc will be equal to the fum of the fides AE, EC, and 1C wxl be equal to their diffe <Irence, But BOOK THE "THIRD. 109 But the re&angle under the fum and difference of the two fides of any triangle, is equal to the reétangle under the bafe and the difference of the fegments of the bafe (Cor. 11. 16.) ; whence the retangle of HC, cr is equal to the rectangle of ac, cD. And, in the fame manner, it may be proved, that the reSangle of Hc, c1 is alfo equal to the reftangle of cB, CF: confequently the re&tangle of ac, c¢p will be equal to the rectangle of cB, cF. Q. KD, PROP. XXIX. TaRrEoRsn. If two right lines be drawn from any point without a circle, the one to cut it, and the other to touch it; the reGtangle of ‘the whole and external part of the one, will be equal to the {fquare of the other, ‘Let c8, ca be any two right lines drawn from the point c, the one to cut the circle Apsc, and the other to touch it; then will the reangle of ¢B, ¢F be equal to the {quare of ca. ' For, '.n' ELEMENTS OF GEOMETRY. For, ﬁnd E, the centre of the circle (IIL. 1.), and through the points E, ¢ draw the line cEG ; and join EA : . Then, fince Ac 1s a tangent to the circle, and EA isa line drawn from the centre to the point of contaét, the angle cAE is a right angle (IIL 12.) A-ﬂd becaufe EA is equal to EG, or ED, the lihe co will be equal to the fum of EA, EC, and cD will be equal to their difference. Since, therefore, the re&angle under the fum and dif- ference of any two lines is equal to the difference of their fquares (IL. 13.), the rectangle of cG, cp will be equal to the difference of the {quares of CE,. EA. But the difference of the fquares of cE, EA is equal to the fquare of ca (Cor.1I.14.) ; therefore the reCtangle of cG, cp will alfo be equal to the fquare of ca. And it has been fhewn, in the laft propofition, that the reCtangle of cc, c©p is equal to the reCtangle of cB, cF; - confequently the reftangle of cg, cr, will alfo be equal to the {quare of ca. : . cvi K dw DD, PRO P, SRS s S s R e s e S e e s s O s e e S iaEb e B AR SR S B e S A e i S B s Sl R R TR e Y BOOK THE THIRD, 111 P R O~P. XXX ’I HEORE M. 1If two rzght lines be drawn ﬁom a pomt without a circle, the one to cut it, and the . ~other to meet it; and the reCtangle of the whole and external part of the one be equal ‘to the {quare of the other, the latter will be a tangent to the circle. Let aB, ac be two right lines, drawn from any point A, without the circle cBp, the one to cut it, and the other to meet it ; then, if the re@tangle of A, A be equal to the {quare of ac, the line ac will be a tangent to the circle. | For, let ¥ be the centre; and from the point a draw AD to touch the circle at » (IIl. 10.) ; alfo join ¥p, FA, FC. Then, fince Ap is a tangent to the circle, and AEB cuts it, the reftangle of aB, AE is equal to the fquare of ap (III. 2q9.) But the reGangle of A3, Ar is alfo equal to the fquare of ac (&y Hyp.) ; whence the fquare of ac is equal to the fquare of Ap, or aAc equal to ap (II. 3.) And, 112 ELEMENTS OF GEOMETRY. And, becaufe Fc is equal to FD, AC to AD, and AF¥ common to each of the triangles Arc, Arp, the angle acF will aifo be equal to the angle aApr (1. 21.) ' But, fince AD touches the circle, and DF is a line drawn ~ from the centre to the point of conta& the angle ADF isa right angle (ill. 12.) The angle acr, therefore, is alfo a right angle 5 and cF produccd is a diameter of the circle. | And fince a right line, drawn from the end of the dia- meter, at right angles to it, touches the circle (III. ro.), ac will be a tangent to the circle cBD, as was to be thewn. il BOOK BOOK THE FOURTH. 119 B OO K 1V DEFINITIONS. 1. One redilineal figure is faid to be infcribed in ano- ther, when all the angles of the one are in the fides of the other. 2. One re&ilineal figure is faid to be defcribed about “another, when all the fides of the one pafs through the angular points of the other. </\ 2l 3. A rectilineal figure is faid to be infcribed in a circle, when all its angular points are in the circumference of - 4. A retilineal figure is faid to be defcribed about a circle, when each fide of it touches the circumference of the circle. ff’w\' < | i dipe 8y " the circle, J14 ELEMENTS OF GEOMETRY. 5. A circleis faid to be infcribed in a rectilineal figure, when its circumference touches every fide of that figure, = o N\ ¥ £ N Gl T30 6 be delithed bout a relisud figure, when its circumference pafles through all the angular points of that figure. ~#. A right line is faid to be placed, or applied, in a cir- cle, when the extremities of it are in the circumference of the circle, 8. All plane figures contained under more than four fides are called polygons; and if the angles, as well as fides, are all equal, they are called regular polygons. 9. Polygons of five fides, are called pentagons ; thofe of fix fides hexagons ; thofe of feven heptagons; and fo on, | - PROP, BOOK THE FOURTH., 11§ PRO P I. ProBLEM. To place a right line in a given circle, equal to a given right line, not greater than the diameter of the circle. AR Let abc be a given circle, and p a given right line, hot greater than the diameter ; it is required to place 2 line in the circle ABe that fhall be equal to p. Find e, the centre of the circle (IIL. 1.); and draw any diameter AB; then if AB be equal to » the thing required is done, But if not, make AE equal to b (I. 3.); and from the point A, at the diftance ax, defcribe the circle FEc, cut- ting the former in c. Join the points A, ¢; and ac will be equal to D, as was required. | For fince a is the centre of the circle ABc, Ac is equal to AE. | But p is alfo equal to aE, by conftruion; whence Ac is, likewife, equal to p. In the circle aBc, therefore, a right line has been, placed equal to p, which was to be done, 3 % : PR QP 116 ELEMENTS OF GEOMETRY. PROP. II. PROBLEM. To infcribe a friangle in a given circle, that fhall be equiangular to a given triangle. Let anc be the given circle, and DEF the given trian- ole ; it is required to infcribe a triangle in the circle ABC, that fhall be equiangular to the triangle DEF. Draw the right line GH to touch the circle ABc in the point ¢ (III. 10.); and, make the angle HcB equal to the angle o (I. 20.), and the angle GcaA to the angle E; and join aB; then will AcB be the tr1angle required. For, fince the right line Gu is a tangent to the c1rcle, , and cg is a chord drawn from the point of contadl, the anﬂle»HC'B will be equal to the angle cAB in the alternate ~ fegment (1II. 24.) " But the angle mcs is equal to the angle p, by con- ftru&ion ; therefore the_ angle cAB is alfo equal to the angle 0. ” ; And, in the fame manner, it may be proved, that the | angle CBA is equal to the angle E. But, fince the angle cae is equal to the anfrle D, and ‘the angle cBA to the angle E, the remaining angle ACB will alfo be equal to the remaining angle ¥ (Cor. 1. 28.), and confequently the triangle Acs is equiangular to the triangle DEF. Q. E. D. 4 i) Yo BOOK THE FOURTH. I7 PROP. IIl. ProELEM. To circumfcribe a triangle about a given circle, that thall be equiangular to a given triangle. ..:\fi Ay D / \}\ I/v”\:\% % Wt A// / \ \\1 // EoaR K TE R £ Let aznc be the given circle, and DEF the given tri- angle ; it is required to circumfcribe a triangle about the n circle aBc that fhall be equiangular,to the triangle DEF. s 3 ~ Produce the line EF to G and H; and, at the centre 1, : make the angles a1B, BIC equal to the angles bEG, DFH (I. 20.) ; and draw the lines MK, KL, LM, to touch t};/e circle in the points A, B, ¢ (IIl. 10.) ; and join AB. Then, fince the angles 1Ak, KBI, are, each of them, a right angle (IlI. 12.), the angles BAK, K8A, taken together, will be lefs than two right angles. | But when a right line interfeCts two other right lines, and makes the two interior angles, on the fame fide, to- gether les than two right angles, thofe lines will, if pro- duced, meet each other (1. 25. Cor.) / “The line MKk, therefore, meets the line KL ; and, if A, C,y CyB be joined, the fame may be proved of the lines k1, LM and MK ; confequently the ﬁgure KLM is a triangle. ' And, becaufe the four angles of the quadrilateral Arpx are equal to four right angles (Gdr. Li 28.), and the angles - La 1AK, 118 | ELEMENTS OF. GEOMETRY. IAK, KBI are each a right angle, the remaining angles a1B, BkA will be equal to two right angles. But the angles DEG, DEF are alfo equal to two right angles (I. 13.); therefore, fince the angle pEG is equal to the angle A1B (y Con/l. ), the remaining angle Bk A will be equal to the remaining angle DEF. | And, in the fame manner, it may be proved, that the angle cLE is equal to the angle DFE. The angle mxtr, therefore, being equal to the angle DEF, and the angle MLK to the angle DFE, the remain- ing angle kML will alfo be equal to the remaining angle EDF ; and confequently the tnangle KLM 1s equiangular to the tnangle EFD, otk B D, 'p R 0P IV, "PROBLEM. In a given triangle to infcribe a circle, Let asc be the given triangle ; it is required to infcribe a circle in it. Bife& the angles cAB, aBc, with the right lines AD, pe (I g.) Then, fince the angles AR, DBA are lefs than two right angles (I. 28.), the lines ap, b3, Wlll if produced, meet each other (L. 25. Cor.) - And, if from the point of interfe&ion b, there be drawn the perpendxculars DF, DG and DE, they will be the radia of the circle required For, BOOK THE FOURTH. 119 For, fince the angle EAD is equal to the angle DaF {4y Conft.), and the angle AED to the angle DFa, (being each of them right angles), the remaining angle Epa will alfo be equal to the remaining angle ApF (L. 28. Cor.) The triangles ADE, DAF, therefore, being equiangular, and having the fide Ap common to both, the fide DE will alfo be equal to the fide or (I. 21.) e And, in the fame manner, it may be proved, that the. fide pG is equal to the fide DF. The right lines bE, DG and DF are, therefore, all equal to each other ; and the angles at the points F, E and G are right angles, by conftruction. ’ If, therefore, a circle be defcribed from the centre D, with either of the diftances DE, BG or DF, it will touch the fides in the points E, G, ¥ (IIL. 10.) and be infcribed in the triangle ABC, as was to be done. PROP, V. ProBLEM. To circumfcribe a circle about a given triangle. | | Let asc be the given triangle; it is required to cir- cumfcribe a circle about it. | | Bife& the fides ac, ce with the perpendiculars DE, EF (I, 10 4nd 11.); and join DF. ra ‘ ‘Then, 120 ELEMENTS OF GEOMETRY. Then, fince the angles EDF, DFE are lefs than two right angles (4y Conft.), the lines pe, EF will meet each other (l.25. Cor.) ! Let g, therefore, be thelr point of 1nterfe’°uon, and draw the lines EA, EC and EF. ' Then, becaufe AD is equal to pc (4y Confl.), DE com=- mon, and the angle ADE equal to the angle Epc (being each of them right angles), the bafe A will alio be equal to the bafe Ec (L. 4.) And, in the fame manner, it may be proved, that Ec is equal to EB ; »confequently EA, EC and EB are all equal to each other. If, therefore, a circle be defcribed from the point E, at either of the diffances Ea, Ec or £B, it will pafs through the remaining points, and circumicribe the triangle azc, - as was to be done, | ER O P Vi PROBLEM.; To infcribe a {quare in a given circle. Let Ascp be the given circle ; it is requxred to in- fcribe a {quare 1n it, Through E, the centre of the circle, draw any two diameters AC, BD at right angles to each other (Lorr,290)s and join AB, BC, ¢D and DA; then will Bcpa be the fquare required. g For R ) BOOK THE FOURT H. 121 - For fince the two fides BE, EA, are equal to the two- fides ED, EA, and the angle BEA to the angle AED, (be- ing each of them right angles), the bafe BA will be equal to the bafe ap (I. 4.) L e And, in the fame manner, it may be proved, that the“ fides BC, cD are each equal to the fides BA, AD; Whence the figure scpA is equilateral. | It is alfo reftangular: for fince BDA is a femi-circle, the angle BAD is a right angle (IIL. 12.) - And, for the fame reafon, the angles ABC, BCD and ‘ CDA are each of them right angles. ‘ The figure BCDA, therefore, being equilateral, and having all its angles right angles is a {quare, and it is infcribed in the circle ABcD, as was to be done, ’ PROP. VIL PropLEM. ; To c1rcumfcr1be a. fquare about a. ngen circle. ‘ | v ot ™ G C H Let aBcb be the given circle; it is required to circume fcribe a {quare about it, Draw any two diameters Ac, BD at right angles to each other (L. 11, 12.) ; and through the points a, B, ¢, D, draw the tangents KF, FG, GH, HK (lIL ro. ), and join AB. Then, 22 ELEMENTS OF GEOMETRY. ~ ‘Then, fince the angles EAF, EBF are, each of them, right angles (Ill. 12.), the angles FAB, FBA will be, to- gethery lefs than two right angles ; whence the lines K, ¥G will meet each other (l. 25. Cor.) | And, if the points A,D, D;c and c;B be joined, it may be proved, in like manner, that all the other lines Fk, XH, HG and GF will meet each other. ~And, fince the angles at the points a, B, ¢, b are right angles (III. 12.), as alfo the angles at the point £ (y Conf?.), the figure FH, and all the parts into which it is divided, will be parallelograms (I.22, 23.) - But the oppofite fides of parallelograms are equal to - each other (I.30.); whence the fides F6, GH, HK and . KF, being each equal to the diameter ac, or BD, the figure Fu will be equilateral. | - Itis, alfo, retangular : for fince FE is a parallelogram, - and BEA is a right angle (by Confl.), the angle ¥ will, “alfo, be a right angle (1. 28. Cor.) | And, in the fame manner, it may be proved that the angles G5 11 and x are right angjes. ‘The figure FH, therefore, being equilateral, and hav- ing all its angles right én_gle% is a fquare ; and it is cire eumfcribed about the circle ABcD, as was to be done. PROP. e BOOK THE FOURTH. 123% PROPD VIII. To infcribe a circle in a given {quare, e e B H C Let aABcD be the given fquare, it is required to in- fcribe a circle in it. Bife& the fides Ap, ABin the points F and ¢ (L 10.), and draw FH, GK parallel to A and ap (I, 27.); then - will the point of interfection E be the centre of the cxrcle required. For, fince AE is a parallelogram (2y Confd.), the fide AF will be equal to the fide GE, and the fide ag to the fide er (I. 30.) | - - But the fide ¥ is equal to the fide ac, (being each of them equal to half the fide of the fquare ap or ag), whence the fide e will alfo be equal to the fide EF. And, in the fame manner, it may be proved, that HE, EK are each equal to GE and EF. The lines £¥, EG, EH and EK are, therefore, all equal to each other ; and the angles at the points r, 6, 1 and k are right angles, by the nature of parallel lines. If, therefore, a circle be defcribed from the point g, at the diftance £F, £G, EH or EK, it will pafs through the remaining points, and be infcribed in the {quare ac, as ‘was to be done, PR O P, 124 ~ ELEMENTS OF GEOMETRY. PROP. IX. *® To circumfcribe a circle about a given fquare. | Let arcp be the given fquare; it is required to cir- cumfcribe it with a circle. ¥ Draw the diagonals Ac, BD, and the point of inter- feé’mon £ will be the centre of the circle required. For, fince the fides DA, Ac are equal to the fides Ba, Ac, and the bafe BcC to the bafe cp, the angle pac will be equal to the angle caB (I .): that is, theangle BAD wxll be bifected by the line Ac. | And, in the fame manaer, it may be proved, that all ;the other angles of the iquare are bifeted by the lines pe and caA. But the angles cDA, DAB, being right angles, are equal to each other; whence the angles Epa, EAD are allo equal to each other; and confequently the lme ED IS equal to the line £a (L. 4 ) And, in like manner, it may be fhewn, that the lines EB, EC are each equal to the lines Ep, EA; whence the lines EA, EB, EC and ED are all equal to each other, _If, therefore, a circle be defcribed from the point E, at either of the diftances EA, EB, EC or ED, it will pafs through the remaining points, and circum{cribe the {quare AC, as was to be done. ' PR O P. "BOOK THE FOURTH. 12§ PROP. X. ProsLEM. To inf{cribe an ifofceles triangle in a given. circle, that fhall have each of the angles at its bafe double the angle at the vertex, Let aBc be the given circle ; it is required to infcribe an ifofceles triangle in it, that fhall have each of the an- gles at its bafe double the angle at the vertex. Draw any diameter cE, and divide the radius DE in the point ¥ fo, that the retangle of DE, EF may be equal to the fquare of rp (lI. 22.) From the point E apply the right lines EA, EB each cequal to rp (IV. 1.), and join AB, AcC, CB; then will azc be the triangle required. For, through the points D, F, B defcribe the cxrclc ‘8pr (III. 18.), and draw the lines Bp, BF. ‘Then, fince the retangle pE, EF isequal to the {quare of Fp, or its equal £, the line EB will touch the circle BDF, at the point B (IIl. 30.) And, becaufe EB is a tangent to the circle, and BF is a chord drawn from the point of conta&, the angle EBF will be equal to the angle Fpe in the alternate fegment (1I1. 24.) And 136 ELEMENTS OF GEOMETRY. And if, to each of thefe angles, there be added the angle rBp, the whole angle DBE or FEB W1ll be equal to the angles ¥DB, FBD, taken together: But the angle DEE is equial to the angle DEB of FEB (1. 5.); 4nd the dngles ¥bB, ¥BD to the angle erB (1. 28.) 3 whence the angle reB will be equal to the angle E¥5, and the fide £ to the fide 57 (I 5.) And fince EB is equal to ¥p, by conftru&ion, sr will alfo be equal to b, and the angle ¥Fpr to the angle ‘reD (l. 5.) Thefe two angles, therefore, taken together, are dous ble the angle ¥pB; whence the angle EFe, or its equal FEB, is alfo double the angle FpB., | ' - But the angle FEB, or CEB, is equal to the angle caB (I11. 14.), and the angle ¥DB, or EDB, to the angle AcB (II1. x4. and L. Ax. 6.); confequently the angle caB is alfo double the angle acs. And, fince EAc, EC are right angled triangles (IIL 16.), having EA equal to £B (4y Con/fl.) and Ec com= mon, the remaining fide aAc will be equal to the remains ing fide cp (IIL 8. Gor.) The triangle ABc, therefore, is ifofceles; and has each of the angles at its bafe double the angle at the vertex ; as was to be fhewn. , 3 PROP. BOOK THE FOVRTH. 127 PROP. XI, PropieEM. In a given circle to infcribe a regular pen= tagon. AV Let cpARBE be the given circle; it is required to in- {cribe a regular pentagon in it. . Make the ifofceles triangle aBc fuch, that each of the - angles cAB, cBA may be double the angle acs (IV. 10.) Bife& the angles caB, cBaA with the lines Ak, zp (1. 9.), and join the points AD, DC, CE, EB; then will ABECD be the pentagon required. For, fince the angles cAB, cBA are each double the angle acB (4y Confl.), and the lines AE, BD bifect them, the angles AcB, CAE, EAB, ABD and DBc are all equal “to each other. S And fince equal angles ftand upon equal circumferences (111. 21.), the arcs ¢D, DA, AB, BE and Ec are alfo equal to each other. ' But equal arcs are fubtended by equal chords (111, 22.) ; confequently the fides cb, DA, AB, BE and Ec are, like- wife, equal. The figure ABECD is, therefore, equilateral ; and it is alfo equiangular. For, fince the arc ¢p is equal to the arc BE, to each of them add pAB, and the arc cpaB will be equal to the arc DABE. | | B But 128 ELEMENTS OF GEOMETRY. ~ Butequal angles are fubtended by equal arcs (III. 21.), whence the angle cEB is equal to the angle pce. And, in the fame manner, it may be thewn, that each of the angles cpA, DAB, ABE are equal to the angle CEB or DCE. / - The pentagon ABECD, therefore, is both eqmlateral and equiangular ; and it is infcribed in the given circle, as was to be done. ' { PROP. XIL About a given circle to defcribe a regular _pentagons. | | E A G Let aBcDE be the given circle; it is required to cir- cumfcribe it with a regular pentagon. Infcribe the regular pentagon peasc (IV.1r1.), and through the points A, B, C, D, E, draw the tangents FG, GH, HX, KL and LF ; alfo join the points 03RS0, B30 018, 0,D and 0, E. | Then, fince the angles OEF, OAF are right angles (1Il. 12.), the angles 0EA, OAE, taken together, are lefs than two right angles ;- whence the lines LF, rG will meet ~each other (I. 25. Gor.) And, in the fame maﬁner, it may be proved, that all the other lines FG, GH, HK, KL and LF Wlll meet each other. And, an i V) BOOK THE FOURTH. 12g And, fince ok, oA and o are all equal to each other, and EA is equal to AB, the angles oEA, OAE, 0AEB and oBA will be all equal to each other (L. 7.) . But the angles at the points E, A, B are alfo equal, being each of them right angles (IIL. 12.) ; confequently the angles AEF, EAF, BAG and aBG are likewife equal ; and the angle F equal to the angle ¢ (I. 21.) And, in the fame manner, it may be thewn, that the angles G, H, K, L and F are all equal to each other. Since, therefore, the triangles EFa, acs, &c. are equiangular, and have their bafes EaA, aB, &c. equal to each other, the remaining fides EF, ra, Aq,, GB, &c, will alfo be equal (L. 21.): And fince LF, FG, &c. are the doubles of £F, Fa, &c. the figure FGHKL isa regular pentagon 5 and it is clrcurm- {cribed about the circle ABcDE, as was to be done, PRO P XHIL PROB\LEM, In a given regular pentagon to infcribe 2 fcxrclc. s S C H- D Let ABcDE be the given regular penfaoon, i re- . quired to infcribe a circle in it. ~ Bife& any two angles BcD, cDE with the right lines co, op (I. g.), and the point of interfeGion o will be the centre of the circle required, 130 ELEMENTS OF GEOMETRY. _For draw the lines oB, oA and oE, and let fall the perpendiculars oH, OK, oL, oF and.oc (L.12.): Then, becaufe cB is equal to c¢p (4y Hyp.), co com- mon, and the angle Bco equal to the angle ocp (4y Confi.), - the angle cso will alfo be equal to the angle opc (I.7.) But the angle opc is equal to half the angle cpE (4y Confl.) and the angle cpE is equal to the angle cBA (4y Hyp.) ; confequently the angle cro ‘is alfo equal to half the angle cBa. ‘ The angle cra, therefore, is bifected by the line 8o ; and, in the fame manner, it may be thewn, that the an- gles at the,fpoints A, E are bifeéted, by the lines a0, oE. Again, becaufe the triangles 0Gc, O€H are equiangu- lar, and have oc common to each, the perpendicular oG ' will be equal to the perpendicular on (l. 21.) And, in the fame manner, it may be fhewn, that on, OK, OL, OF and oG are all equal to each other. If, therefore, a circle be defcribed from the centre o, at either of thefe diftances, it will pafs through the re- maining points, and be infcribed in the pentagon ABcDE, ~as was to be done, PROP, e et St R A R . b e 5 b R S A R R s e R R R st 3 4l Or BCORYTHE ROURTH. 131 PROP XV, Prosl . To defcribe a circle about a given regular D ) ~ Let aBcDE be a given regular pentagon ; it is required to circumfcribe it with a circle. . Bife& any two angles Bcp, cDE, with the right lines co, op (l. g9.), and the point of mterfeéhon o will be the centre of the circle required. For, draw the lines o8, 0A and ok : | Then, becaufe cB is equal to ¢p (4y Hyp.), co com- mon, and the angle Bco equal to the angle ocp (4y Gonft. ), the angle cso will alfo be equal to the angle opc (I 4.) But the angle opc is equal to half the angle cpE, ([,] pentagon, - Conft.), and the angle cDE is equal to the angle cpa (by Hyp.) 5 whence the angle cBo is alfo equal to half the angle cBA. ' The angle cBa, therefore, is bifected by the line po ; and, in the fame manner, it may be thewn, that the an- gles at the points A, E are bife&ed, by the lines a0, OE. Since, therefore, the angle ocp is equal to the angle opc (by Hyp. and Ax. 7.), the ﬁde oc will alfo be equal to the fide op (I. 5.) And, in the fame manner, it may be fhewn, that op, OF, 0A, oB and oc are all equal to each other, Ka | If 132 ELEMENTS OF GEOMETRY. If, thercfore, a circle be defcribed from the point o0, at either of thefe diftances, it will pafs through the remain- ing points, and circumicribe the pentagon ABCDE, as was “to be done, PROP. XV. PROBLE M. In a given circle to infcribe a regular hexagon. Let Acer be a given circle; it is required to in- fcribe a regular hexagon in it. 2% hrough the centre o draw the dmmeter AD, and make pe equal to po (IV. 1.), and it will be the fide of the hexagon required. For, draw the diameter cF, and make BE parallel to ¢p (I.27.) ; and join the points DE, EF, FA, AB and BC: ' Then, fince poc is an equilateral triangle, the angles oDC, OCD and poc will be all equal to each other (L. 5. Cor.) - | And, becaufe o is parallel to cp, the angle Eop will be equal to the angle onc (L. 24.), and the angle FOE to the angle ocp (L. 25.) ; But the angles obc, ocb are each equal to the angle poc ; therefore, the angles, poc, EoD and FOE are all equal to each other ; as are alfo the. oppofite angles FOA, A0B and BOC, ~ Since, ¢l BOOX "T HE FOUR'TH.- 133 Since, therefore, the triangles cop, Dok, &c. have two fides, and the included angle of the one equal to two {ides and the included angle of the other, they will be equal in all refpeéts (L &) The fides ¢cp, DE, EF, &c. are therefore all equal to each other, as are alfo the angles Bcp, cpE, &c. whence ABCDEF is a regular hexagon; and it is infcribed in the circle Acer, as was to be done. ScHOLIUM. Befides the figures here conftruced, and thofe arifing from thence by continual bifeéions, or taking the differences, no other regular polygon can be defcribed, by any known method, purely geometrical. It may alfo be obferved that fome of thefe figures, as well as feveral others, in the former part of the work, may often be defcribed in a much eafier way, for pra&ical purpofes ; but the principles upon which they depend can only be obtained from the following bocks ef the Ele~ ments. Nt BOOK 134 ELEMENTS OF GEOMETRY., B OOk N DEEFINIT PO NS 1. Alefs magnitude’is faid to be a part of a greater, when the lefs is contained a certain number of times in the greater. 2. A greater magnitude is faid to be a multiple of a lefs, when the greater is. equal to a certain number of times the lefs. e _ 3. Ratio is a certain mutual relation of two magnituﬂes of the fame kind, which arifes from confidering the quan- tity of each. A 4. When four magnitudes are compared together, the firft and third are called the antecedents, and the fecond and fourth the confequents. - 5. Four magnitudes are faid to be proportxonal when any equimultiples whatever of the antecedents, are, each of them, either equal to, greater, or lefs, than any equi- multiples whatever of their confequents. 6. Inverle ratio is, when the confequents are made the antecedents, and the antecedents the confequents. n. Alternate ratio is, when antecedent is compared with antecedent, and confequent with confequent. 8. Compounded ratio is, when each antecedent and its eonfequent, taken as one quantity, is compared, either with the confequents, or the antecedents. 9. Dix}ided ratio 1s, when the difference of each ante- cedent and its confequent, is compared, elther Wlth the confequents, or the antecedents. PR‘O P, p = BOOK -THE FIFTH. 133 ¥ RO 110 oREN If any number of magnitudes be equimu]- tiples of as many others, each of each ; what- ever multiple any one of them 1s of its part, the fame multiple will all the former be of ~all the latter. - B ¥ *?-""""' - | . b A G B ERRARA | D ‘Let any number of magnitudes AB, cp be equimulti- ples of as many others E, F, each of each; then what- ever multiple AB is of £, the fame multxple will aB and D together, be of E and F together. For fince aB is the fame multxple of E that cD is of - ¥ (by Hyp.), as many magnitudes'as there are in AB equal to E, fo many wili there be in ¢D equal to F. Divide aB into magnitudes equal to E (I. 35.), which let be 4G, GB; and CD 1nto marrmtudes equal to F, which Tet be cH, HD. Then the number of magmtudﬁs CH, HD, in the ong, will be equal to the number of magmtudes AG, GB, in the other. And becaufe AG is equal to E, and cH to F (by Confl.), AG and cH, taken together, will be equal to E and ¥ - taken together, For the fame reafon, becaufe GB is equal to E, and HD to F, 6B and HD taken together, will be equal to £ and ¥ taken together, K 4 : As 136 ELEMENTS OF GEOMETRY. As many magnitudes, therefore, as there are in a® equal to E, fo many are there in AB and cbp together, equal to E and F together. | And, confequently, whatever multiple AB is of E, the fame multiple will AB and cD together be of E and F to- gether, ‘ VIBID, PROP. II. TucorEeM, If any number of magnitudes be multiples of the fame magnitude, and as many others be the fame multiples of another magnitude, each of each, the fum-of all the former will be the fame multiple of the one, as ‘the fum of all the latter is of the other, A B : " OF ’ ¥ s SO D 4 27! Tl Let any number of magnitudes AB, BE, be multiples of the fame magnitude c, and as many others Dg, cH, the fame multiples of another ¥, each of each ; then will the whole AE, be the fame multiple of c, as the whole b, is of F, For fince aB is the fame multlple of c that pGg is of F (by Hyp.), there will be as many magnitudes in AB equal to ¢, as there are in bG equal to F. And becaufe =k is the fame multiple of ¢ that GH is of F (by Hyp.), there will be as many magnitudes in BE cqual to ¢, as there arcin GH equal to F. - | : As AB \ EDOR LT HE FIRES, 13% As many magnitudes, therefore, as there are in the whole AE equal to ¢, fo many will there be in the whole DH equal to F. s The whole AE, therefore, is the fame multiple of ¢, as the whole pu is of F, % Qik, B P,R'O Sl T HEOREM, If the firft of four magnitudes be the fame multiple of the fecond as the third is of the fourth ; and if of the firft and third there be taken equimultiples, thefe will alfo be equi- multiples, the one of the fecond, and the other of the fourth. E - IS —F .G Lo H . C ey B e P sl Let a the firft, be the fame multiple. of B the fecond, as c the third, is of D the fourth; and let EF and cu be equimultiples of A and c; then will £F be the fame mul- tiple of B, that cH is of D. | For fince EF is the fame multiple of A that gu is of ¢ (by Hyp.), there will be as many magnitudes in £¥ equal to A, as there are in GH equal to c. Divide EF into the magnitudes Ex, XF each equal to « a (L. 35.); and GH into the magnitudes 6L, LH, each equal to c. ‘Then 7138 ELEMENTS OF GEOMETRY. Then will the number of magnitudes Ex, KF in the one, be equal to the number of magmtudes Gin LH in the - other. ' And becaufe A is the fame multiple of B that cis of D (by Hyp.), and EK is equal to a, and 6L to ¢ (&y Conft.), £k will be the fame multiple of ® that 6L is of b. In like manner, fince KF is equal to A, and LH to c, kF will be the fame multiple of B, that LH isof D, And fince £k, KF are each multiples of B, and 6L, LH are each the fame multiples of p, the whole eF will be the fame multnple of B, as the whole gn is of D (V. 2.) i QuE S P RO P IV THEOREM. If the ﬁrﬁ: of three magmtudes be greater than the fecond, and the third be any mag- nitude whatever, fome equimultiples of the firft and fecond may be taken, and fome multiple of the third fuch, that the former fhall be greater than that of the third, but the latter not greater, LI S e N T ».]2.._.{ Let as, BC be two unequal magnitudés,: and p any other magnitude whatever ; then there may be taken fome equimultiples of AB, BC, 2nd fome multiple of b fuch, that <® BOOK THE FIFTH. 139 that the multxple of aB fhall be greater than that of D, but the multiple of BC not greater, | | For of BC, €A take any equimultiples GF, FE fuch, that they may be each greater than p; and of D take the multiples k and L fuch, that L may be that which is furfk ~ greater than GF; and K that which is next lefs than 1. Then, becaufe L is that multiple of D which is the firlt that becomes greater than GF, the next preceding multi- ple k will not be greater than GF 3 that is 6F will not be lefs than k. ‘ And, fince FE is the fame multiple of ac that GF is of BC (by Confl.), GF will alfo be the fame multvlple of BC that Ec is of AB (V. 1.) The magnitudes EG and GF are, therefore, eqmmul- tiples of the magmtudes AB and Bg, and 1 is 2 multiple of - D, And, fince GF is not lefs than x, and 'Ef is greater than D (&y Confl. ), the whole EG will be greater than and D taken together. : But x and b, taken together, are equal to L (by Confl.) 3 ‘therefore EG will be greater than £, and FG not greater than L, as was to be thewn, I40 ELEMENTS OF GEOMETRY, PROP V. THEOR;E_M.. . If four magnitudes be proportional, any equimultiples whatever of the antecedents will be proportional to any eqmmultlples whatever of the confequents. x K r ¥ 1 R A o ‘ B e v, b ‘ . e —iS He N ' i Tet A be to B ascis' to b, and of A and ¢ take any equimultiples Ex, FL; and of B and D any eqmmultlples GM, 1N ; then will Ex be to 6M, as FL isto HN. - For of £x and ry take any equimultiples whatever ep, FR ; and of GM and HN any. equnnulnples whatqvér 054 HT ! iy T | Then, ﬁnce EK -1s the fame multiple of A, that ¥t is of ¢ (by Confl.), and of EX, FL have been taken the equi- multiples Ep, FR, EP will be the fame multiple of A, that FrRisof . (V.3i) And, in the fame manner, it may be thewn, that Gs is the {ame multiple of B, that uT is of b. But A has the fame ratio to B that ¢ has to p (4y Hyp.); and of A and ¢ have been taken the equimultiples Ep, FR ; and of B and p the equimultiples Gs, HT. If, therefore, Ep be greater than s, Fr will alfo be greater than ur ; and if equal, equal; and if lefs, lefs {ViDef. %) And, Y "BQOK T HR F B N, 14% And, fince Ep, FR are any equimultiples whatever of EK, FL; and Gs, HT are any equimultiples whatever of GM, HN ; EK will have the fame ratio to cm, that FL has to HN (V. D¢f. 5.) i Q. EaD. PROP. VIL THEOREM. If four mégnitudes be proportional, and the firft be greater than the fecond; the third will alfo be greater than the fourth; and' if equal, equal; and if lefs, lefs. | P . g G 4 Ay R Bt oo e ¥ y H; 4 Let A have to B the fame ratio that ¢ hasto p ; thenif A be greater than B, ¢ will alfo be greater than p; and if equal, equal ; and if lefs, lefs. For, of A and c take any equimultiples & and ¢, and of B and D the fame equimultiples ¥ and n.- Then, becaufe A is to B, as c is to p (&y Hyp.), if = be greater than ¥, ¢ will alfo be greater than 1 ; and if equal, equal; and if lefs, lefs (V gl And, fince E, ¥, G, H are the {ame mmtip’es of A, B, ¢, D, each of each, thefe laft magnitudes will zlfo obferve the fame agreement of equality, excefs, or defet with their equimultiples. ‘ If, therefore, A be greater than B, c will alfo be greater than D ; and if equal, equal ; and if lefs, J8(s, oo ,, ‘ Qo Bl . PR.O P, 142° ELEMENTS OF GEOMETRY.« PRO P - Vil.- LHEOREM. If four magnitudes be proportional, they will be proportional alfo when taken in- verfely. At C et 3 ' ST s o Dt If A has to B the fame ratio that ¢ has to p; then, in- verfely, B will have to A the fame ratio that b has to c. For, of B and D take any equimultiples whatever E and ¥ and of A and c any equimultiples whatever G and H: Then, fince A is to B as ¢ is to D (&y Hyp. ),and G, i are equimultiples of 4, ¢, and E, F of B, D (&y Conft.), if G be greater than £, H will be greater than F; and if equal, equal; and if lefs, lefs (V.. D% 5.) And, becaufe ¢ has with E the fame agreement of equality, excefs, or defet, that i has with F, E will have with G the fame agreement of equality, excefs, or defe&, that ¥ has with H. If, therefore, E be greater than G, F will alfo be greater than H ; and if equal, equal; and if lefs, lefs. ‘But E and F are any equimultiples whatever of sand D (29 Confl.) ; and G and H are any equimultiples whatever of Aand ¢ ; therefore, Bisto Aaspistoc (V. Defe s5.) Q. E. D. PR OD. Ia fo BOOK THE FIBDH, . 143 PROP VI TurorRem - If the firft of four magnitudes be the fame multiple or part of the fecond as the third is of the fourth; the firft will have the fame ratio to the fecond as the third has to the fourth, | | | E» —t & - A ey | B D vt F X; Let A the firft, be the fame multiple of B the fecond, that ¢ the third is of b the fourth; then will A have to B . the fame ratio that c¢ has to b. For of A and c take any equimultiples whatever E and G ; and of B and D any equimultiples whatever F and 1 Then, becaufe A is the fame multiple of B that ¢ is of D (by Hyp.), and E is the fame multiple of A that G is of ¢ (&y Gonfl.), E will alfo be the fame multiple of B that cisof o (V. 3.) | And, fince E is the fame multiple of B that G is of b, and F is the fame multiple of B that H 1s of D (4y Confl.), if £ be greater than F, G will be greater than H ; and if equal, equal ; and if lefs, lefs. But E and G are any equimultiples whatever of a and ¢; and F and H are any equimultiples whatever of B and D ; therefore, a will have to » the fame ratio that ¢ has top (V. Defis.) Again, 144 ELEMENTS OF GEOMETRY. Again, let the firft B, be the fame part of the fecond A, - as the third b, is of the fourth ¢ ; then will 5 have to A the fame ratio that o has to c. ‘ For a is the fame multiple of 8 that ¢ is of D (4y Hyp.) 3 therefore Ao will have to B the fame ratio that ¢ has ton (V.8.) And, fince A is to B as ¢ is to D, therefore, alfos inverfely, Bis to A as pistoc (V.7.) ' P R T atonewm. Equal magnitudes have the fame ratio to the fame magnitude, and the fame has the fame ratio to equal magnitudes. Let a and B be equal magnitudes, and ¢ any other magnitude whatever ; then A will have to ¢ the fame ratio that B has to c. ; For of A ard B take any equimultiples whatever b and E; and of ¢ any multiple whatever F : Then, becaufe » is the fame multiple of A that & is of B, and A is equal to B, D will alfo be equal to E. And, fince » and E are equal to each other, if p be greater than ¥, £ will alfo be greater than r; and if equal, equal ; and if lefs, lefs. ; 5 | But thy toAascistoB (V. Def.5.) BOOK THE FIFTH: 144 Biit and E are any equimultiples whatever of A and B, and F is any multlple whatever of c; therefore Aisto c as B is to ¢ (V. Def. 5.) Again, let A and B be equal magmtudes, and ¢ any other magnitude whatever then ¢ has to a the fame ratio that it has to =. \ | For, havmg made the fame con{’crué’uon as before, D ay, in like manner, be fhewn to be equal to E : And, fince D is equal to E, if F be greater than D, it will alfo be greater than E; or if equal, equal ; or if lefs, lefs. But F is any multiple whatever of ¢, and p and £ are any equimultiples whatever of A and B; therefore, c is BROP. X Tisorse, Magnitudes which have the fame ratio to the fame magnitude are equal to each other ; and thofe to which the fame magnitude has the fame ratio are equal to each other, S| e | /" 3 e | D i ’.__._.......'_t;..‘.__..._.q ¥ Let & have fo ¢ the fame ratio that & has to ¢; then will A be equal to 8. Fory if they be not equal, one of theta muft be greater than the other, L 5 Let 146 ~ ELEMENTS OF GEOMETRY. Let A be the greatet ; and of A and B take the equi- “multiples » and E; and of ¢ the multiple ¥ fuch, that p ‘may be greater than F, and E not greater than r (V. 4.) Then, fince A is to ¢ as B is to ¢, and D and E are equimultiples of A and B, and ¥ 15 a multiple of ¢, D be- ing greater than ¥, E will alfo be greater than ¥ (V. Def.5.) e But, by conftruction, E isnot greater than F § whence it is greater and not gxeater at the fame tune, which 1s abfurd. ‘ The magmtude A is, therefore, not greater than 3 ; s and in the fame manner it may be fhewn that it is not lefs ; confequently they are equal to each other. Again, let ¢ have to A the fanie ratio that it has to B ; 3 then will A be equal to B. For, fince c is to A as c is to B, therefore, alfo, in- verfely, a‘will'be toc as Bistoc (V.7.) But magnitudes which have the fame ratio to the fame magnitude have been fhewn to be equal to ‘each other ; \ therefore Ais equal'to B, | o Qs . | PR OP. 't 0B; OF ABOOK THE FIFTH, 147 P RO P 5 THEORAE\M,.. Ratios whxch are the fame to the fame ratto, are the fame to each other. A —— Cr— ¢ Er— B— Dy Fr— : 3% M ~ Nr 4 Let A be to B as cis to D, and CtoD as E is to Fj then will A be to B as E is to F. For, of A, c and E take any eqmmultlples whatever G, Hand K; and of B, D and F any equimuitiples what= ever Ly Mmand N: Then, fince A isto B as ¢ isto D (by‘Hyp.), and ¢ ~and H are equimultiples of A and ¢, and L and M of & and D, if G be greater than 1, u will be greater than M3 and if equal, equal ; and if lefs, lefs (V. Dief::5.) And, becaufe cis to p as E is to F (&y Hyp.), and.u and K are equimultiples of ¢ and g, and M and N of p and F; if H be greater than M, k will be greater than ~ ; and if equal, equal; and if lefs, lefs (V. Def. 5.) But if G be greater than L, it has been fthewn that u will alfo be greater than m; and if equal, equal; and if lefs, lefs ; whence, if G be greater than 1, x will alfo be greater than N; and if equal, equal ; and iflefs, lefs. And fince G and K are any equimultiples whatever of A and E, and L and N are any equimultiples whatever of 8 and F, o will have to 2 the fame ratio that E has.- to ¥ [V. Def. 5.) W $Q, BB L » 2R O% e f4.8 ELEMENTS OF GEOMETRY, PROP. XII. THEOREM, If any number of magnitudes be propor= tional, either of the antecedents will be to its confequent, as the fum of all the antece-~ dents is to the fum of all the confequents. Gt M p————y K ——— Aty Caeq y R ; B et Dr—t X Lip— — M —t N} < Let A be to B as c is to D, and as E is to F; then will A be to B, as A, ¢ and E together, are to 8, D and F to- gether, | V“ For, of 4, cand E tak\, any equimultiples whatever G, Hand K ; and of B4 D and ¥ any equxmultxples what- ever Ly M and N ¢ | Then, fince A is to B, as C is to D (by Hyp.), and G, i1 are equimultiples of 4, ¢, and 1, M of B, Dy if G be greater than 1y ‘H will be greater than M, and if equal, equal 3 and if lefs, lefs (V. Def. 5.) And becaufe A is alfo to B as Eis to F (4y Hyp ), and G, K are equimultiples of A, E; and L, N of B, F, if G be greater than 1, K will be greater than w; and if equal, equal ; and if lefs, lefs (V. Def, g3 From hence it follows, that if G be greater than L, G, 1 and K together, will be greater than L; M and N to- gether 3 and if equal, equal ; and if lefs, lefs. 5 " Bt n [ But ¢, and G, H, K together, are any equimultiples whatever of A, and A, ¢, E together {#y Gonfl.); and L, and L, M, N together, are any equimultiples whatever of B,and B, D, F together ; whence, as A is to B, {ois 4, ¢ - and E together, to B, D and F together (V. D¢ 5.) Q. E. DD, PROP. XII. THEOREM. Equimultiples of any two magnitudes have the fame ratio as the magnitudss them- {elves, Let cp be the fame multiple of A that EF is of B; then will cp have the fame ratio to EF that A has to E. For, fince cp is the fame multiple of A that £F is of ra there are as many magnitudes in cp equal to A, as there are in EF equal to B. Let cp be divided into the matrmtudes CG, GH, HD, cach equal to A (I. 25.); and EF into the magmtudes EK, KL, LF, each equal to B, | ’ Then, the number of magnitudes c¢G, GH, HD in the one, will be equal to the number of magnitudes EK, KLj LF in the other. ' And, becaufe pH, HG, Gc are all equal to each other, as are alfo FL, LK, KE, DH will be to FL as HG to LK, and as Gc to KE (V. g.) L 3 Andy 2 150 ELEMENTS OF GEOMETRY. And, fince ény antecedent is to its confequent, as all the antecedents are to all the confequents (V. 12.), Fr will be to pH, as FE is to pC. But pH is equal to A (by Confl.), and FL is equal tQ B3 therefore B will bé to A, as FE to bcy and, inverfely, DC to FE as A to B. - i Qs B D Y RO VXV, 'THEQ‘REM, If four magnitudes of the fame kind be pmportzonal and the ﬁxf’c be greatur than the third, the fecond will alfo be greater than the fourth; and if equal, equal; and if 1efs, lefs | ‘ X sy G) A e : Ci A g D bty v righiie o Let a'be to B as € is to D3 then if a be greater than é, g will alfo be greater than D ; and if equal, equal ; and if lefs, lefs. / - | | Firft, let A be greater than c; then s will be greater than p. | | i : For, of a, c take the equimultiples F, ¢ ax1d_'bf5 the mult‘ip“iei F {uch, that E may be greater than F, but G ot greater (V. 4.); and make ® the fame multiple of D thag Fis of B : Then, becaufe A is to B as cisto D (by Hyp. ) and E, G are equimultiples of A, ¢, and ¥, # of B, D (by Confl.), E being greater tbap F, ¢ will alfq be gfeater than 5 (V. D 5.) b And, Sy W BOOK.. THE FIF T H, 181 And, fince, by conftrution, F is not lefs than G, and G has been proved to be greater than u, F will Lhikewife - be greater than H. But ¥ and H are equimultiples of 8 and p- (&y Ca;yi 1s therefore, fince F 1s greater than H, B will alfo be greater than p. ‘ Secondly let A be equal fQ C; thenv m’ill B be equal TR | £ For, A is to B as C is to D (Zzy Hyp.) 5 or, fince A is equal to ¢, A isto Bas Ais to D; therefore B is equal top (V. 10.) | Thirdly, let A be lefs than c; then will B be lefs than p. | For, c is to b as A is to B, by the pmpoﬁtion; there._- fore ¢ being greater than a, D will alfo be greater than B, by the firft cafe, G R R PROP. XV, THEOREM. if four magﬁitudes ‘of the fame kind be proportional, they will be proportional alfo when taken alternately, ok i Of — E : ] B r— I psmiied G ) H 2 4 Let A be to B as c is to D ; then, allo, alternately, a will be to c as B is to D. ' ; For, of A and B take any equimultiples whatever & and G ; and of ¢ and D any equimultiples whatever r and & : S 4 ' "Then, < 152 ELEMENTS OF GEOMETRY. Then, fince E is the fame multiple of A that G is of B, and that magnitudes have the fame ratio as their equi- multiples (V. 13.), AistopasEis to G. But A istoBasc is to b, by the propofition ; whence cistopaseistoc (V.rr1.) In like manner, becaufe F is the fame multiple of ¢ that " Hisofp, cwillbetopas Fiston (V. 13.) But ¢ has been fhewn to be to D as E is to G ; confe= quently, Ewillbeto G as Fisto n (V,11.) Since, therefore, E has the fame ratio to G that ¥ has to H, if E be gréater than F, ¢ will alfo be greater than H; and if equal, equal ; and if lefs, lefs (V. Dt g} - ButE and G are any equimultiples whatever of A and B; and F and H are any eqmmultxples whatever of ¢ and therefore AistocasBistoD (V Def. 5.) Q. E. D, | PROP. XVI Turonen, If four magnitudes be proportional, ths fum of the firft and fecond, will be to the frft or fecond, as the fum of the third and fourth, is to the third or fourth, Let AE be to EBas ¢F is to FD ; then will AB be to BE, ®r AE, 2s CD is to DF, OF CF. , For, RS S e e P v | " AB, or EB, as CF is to CD, Or FD, BOOK 'THE FRELTH. © . 153 For, fince AE is to EB as cF isto FD (&y Hyp.) ; there- fore, alternately, A will beto cras EBto Fp (V. 15.) And, fince the antecedent is to its confequent as all the antecedents are to all the confequents (V. 12. ), AE will be to cF as AB is to CD. , But ratios which are the fame to the fame ratio are the fame to each other (V. 11.); whence ap will be to ¢cp as EB is to FD ; and, alternately, AB to EB as ¢D to DF (V. 15.) Again, fince AE has been thewn to be to cF as AB is to cp, therefore, by alternation, AE will be to AB as cF Jstocp (V.15.) But quantities which are dire&ly proportxonal are alfo proportional when taken inverfely ; whence AB wxll be to Q. E. D. . AEas cpisto cF (V. 7) PROP. XVIIl. THEOREM. If four magnitudes be proportional, the difference of the firft and fecond, will be to the firft or fecond, as the difference of the third and fourth, is to the third or fourth, | o e P A—F—B c—-D . i " e o Let AB be to BE as cD is to DF ; then will AE be to Far, 154 ELEMENTS OF GEOMETRY. For, of AE, EB, CF, FD take any equimultiples what- ever GH, HK, LM, MN; and of EB, FD any other equi- multiples whatever Kp, NR: Then, becaufe GH is the. fame multiple of AE that HK 1s of EB (by Con/l.), GH will be the fame multiple of AE that ¢k is of aAB (V. 1.) But cH is the fame multiple of AE that LM is of cF (b Conft.} ; therefore Gk is the fame multiple of AB that Lm s of-Cx. | Inlike manner, becaufe LM is the fame multiple of CF that mn is of Fp (byConfl.), LM will be the fame multi- ple of cF that LN is of cp (V. 1.) But tm has been fhewn to be the fame multiple of CF that 6K is of AB; therefore GK is the fame multlple of AB that LN is of cD. | Again, becaufe HK, KP are the fzime multiples of B, that MN, NR are of FD (by Confl.), HP will be the fame multiple of EB, that MR is of FD (V. 2.) And, fince AB is to BE as cb to DF (4y Hyp.), and GK, EN are eqUimuk‘iples of AB, cD, and HP, MR of BE, DF, if ok be greater than up, LN will be greater than ; and if equal equal ; and if lefs, lefs (V. Def. 5.) brom the two former of thefe take away the common part Hi, and from the two latter, the commeon part MN ; - then if ¢H be greater than xp, LM will be greater than NRr, and if equal, equal; and if lefs, lefs. But 6, LM are any equimultiples whatever of AE, CF (8y Conf?.), and KP, NR are any 'equimultiples whatever ~of £B, FD; whence AEis to CFas EB to FD (V. Def. 5. s and, alternately, AE to EB as CF to ED. | BOOK THE FIFTH. 155 And fince aB is the fum of AE, EB, and cp of ¢cF, FD, - Ae will be to AE or B, as cD is to cF or FD (V. 16.); and, inverfely, AE or EBto AB, as CF or FD to CD. Qi Scmorrum. When thg confequents are greater than the antecedents, the fame demonftration will hold, if the terms be taken inverfely, PROP XVIH. TueoRrEM If four magnitudes of the fame kind be proportional, the greateft and leaft of them, taken together, will be greater than the other two, " Let aB, cp, E, F be the four proportional magnitudes, of which aB is the greateft and r the leaft ; then will AB and F together, be greater than cp and & together. For in AB take AG equal to E, and in cD take cu equal tor (L. 3.) ‘Then, becaufe aB is to cp as E to F (Jy £yp.), and AG isequal to Eapd cH to F (by Con/i )y AB will be to CD as AG to CH. -But magnitudes which are proportional, are alfo pro- portional when taken alternately (V. 15.); therefore as will be to AG as cp to cH. And, fince &G is the difference of AB and AG, and by of cp and cH, BG will be to AB as DH to ¢p (V. 17 and, inveriely, AB to BG as ¢D to DH, Bu? 156 ELEMENTS OF GEOMETRY. But Ap is greater than c¢p (by Hyp.); whence G& is alfo greater than. HD (V. 14). And, becaufe ac is equal to £, and cH to F, AG and “F together, are equal to'cH and E together, To the firt of thefe equals add cB, and to the fecond HD, then will AG, GB and F together, be greater than cH, uD and E together. _ But AG, GB are equal to AB, and cH, HDto CD; cone fequently AB and F together are greater than c¢p and E together. Qi E: D, Scuorrum. That F muft be the leaft of the four mag- nitudes when A is the greateft, appears from propofitions VI, and XIII BOOK BOORK THE 'SIXTH.: 157 B O O X VviI DEFINITIONS. 1. Similar retilineal figures, are thofe which are equi- angular, and have the fides about the equal angles pro- - portional, 2. The homologous, or like fides, of ﬁmx]ar figures, are thofe which are oppofite to equal angles. 3. Two figures are faid to have their fides reciprocally proportional, when the firft confequent, and fecond ante- cedent, of the four terms, are both fides of the fame . figure, ' | 4. Of three proportional quantities, the middle one is faid to be a mean proportional between the other two; and the laft a third proportional to the firft and fecond. . 5. Of four proportional quantities, the laft is faid to be a fourth proportional to the other three, taken m order. 6. If any number of magnitudes be contmually propox-» tional, the ratio of the firlt and third is faid to be duphcate that of the firft and fecond; and the ratio of the firft and fourth, triplicate that of the firft and fecond. 7. And of any number of magnitudes, of the fame kind, taken in order, the ratio of the firft to the laft, is faid to be- compounded of the ratio of the firft to the fecond, of the ‘ fecond to the third, and f{o on, to the laft. 8. A right line is faid to be divided in extreme and mean ratio, when the whole line is to the greater fege " ment, as the greater fegment is to the lefs, PR QP 758 ELEMENTS OF GEOMETRY. FRaoP L Triangles and parallelograms, having the fame altitude, are to each other as their bafes. E A ¥ /TR BIC D K L Let the triangles ABC, AcD, and the parallelograms vc, cF have the fame altitude, or be between the fame parallels 8D, EF ; then will the bafe Bc be to the bafe - - €D, as the triangle ABC is to the triangle ACD, oras the parallelogram Ec is to the parallelooram CF. For, in 8D produced, take any number of parts what- | ever BG, GH, each equal to 8¢ ; and DK, KL, any num- ber whatever, each equal to cD ; and join AG, AH, AK and AL: | Then, becaufe ¢B, 8G, GH are all equal to each other, the triangles AHG, AGB, Asc will alfo be equal to each other (II. 5.); and whatever multiple the ‘bafe mc is of the bafe Bc, the fame multiple will the triangle AHC be of the triangle ABC. And, for the fame reafon, whatever multiple the bafe rc is of the bafe cp, the fame multiple will the trxangle arc be of the triangle Apc. , If, therefore, the bafe Hc be equal to the bafe cL, the triangle anc will be equal to the triangle aLc; and if greater, greater ; and if lefs, lefs. 3 ,. But I BOOK THE SIXTH. 159 But the hafe xc, and the triangle anc, are any equi- multiples whatever of the bafe Bc, and the triangle aBc ; “and the bafe cL and the triangle arc are any equimulti- ples whatever of the bafe cp and the triangle apc; whence the bafe BC is to the bafe cp, as the triangle ABc is to the triangle acp (V. Def. 5.) Again, becaufe the parallelogram ck is double the tri- angle aBc (. 32.), and the parallelogram cF is double the triangle ADc, the triangle aABc will be to the triangle Apc as the parallelogram cE is to the parallelogram CE (Vi 1d) But, it has been fhewn, that the bafe Bc is to the bafe cp, as the triangle ABC is to the triangle Apc; there- fore the bafe Bc is alfo to the bafe cD, as the parallelogram CE is to the parallelogram cr. % Q.. D Corovrr. Triangles and parallelograms, having ¢ qual altltudes, are to each other as thelr bafes. P R O Pl T'riangles and parallelograms, having equal bafes, are to each other as their altitudes. 35 0. F\ / : LN /l\ \ v G B D 1 X Let arc, DEF be two triangles, having the equal bafes AB, DE, and whofe altitudes are cH, F1; then will the : triangle 160 ELEMENTS OF GEOMETRY. triangle aBc have the fame ratio to the triangle DEF; a® cH has to FI. | / | For make Bp perpendicular to AB, and equal to cH (I. 11 and 3.); and in BP take BQ_equal to FI1, and join AP, AQand cP. Then, becaufe ep is equal to cH, and the bafe AB is common, the triangle Azp will be equal to the triangle anc (1. 5.} And, becaufe AB is equal to pE, and B(L to FI,' the trxangle aQ_will alfo be equal to the triangle peF (1L 5.} But the triangle ABP is to the triangle ARQ as BP i5 tO 8q_ (VL. 1.); therefore the triangle Arc is alfo to the triangle DEF as BP is to BQ, Or as CH to r1 (V. 9. ) - And, fince paralleloggms, having the fame bafes and altitudes, are the doubles of thefe triangles, they will, like- wife, liave to each other the {ame ratio as their altitudess W8 Tl A Coxr. 1. If the bafes of equal triangles are equal, the altitudes will alfo be equal ; 4nd if the altitudes are equaly the bafes will be equal. Cor. 2. From this, and the former propofition, it alfo appears, thatretangles which have one fide in each equal are proportional to their other fides. PROP. BOOK THE SIXTH. 161 PR OP IR Tazoasw If a right line be drawn parallel to one of the fides of a triangle, it will cut the other . fides proportionally : and if the fides be cut proportionally, the line will be parallel to the remaining fide of the triangle. L Let axc be a triangle, and pE be drawn parallel to the, fide Bc; then will ap be to DB, as AE is to Ec. ForJom the points B, E, and ¢, D Then, becaufe the triangles DBE, DCE are upon the fame bafe DE, and between the fame parallels DE, BC, they - will be equal to each other (I. 31.) And, fince equal magnitudes have the fame ratio to the fame magnitude (V. g.), the triangle DB will be to the triangle DAE, as the triangle DCE is to the triangle pAE, But triangles of the fame altitude are to each other as their bafes (VI. 1.); whence the triangle DBE will be to the triangle DAE as DB is to DA. For the fame reafon, the triangle DCE will be to the triangle DAE, as EC is to EA. M A A 1l d » 162 ELEMENTS OF GEOMETRY. And, fince ratios which are the fame to the fame ratio, are the fame to each other (V. 1 1.), pB will be to DA as EC is to EA ; or, inverfely, AD to DE as AE to EC. Again, let the fides AB, Ac be cut proportionally, in the points D and £; then will the line pE be parallel to BC. : For, the fame conflru@ion being made as before, the triangle pen will be to the triangle pEA, as DB is to DA (VI. 1.); and the mangle EDC to the tnangle DEA as Ec to Ea (VL. 1.) | And, fince DB is to DA as Ec is to EA (by Conft.), the triangle DEB will" be to the triangle DEA as the triangle EDC 1§ to the triangle pEa (V. 11.) But magnitudes which have the fame ratio to the fame magnitude are equal to each other (V. 10.); whence the triangle DEB is equal to the triangle Epc. And fince thele 'triémgles are equal to each other, and are upon the fame bafe DE, they will have equal altitudes (V1. 2. Cor.), or ftand between the fame paralw lels ; whence DE is parallel to Bc. Q. E.D. Cororr. In the fame manner it may be fhewn, that the fides of the triangle are proportional to any two of the parts into which they are divided ; and that the like parts of each are alfo propomonal PROP. BOOK THE SIXTH. 162 PROP. IV. THEOREM. i If the vertical angle of a triangle be bifected, the fegments of the bafe will have the fame ratio with the other two fides: and if the fegments have the fame ratio with the other two fides the angle Wﬂl be bife&ed, Let the angle BAc of the triangle asc. be bzfe&ed by the right line ADp ; then will BD be to Dc as BA is to Ac. | ' For through the point ¢ draw cE parallel to pa (I, 27.), and let BA be prgduced to meet CE in E : ~ Then, becaufe the right line Ac cuts the two parallel right lines Ap, Ec, the angle ace will be equal to the alternate angle cap (I. 24.) But the angle cap is equal to the angle Bap, by the propofition ; therefore the angle BAD is alfo equal to the angle AcEk. . And, in like manner, becaufe the right line BE cuts the two parallel right lines Ap, Ec, the outward angle BaD will be equal to the inward oppofite angle aec (1. 25.) M 3 , But 164 ELEMENTS OF GEOMETRY. But the angle AcE has been fhewn to be equal te the angle aD ; whence the angle AcE is alfo equal to the angle AEC ; and the fide AE to the fide' ac (L. 5.) And, fince BEC is 2 triangle, and AD is drawn parallel to the fide Ec, BD will be to pc as BA is to AE (VI.3.); "or, becaufe AE is equal to Ac, B> will be to bc as Ba is to AC. Again, let BD be to D¢ as BA is to Ac 3 then will the ~ angle BAc be bifected by the line Ap. | For, let the fame conftruction be made as before : Then fince BD is to DC as BA is to AC, (by Hyp.), and Bp to pc as BA to AE (VI. 3.), therefore, allo, BA will be to Ac as BA is to AE. And fince magnitudes which have the fame ratio to the fame magnitude are equal to each other (V. 10.), AC ‘will be equal to AE, and the angle AEc to the angle ace (L. s.) " But the angle aEc is equal to the outward oppofite ~ angle 8aD (L 28.); and the angle ACE is equal to the alternate angle cap (L. 24.) ; whence the angle Bap will ~alfo be equal to the angle cap. @ BaD. & R OP, BOOR® THE S§IXTH. 165 P ROP. . Voo T ne Gl The fides about the equal angles of equi- angular triangles are proportional ; and if the fides about each of their angles be pro- portional, the trianOIes will be equxangular. D E Let ABC, DEF be two equiangular triangles, of which BAC, EDF are correfponding angles; then will the fidle aAB be to the fide Ac, as the fide DE is to the {ide pF. ' - For make AG equal to pE, and an to oF (1. 3. ), and join GH : Then, fince the two ﬁdes AG, AH of the triangle AHG, ~ are equal to the two fides DE, pF of the triangle DFE, and the angle A to the angle p, the angle acn will alfo be equal to the angle pEr (I. 4.) But the angle DEF is equal to the angie ABE (by Hyp.) ; confequently the angle Acu will alfo be equal to the angle ABc, and G will be parallel to Bc (1. 23.) And, fince the line cH is parallel to the line Bc, the fide Az will be to the fide ac, as the fide AG is to the " ide an (VI 3.) ' But AG is equal to pE, and AH to DF; whence the fide ap will be to the fide aAc, as the fide DE is to the fide DF. ‘ M 3 Again, i 166 ~ ELEMENTS OF GEOMETRY. Again, let AB be to Ac, as DEisto DF; and AB to BC, as DE to EF; then will. the triangle ABc be equi- angular with the triangle peF, : For, let the fame conftruction be made as before : ‘Then, fince AB is to Ac as AG is to au (by Hyp.), the line cu will be parallel to the line Bc (VL. 3.), and the triangle AGH will be equiangular with the triangle aBc. And fince the f{ides about the equal angles of equi~ angular triangles are proportional (VI. 5.), the fide A= will be to the fide BC, as the fide AG is to the fide GH. But the fide aB is to the fide Bc, as the fide DE is to fide EF (by Hyp.); therefore, alfo, the fide ac will be to the fide 6H, as the {ide DE is to the fide EF (V. 11.). - And, fince the fide A is equal to the fide pE (4y Conf1.), the fide cu will alfo be equal to the fide F (V. 10.), and confequently the trian&g:le DEF will‘ be equiangular with the triangle acn (L. 7.) or aBc, as was to be thewn. PR 0P VL \THEOIREM If two trxangles have one angle of the one equal to one angle of the other, and the fides about the equal angles proportional, the tr1-‘ -~ angles will be equlangular. A B o8 D = :'"lLet_ ABC, DEF be two triangles, having the angle a . #qual to the angle p, and the fide AE to the fide ac, as the | B RIT MEL SRXT H. 167 the fide DE 15 to the pF; then will the trlangle ABC be €quiangular with the triangle DEF. - For, make a¢ equal to DE, and AH to DFj; and join GH Then, fincé the fides AG, AH are equal to the fides DE, DF, and the angle A to the angle b (&y Hyp.), the {ide cu will alfo be equal to the fide EF, and the tuangle AGH to the triangle prr (I. 4.) And, fince AB isto Ac as AG is to AH (by Hyp.), the line cu will be parallel to the line c (VI. 3.); and con- fequently the énole AGH is equal to the angle ABC, and the angle aHG to the angle ace (L. 25.) The triangle aBc is, therefore, equiangular thh the triangle AGH, and confequently it will alfo be equiangular with the triangle DEF. Q: E: D, PROP Vil Fpienemwm. In a right angled triangle, a perpendicu- lar from the right angle, 1s a mean propor- tional between the f{egments of the hypo- tenufe ; and each of the fides is a mean proportional between the adjacent {egment and the hypotenufe. C o D B | Let asc be a right angled triangle, and cp a perpen- dicular from the right angle to the bypotenufe ; then will { Mg . cD 168 ELEMENTS OF GEQMETRY., cD be a mean proportional between AD and pB; AcC a mean proportional between AB and AD ; and Bc between AB and BD. | . , For, fince the angle BpC is equal to the angle Acs (1. 8.), and the angle B is common, the triangles psc, aABc will be equiangular (I 28. Cor.) ‘ And, in the fame manner, it may be fhewn, that the triangles Apc, ABc are equiangular; whence the tri- - angle apc is alfo equiangular with the triangle pzsc. But the fides of equiangular triangles are proportional (VI. 5.); therefore the fide AD is to the fide cp, as the fide cp is to the fide pB. | In like manner, fince the triangles Apc, DBc are each of them equiangular with the triangle ABc, aB will be to AC as AC is to AD; and AB to BC 2s BC is to BD (VI.5.) , | v Q. E. D. CoroLr. If aAcs be a femicircle, and c¢p a perpendi- cular let fall from any point ¢ ; then will ADXDB=DC? ABXAD=AC? and ABXBD=BC™ PR O P BOOK THE SIXTH, 169 PR OP VI Prosien L To find a third proportlonal to two glven right lines, , Let A, B be two given right lines; it is requlred to find 2 third proportional to them. Draw the two indefinite right lines cr, cg, makmg any angle c, and take cp equal to A, and cx:, DF each equal B (I. 3.) Alfo join the points D, E, and make Fg parallel to pe (I 27.); and Ec Wﬂl be the third proportional re- ~quired. For, fince cFG is a triangle, and bk is parallel to rg (by Conft.) cD will be to DF as ck is to EG (V1. 3:) But pF is equal to cE, by con{‘cru&mn therefore cp is to CE as CE is to EG. | And, fince cp is equal to 4, and cE to g (4 Confl.), A will be to B as B is to kG, Q B ScHorLiuM. A third proportlonal to two given right lmes, may alfo be found by means of the laft propo- fition. ; | Eor let AD, DC be two given right ]ings, 6ne of which ;)C is perpendicular to the other. | Then 170 EL ELMENTS OF; GEOMETRY. “IT'hen if cB be drawn at right anorles to AC, and AD be produced till it meets the former in 8, DB will be a third proportional to Ap and pc as was required. Again, let D, DC be the two given right lines, placed at right angles to each other, as before: Then, if ca be drawn at right angles to Bc, and 8D be produced till it meets the former in 14, DA will be a fourth proportional to 8p and oc. PROP 15 PBEOoRLEN. To find a fourth proportional to three given right lines. | Let a, B, c be three given right lines ; it is required tq find a fourth proportional to them. - ‘Draw the two indefinite right lines pG, pH, making ~any angle b ; and take DE equal to A, DF to B, and 5 to o (1.3 ) | Join the points E, Fyand make GH parallel to EF (I 27 i )5 and FH will be the fourth proportional required. For, fince DGH is a triangle, and EF is parallel to ¢ {by Conft.) DE will be to bF as G is to Fu (V1. 3.) But pE is equal to A, DF to B, and EG to c (&y Conft.) 5 tberefore a willbe to 8 as ¢ is to FH (V. g.) Q E. D PROP | BOOK THE SIXTH. 191 PRCOY P X, PROB’LV.B M. To find a mean proportional'between two given right lines. B"""‘""C,‘\EDG Let a, B be two gwen right lines; it is requlrcd to find a mean proportional between them. Draw the indefinite right line cg, in which take cg equal to A, and ED equal to & (I. 3.) | | Upon c¢p defcribe the femi-circle cFp, and from the point E erect the perpendic’u]ar eF (l.11.); and it will be the mean proportional required. For join the points cF, FD : Then, becaufe pFc is a femi-circle, the angle cFp is a right angle (IIl. 16.), and confequently the tnanglcw CDF is rectangular, And fince a perpendicular from the right angle. is a mean proportional between the fegments of the hypote- nufe (VI. 7.), £ will be a mean proportional to CEy ED. | it But cE is equal to a, and ED to B (& Con/l.) ; there- fore EF will be a mean proportional to A and B, as was to be fhewn. Scuorium. If cp, pE be the two given lines, or will be a mean proportional to them, And if cp, cE be the given lines, c¥ will be a2 mean proportional to them (V1. 7.) s | PROP \ i72 ELEMENTS OF GEOMETRY. PROP. XI. ProsLEM. To divide a given right line into two parts which fhall have the fame ratio with two ngen rxght lines, E | il . € D ~ Let cp be a given right line, and a, B two other given right lines ; it is required to divide cp into two parts which fhall be to each other in the ratio of A to B. Draw the indefinite right line cG, making any angle with cp; and make cF equal to A, and G to B (1. 3.): Jom the points G ; and make FE parallel to 6o (I.27.); _and it will divide cp in the point £ as was required. For, fince cpG is a triangle, and FE is parallel to GD (b Confl.), cE will be to ED as CF is to FG (¥l 34 But cF is equal to A, and FG to B (by Confl.) ; there- fore cE will beto Epas aisto & (V.q.) Scuorium. In nearly the fame manner may a third, fourth, or any other part be cut off from 2 given line cp. For if cc be made the fame multiple of cF that cp is | of the part, and c¢p, FE be drawn as above, CE will be the part required. PROQZ i i i ] i g I% ; A ::- s B BElr e gk (3R ottt et e i BOOEy TRHE GRNS B 173 EROT XH Torsk: If four right lines be proportional, the rectangle of the extremes will be equal to the rectangle of the means: and if the rect- angle of the extremes be equal to the ret- angle of the means, the four lines will be proportional. M G N W K A B C D R g T Er v Let AB be to cp as £ is to F; then will the reGangle of AB and F, be equal to the re&angle'of cD and E. , For make Bu perpendicular to AB, and equal to r {I. 10. 3.), and pG perpendicular to cp, and equal to E; and in pG take px equal to B (. 3.), and draw KL parallel to ¢cp (I. 27.) Then, fince parallelograms of equal altitudes, are to each other as their bafes (VI. 1.), the parallelogram an will be to the parallelogram cx, as Az is to cp. And fince AB is to ¢D as E is to F (4y Hyp.), or as pg is to DK (&y Conft. ), the parallelogram an will al{o be to the parallelogram ck as oG is to bk (V. 11.) But parallelograms of the fame bafe, are to each other as their altitudes (VI. 2.); whencé the parallelogram cG will alfo be to the parallelogram ¢k as pg is to px. And 174 ELEMENTS OF GEOMETRY. And becaufe ratios which are the fame to the fame ratio are the fame to each other (V. 11.), the parallelo- gram AH will be to'the parallelogram ck, as the paral- lelogram cG is to the parallelogram ck. But magnitudes which have the fame ratio to the fame magnitude are equal to each other (V. 10.); whence the reGangle A is equal to the reCtangle cG. Again, let an, the reGangle of the extremes, be equal to CG, the reGtangle of the means ; then will AB be tocp as E 18 to F. ~ For, let the fame conftruction be made as before : Then, fince reGtangles of equal altitudes are to each other as their bafes (VI. 2.), the re¢tangle an will be to the reGangle ck as AB is to CD. And, becaufe the retangle aH is equal to the reftangle cG (by Hyp.), cc will alfobe to ck as ABistocp (V.q.) But cG is to cK as DG s to DK or gu (VI. 2.) con- fequently aB will be to cp as pG is to BH (V. 11.) And fince DG is equal to E, and BH to F, AB will be tocpasEistoF (V.q.) | Q. E: D PROP \ JBOOE THE SI G, 175 PROP XHIL Tsaronem If thige right lines be proportional, the rectangle of the extremes will be equal to the {quare of the mean: and if the rectangle of the extremes be equal to the fquare of the mean, the thxee lines will be propor- tional.. | H G TR S S f | | | o | | Al iR o RIS 5 A e ? Ernomacor Let aABbe to cp as cp is to £ ; then will the re&angle of AB and E be equal to the fquare of cp. o For make Bx perpendicular to as (I. 10.) and equal to E (L. 3.); and upon cp deferibe the fquare ¢cc G0 o ¢ and make F equal to cp. Then, fince cp is equal to F, and AB is to cD as cp is to E (by Hyp.), aB will alfo be tocp as ¥ is to E (V.q.) And, fince thefe four lines are proportional, the rectan- gleof AB and E will be equal to the reftangle of cp and F (VI. 12.) - But the re&tangle of ¢p and r is equal to the {quare of ¢D, becaufe cp is equal to F ; therefore, alfo, the rect- angle of AB and & will be equal to the {quare of cp. Again, if the retangle of aB and E be equal to the {quare of cp ; AB will beto cpas ecpisto k. For let the fame conftru@ion be made as before : 4 | Then, 176 ELEMENTS OF GEOMETRY. Then, fince the reGtangle of aB and E is equal to the fquare of cp (by Hyp.), and the fquare of cp i5 equal to thereGtangle of cp and ¥ (IL 2.), the reCtangle of Ap and E will alfo be equal to the rectangle of cp and 7. But if the reétangle of the extremes be equal to that of the means; the four lines are proportional (VI. 12.); whence AB is to cD as F is to E. And fince cp is equal to F, by conﬁruéhon, AB will be to cb as cp is to E. i g o PROP. XIV. THrorEM Equal parallelograms and triangles have their fides about equal angles reciprocally proportional ; and if the fides about equal angles are rec1procally proportional, the pa- ‘rallelograms and triangles will be equal. o N D A E . I.et AB, Ac be two equal parallelograms, and pra, AEG two equal triangles, having the angle DAF equal to the angle GAE ; then will the fide DA be to the fide AE, as the fide, aG is to the fide AF. . | For let the fides pA, AE be placed in the fame right line, and complete the parallelogram Ak. Theng BOGK TIE S1Tu. . 1% Then, becaufe the angles paF, FAE are equal to two right angles (1. 13.), and the angle FAE is equal to the angle pac (I. 15.), the angles DAF, DAG are alfo equal to two right angles ; and confequently FAG is a right line. And fince the parallelogram AR 1s equal to the parallel- ogram AC (4y Hyp.), and aK is another parallelogram, AB IS to AK as AcC is to ak (V. g.) But Az is to Ak as pa to Ak (VI. 1.), and aAcC to AK as AG to AF; whence DA is to AE as AG is to AR (Vori)) | And, if FE be joined, it may be fhewn, in like man- ner, that the triangle DAF is to the triangle AFE as the triangle GAE is to the triangle AFE; and DA to AE as AG tO AF. | ‘ e A'gain, let the angle DAF be equal to the angle GaE, and the fide DA to the fide AE as the fide aG is to the fide AF ; then will the parallelogram Ag be equal to the paraf lelogram AC, and the triangle DFA to the triangle GAE. For fince DA is to AE as AG to AF (by Hyp. ), zgnd DA to AE as AB to AK (VI. 1.), ac will be to AF as aB to Ak (V. 11.) | , But AG is to AF as ac to Ak (VI. 1.); whence aB i to AK as AC to AK (V. 11.}; and confequently the parallelogram AB is equal to the parallelogram ac (V. 10,) | And fince triangles are the halves of parallelogranis, which have the {fame bafe and altxtude, the trmngle DFA will be equal to the triangle GAE. Q@ BoD. Cororr. The fides of equal reGtangles are reciprocally proportional ; and if the fides are reciprocally propor- tional, the reftangles will be equal, N PR OP, 178 ELEMENTS OF GEOMETRY. PR Pi- XYV ' Prosreyu Upon a given right line to defcribe a rec- tilineal figure, fimilar, and fimilarly fituated, to a given retilineal figure, Let ap be the given right line, and cperc the given reGilineal figure ; it is required to defcribe a retilineal figure upon aB, which fhall be fimilar, and fimilarly fitus ated to CDEFG, : | Join pG, DF; and at the points A, B, make the angles BAL, ABL equal to the angles nce, cpc (L. 20.) In like manner, at the points B, L, make the angles BLK, LBK equal to the angles DGF, GDF; and BKH, KBH equal to DFE, FDE. Then, becaufe two angles in one triangle ar equal to two angles in another, each to each, the remaining angles in each of the correfponding triangles, will alfo be equal (1. 28. Cor.) < And fince the angles ALB, BLK, are equal to the an- gles ¢GD, DGF, and LKB, BKH to GFD, DFE, the angle aLk will be equal to the angle cGF, and the angle LKH to the angle GFE. And, in the fame manner, it may be fhewn that the an- gles kuB, HBA and BAL are equal to the angles FED, EDC and DCG. The [ BOOK THE SIXTH. 179 The figures ArukL and cpErG are, therefore, equi- angular : and they have their fides about the equal angles, alfo, proportional. For, fince the triangles ALB, cGD are equiangular, AL will be to 1B as ce to 6D (VI. 5.); or aL to CG as LB to 6D (V. 15.) ,_ ' And, in like manner, Lx will be to 18 as cF to Gb (VI 5.)5 or Lk to 6F as 15 to ¢b (V58 But ratios which are the fame to the fame ratio are the fame to each other (V. 11.); whence ar will be to ca RS LIT0GF 5 ‘Or-AL 10'LK 88 €6/ to ¥ (Vi s, ) And, in the fame manner, it may be fhewn, that the fides about the angles x, H, B, A, are proportiénal to the fides about the angles F, E, D, C. The figure ABHKL is, therefore, fimilar, and fimilarly fituated with the figure cpEFG (VI. Def> 1.) ; and it is defcribed upon the right line aB, as was to be done. EROP- W raaili s Equiangular, or fimilar triangles, are to each other as the {quares of their homologous fides. | S Let arc, DEF be two fimilar triangles, of which the fides AB, DE are homologous ; then will the triangle ABc | N 2 ‘ be 130 ELEMENTS OF GEOMETRY. - be to the triangle DEF as the fquare of aB is to the fquare of DE. ‘ For, on AB, DE defcribe the fquares AL, DN (1. 1.), and let fall the,perpendicul‘a}rs cg, Fu (1. 12.) Then, fince the triangles ABc, DEF are fimilar (4y Hyp.), ac will be to AB as DF to DE (V1. Defi 1.) 5 oF AC to DF as AB to DE (V. 15.) And, becaufe the triangles AGC, DHF are equiangular, Ac will be to cc as pF to.ru (V1. 5.); or AC to DF. a3 cG to FH (V. 15.) But fatios which are the fame to the fame ratio, are the fame to each other (V. 11.) ; therefore ¢G 1s to FH as AB to DE; Or CG to AB as FH to DE (V. 15.) And fince triangles which have the fame bafe, are to each other as their altitudes (V1. 2.), the triangle ABcC is to the triangle AKB as ¢G is t0 AK, OF AB. In the fame manner it may be thewn, that the triangle PEF is to the triangle DME as FH is to DM, Or DE. But cc has been fthewn to be to AB as FH is to DE ; therefore the triangle ABc is to the triangle AKB as the triangle DEF is to the triangle DME (V. 11:) And fince the fquare AL is double the triangle AKB (k.22. and the {quare DN is double the triangle DME, the triangle ABC will be to the triangle DEF as the fquara AL is to the fquare pN (V. 13 and 15.) PROP. = e ‘BOOK T HE SEXEH, 181 PROP XVIL ThroREM Similar polygons are to each other as the {quares of theitr homologous fides. F /,,m A B _ ¥ G Let ABcDE, FGHIK be fimilar polygons, of which AB, FG are homo]"ogous fides ; then will the pdlygon ABCDE be to the polygon rcuik as the fquare of ABis to the - fquare of Fa. For join the points BE, BD, GK and GI : Then, fince the angle A is equal te the angle r, and AB.is to AE a5 FG is to FK (VI. Def. 1.), the triangles EAB, KFG will be equiangular, or fimilar (VI. 6.) And if, from the equal angles AED, FKI, there be taken the equal angles AEB, FKG, the remaining angles BED, ;k1 will alfo be equal to each other. But Ep is to KI as EA is to KF (V1. Def. 1.and V. 15.), and EA is to KF as EB to KG (Vi ciomi Vorgirs whence Ep will be to k1 as EB is to KG (V. 11.) Since, therefore, the angles BED, GKI are equal to each other, and the fides about them are proportional, the triangles BED, GK1 will, alfo, be equiangular, or {imilar (VL6.) - And, in the fame manner, it may be fhewn, that the triangles BCD, GHI are equiangular, or {imilar, N 3 But 182 ELEMENTS OF GEOMETRY. But fimilar triangles are as the fquares of their like fides (VI.16.); whence the triangle EaB is to the tri- angle K¥G as the fquare of EB is to the fquare of KG. And, for the fame reafon, the triangle EBD is to the triangle K 61 as the fquare of £B is to the fquare of KG. But ratios which are the fame to the fame ratio, are the fame to each other (V. 11.); whence the triangle EAB is to the triangle KFG as the triangle E8D is to the tri- angle KG1. ' | And in the fame manner it may be fhewn that the tri- angle EBD is to the triangle KG1 as the triangle pBc is to the triangle 1cH. , - The triangle EAB, therefore, is to the triangle KFG, ~as the triangle EBD is to the triangle xG1, and a"s' the triangle DBC is to the triangle 161 (V. r1.) ~ And fince the fum of the antecedents is to the fum of the Eonfequents as the firft antecedent is to its confe- quent (V. 16.), the polygon aABcDE will be to the polygon FGHIK as the triangle EAB is to the triangle KFG. But the triangle EaB is to the triangle KrG as the fquare of AB is to the fquare of F¢ (VI. 16.); whence the polygon ArcDE is alfo to the polygon FGHIK as the fquare of AB is to the fquare of FG. Q. E. D. FROP i e S BOOK THE SIXTH. 183 PR O, XN, THEORE W Parallelograms and triangles, having two equal angles, are to each other as the rect- angles of the fides which are about thofe angles. B‘ R g Vs G C Let AB, Ac be two parallelograms, having the angle DAF equal to the angle GAE ; then will AB be to Ac as the reftangle of DA, AF is to the re@angle of ca, AE. For let the fides DA, AE be placed in the fame right line, and complete the parallelogram ax. Then, becaufe the angles baF, FAE, are equal to two right angles (I. 13.), and the angle FaE is equal to pac (1. 15.), the angles DAF, DAG are alfo equal to two right angles ; whence FG is a right line (I. 14.) . ‘And fince parallelograms, of the fame altitude, are to each other as their bafes (VI. 1.}, the parallelogram as is to the parallelogram Ak as AD is to AE. But AD is to AE as the reltangle of Ap, AF is to the reGtangle of Ax, aF (V1. 2. Gor. 2.); therefore apisto - aK as the rectangle of AD, AF is to the reftangle of aE, AF (V.11.) ' : And in the fame manner it may be thewn, that ac is to AK as the reCtangle of AG, AE is to the reGtangle of N 4 AE, 184 ELEMENTS OF GEOMETRYs AE, A¥ ; whence AB is to ac as the re&ang}e of AD, AF 1s to the rectangle of ac, AE (V 11 and 15. Ay Again, let pra, AEG be two tnangles, having the angle PAF equal to the angle cAE, then will DFA be to AEG as the reftangle of pa, AF is to the reQangle of GA, AE. For lct the {ides pA, AE be placed in the fame right line ; and complete the parallelograms a®, ac, AK. Then, as before, AB is to Ac as the re@angle of pa, AF is to the re&: angle of AG, AE. But the triangles DFA, AEG are half the parallelovramsr AB, ac (l. 30.); whence DFA is to AEG as the retangle of DA, AF is to the rectangle of Ga, AE. ! Q. D, Scuorium. If the hne EF¥ be drawn, the latter part of thxs propofition may be proved from the triangles, inde- ‘ pendently of the former. ' - . PROP. XIX. TuEorem. The re@:ang]es under the corre{nondmg hnee of two ranks of proportionals, are them- f@lvf“s PIO"’Q"U&)D&ISQ ke o X s | | ; | AT T TR et - : ¢k Yot A be to ¢B as DE is to FE, :md GH tO GE'as LM - to 1805 then will the reangle of ae, 6B be to that of CB,s BOOK TR SR T, 183 €3, GK as the reftangle of pE, 1m is to that of FE, LN. | e For draw BqQ, Es at rlght angles to- aB, DE (L. 11.), and make BP equal to GH, BQ_to GK, ER to LM, and Es to LN (I. 3.); and complete the parallelograms ap, €Q, DR and Fs. Then fince parallelograms, of the fame altitude, are to each other as their bafes (V1. 1.), ap will be to cp as AB to CB; and DR to FR as DE to FE. But ap is to cB as DE to FE (by Hyp.); whence AP will be to cp as DR to FR (V. 11.); or AP to DR as cp to FR (V. 15:) And fince parallelograms of the fame bafe are to each other as their altitudes (VI. 2.), cp will be to cq_as BP to BQ; and FR to Fs as ER to Es. Or, becaufe BP, BQ are equal to GH, GK, and ER, ES to LM, LN {(by Gonfl.), cp will be to cq as GH to GK ; and FR to Fs as LM to LN (V.0q.) | But guisto Gk as LM to LN (&y #yp.), thercfore cp will be to cq as Fr is to Fs (V. 11. )> Or CP to FR2s CQ_ to Fs (V. 15.) And it has been before fhewn that AP is to DR as cP to FR; whence AP is to DR as ¢Q to Fs (V. I1.), or AP to cQ as DR to Fs (V. 15.) | But ap is the re&angle of AB, 6H; cQ of cB, GKj; DR of DE, LM ; and Fs of FE, LN (b Cmﬁ.) ; therefore the reftangle of aB, GH is to that of cB, GK as the rect- angle of DE, LM is to that of FE, LN. . Qv B D, Cororr. The fquares of four proportional lines are themfelves proportionals. “ PR OM 186 ELEMENTS OF GEOMETRY. PR O Vo> XK. T @t oR iM. The fides and diagonals of four propor- ‘tional {quares, are themfelves proportional. 3 7, PR it A i s i e 17 EONG ; 2 % | o 5 N ! £ i SN ¥ ; e , ! ! G i.‘__ﬂ i,‘______m \ TL' N é Ak B - Let ac, EF, HL and QN be four proportional fquares; then will their fides and diagonals be alfo proportionals, For, make s a fourth proportional to AB, EB and HK Viigv); and draw the diagonals 8D and kM. Then, fince AB is to EB as HK is to. s (&y Confl.), Ac will be to EF as BL is to the fquare of s (VL. ‘Ig.,Cor.) And becaufe Ac is to EF as HL is to QN (&y Hyp.), HL will be to the {quare of s as HL is to QN, or the fquare of gk’ (Vi 11.) But magnjtudes which have the fame ratio to the f'tme magnitude are equal to each other (V. 10.) ; whcnce the fquare of s is equal to the fquare of QX. And fince equal fquares have equal fides (II. 3. ), s is equal to @k ; and confequently AB is to EB as HK to 10 QK. Again, becaufe the triangles ABD, EBG are equiangu- lar, aB will be to £8 as 8D to BG (VI. 5.) ‘ And becaufe the triangles HXM, gkPp are alfo equi- angular, Hk willbe to Qx, as xmto xp (V1. 5.) But ap has been fhewn to be to EB as HK is to QK ; confequently BD is to BG as KM is to kP (V. 11.) (L D. PrROP A BOOK THE SIXTH. 187 PROP. XXI. PromLEM. 'T'o cut a given rloht hne in extreme and mean propoxtlon. ¥ A E D Let Az be the given right line; it is requlred to cut it in extreme and mean proportion, Upon aB defcribe the {quare ac (II. 1.), and bifect the {ide aApin E (I. 10.); and join BE. In A produced, take EF equal to EB (I. 3.); and upon AF defcribe the fquare Fu (II. 1.); then will AB be divided at the point H as was required. For fince pr is the fum of EB, ED, or EB, EA, and AF is their difference, the rectangle of DF, Fa is equal to the difference of the fquares of £B, £a (IL. 13.) But the rectangle of pF, FA is equal to ng, becaufe FA is equal to FG; and the difference of the fquares of EB, EA is equal to thefquare of aB (IL. 14. Car.) ; whence DG is equal to Ac. And if from each of thefe equals, the part Ak, which is common, be taken away, the remainder ac will b= equal to the remainder Hc. | | But equal parallelograms have the fides about equal angles reciprocally proportional (VI. 15.); whence uk is to HG as HA to HB. | SR And {ince HK is equal to AD, or AB, and HG to HA, AB will be to HA as HA is to HB, Q. E. D, PR OP B 4 188 ELEMENTS OF GEOMETRY. P RO P.v XXH.. PROBLEM. To divide a given right line into two fuch parts, that their rectangle may be equal to a given {quare, the fide of which is not greater than half the given line. : / D, D ¥ & g [ . ‘ “H D s ¢ # Let aB be the given line, and c the fide of the given fquare; it is required to divide AB into two fuch parts that their re€tangle may be equal to the {quare of c. Upon as defcribe the femicircle BpA, and make BF perpendicular to aB (I. 11.), and equal to ¢ (1. 3.) - Through ¥ draw ¥p parallel to as (I. 27.); and from the point D where it cuts the circle, let fall the perpendi- cular pE (I.12.); and AB will be divided at E as was required. " For fince BpA is a fimicirclé (4y Confl.), and DE is perpendicular to the diameter AB (&y Confl.), the rect- angle of AE, EB will be equal to the {quare of e (V1. 7, Cor.) ; G But ED is équal to 8 (I. 30.) or c; whence the rect- angle of AE, EB will be equal to the {quare of ¢ as was to be thewn., - ; | ScuorruMm. When BF, or C, is equal to half AB, FD ‘will be a tangent to the circle, and the reCtangle of AE, £8 will be the greateft poffible. PROP Sook THE Giwru. . 2B P& O XXHL PROB:ZE.EM. To a given right line to add another right line fuch, that the re@angle of the whole and the part added fhall be equal to a gwen ﬁquaxc § Let as be the given line, and c the fide of the given fquare ; it is required to add a line to ap fuch, that the rectangle of the whole and the part added fhall be equal to the {quare of c. Make Be perpendicular to ap (L. Ix.)f, and equal to ¢ (L. 3.) ; alfo bife& a3 in ¢ (I. 10.), and join GE. Then, if As be produced, and D be taken equal to GE (1. 3.), the part BD will be added to AB,as was required. For on aB defcribe the femicircle BFA, cuttmg GE in ¥, and join FD. Then, fince the two fides 6B, GE of the triangie GEB, are equal to the two {ides GF, GD, of the triangle GDF, and the angle G 18’ common, the angle GBE will be equal to the angle GFp, and the fide Fp to the fide BE (1. 4.) , But the angle GBE is a right angle (y C’azﬁ ) 3 whence the angle ¥ is alfo a right angle; and confequently Fp is a tangent to the circle at (IIL. 10.) And 1O ELEMENTS OF GEOMETRY. And fince DF is a tangent to the'circle, and DA is drawn to the oppofite part of the circumference, the rectangle of AD, b will be equal to the {quare of pr (III. 29.) But DF has been fhewn to be equal to BE,or ¢ ; whence the rectangle of AD, DB wxll alfo be equal to the fquaxe or . i | ! Q.'E.D,'F PR DXV Tusonew. Angles at the centres or circumferences of equal circles, have the fame ratio with the arcs on which they ftand. Pl Hi ) J Let aBc, DEF be two equal circles, in which Bec, gHF arc angles at the centre, and BAc, EDF angles at the circumference ; then will the arc Bc be to the arc EF as the angle BGC is to the angle EHF, or as the angle Bac to the angle EDF. For on the c1rcumference of the circle Arc take any number of arcs whatever cK, KL each equal to Bc; and on the circumference of the circle pEF any number of arcs whatever FM, MN, each equal to EF; and join GK, GL, HM, HN. ‘Then, becaufe the ares Bc, CK, KL are all equal to each other, the angles BGc, cGK, KGL will alfo be equal to each other (11l 21.) 0 eac ( il 2 Ead BOOK THE SIXTH. 192 And, therefore, whatever multiple the arc BL is of the arc Bc, the fame multiple will the angle BGL be of the angle BGC. | , | _ For the fame reafon, whatever multiple the arc EN is of the arc £F, the fame multiple will the angle Ern be of the angle EnF. | ‘ ' If, therefore, the arc BL be equal to the arc Ewn, the angle BGL will be equal to the angle Enn; and if equal, equal ; and if lefs, lefs. | But BL and BGL are any equimultiples whatever of Bc and BGc, and EN and Eun of EF and EaF; whence the arc BC is to the arc EF as the angle BGc is to ‘the an- gle guF (V. 5.) o And fince the angle BGc is double the angle sac, and the angle En¥ is double the angle epr (IIL 14.), the “arc B¢ will alfo be to the arc EF as'the angle Bac is to the angle epr (V. 13.) | P RGO-Po o XXV, TaroRem: ‘The reCtangle of the two fides of any tri- angle, is equal to the retangle of the feg- ments of the bafe, made by a line bifecting the verticle angle, together with the fquare of that line. Let aBc be a triangle, having the angle acs bifected by the right line ¢p ; then will the re@angle of ac, cs . be 102 ELEMENTS OF GEOMETRY. be equal to the re&angle of AD, DB, together with the fquare of cp. For, about the trlangle ABC, defcmbe the circle AEC (1V. 5.), cutting cp, produced, in E; andJom EB. Then, becaufe the angle acp is equal to the angle £cB (by Hyp.), and the angle caD to the angle ces (III. 15.), the remaining angle anc will be equal to the re- maining angle csE (I. 28. Cor.) | The trlangles CAD, CEB being, therefore, equzangular, cA will be to ¢p as cE to ¢B (VI. 5.) ; and confequent- ly the reCtangle of ca, cB is equal to the rectangle of cE, co ¥l 18] But the reétangle of cE, cp is equal to the rectangle of ED, DC, together with the {quare of cp (Il.10.); whence the rectangle of ca, ceisalfo equal to the rg:&angle of ED, DC, together with the fquare of cp. : And fince the rectangle of Ep, Dc is equal to the rect- angle of ap, b (IIl. 27.), the re¢tangle of ac, csisallo equal to the re¢tangle of Ap, DB, together with the fquare of cD.. Q. E. D. PO P. BOOK {THE SIXE N 193 PROP XXVI. Turonel The rectangle of the two fides of any tri- angle, is equal to the retangle of the per- pendicular, drawn from the vertical angle to the bafe, and the diameter of .the circum- fcribing circle. | Let aBc be a triangle, having c¢p perpendicular to AB then will the rectangle of Ac, cB be equal to the retan- gle of c¢p and the diameter of the circumf{cribing circle. For, about the triangle aBc, defcribe the circle aEc (IV. 5.); in which draw the diameter cE ; and join EB. Then, fince the angle cAD is equal to the angle cEes (1IL. 15.) and the angle Apc to the angle Esc, being each of them right angles (Confl. and 111. 16.), the re- maining angle acp will be equal to the remainjng an- gle ece (I. 28. Cor.) The triangles acp, Ecs are, therefore, equiangular ; whence Ac is to e€p as cE is to ¢B (VI. 5.) ; and confe- quently the re&tangle of Ac, c¢B is equal to the reftangle of cby el (V1. 12.) Q. E. D. Scamorium. When ABc is an obtufe angle, the per- pendicular cp falls without the circle; but the fame de- monftration will hold. ' O PR O P, 194 ELEMENTS OF GEOMETRY. PR O P XXV THror M. | "The rectangle of the two diagonals of any quadrilateral, infcribed in a circle, is equal to the fum of the re@angles of its oppofite hdes. Let aBcp be any quadrilateral infcribed in a circle, of which the diagonalé are AC, BD ; then will the reftangle of AC, ED be equal to the retangles of AB, pc and AD, BC. For make the angle ¢DE equal to the angle aps (1. 20.) ; then, if to each of thefe angles, there be added the common angle EDB, the angle ADE will be equal to the angle cps. The angle DAE is alfo equal to the angle pzc, being angles in the fame fegment, whence the remaining angle AED is equal to the remaining angle Bcp (L 28. Cor.) Since, therefore, the trianoles ADE, BDC are equian- gular, AD is to AE as BD is to BC (VI.5.); and con- fequently the reCtangle of Ap, BC 1s equal to the reGtangle of ar, BD (VL. 12.) " Again, the angle cDE being equal to the angle ApB (by Conjl.), and the angle EcD to the angle ap (III.- 15.), the remaining angle cEp will be equal to the re- maining angle 8ap (l. 28, Cor.) The [+ ,_4. 9 vl BOOK THE SIXTH. 195 The triangles cEp, ADB are, therefore, alfo equiangu- lar ; whence AB is to BD as Ec is to pc (VI.s.); and confequently the re@angle of AB, Dc is equal to the ret- angle of Ec, 8D (VI. 12). And if, to thefe equals, there be added the former, the retangle of AR, Dc together with the re@angle of AD, Bc will be equal to the re®angle of Ec, BD together with the rectangle of AE, BD. \ , But the rectangles of AE, BD, and Ec, BD are equal to the rectangle of ac, Bp (1I.8.); whence the reftangle of Ac, BD is alfo equal to the re&tangles of AB, pc and AD, BC. | Q. E=lk O 2 = BOOK 196 ELEMENTS OF GEOMETRY. . B 0K IYIL LA N ELON.S 1. The common fection of two planes, is the line in which they meet, or cut each other. 2. A right liné is perpendicular to aplane, when it is perpendicular to every right line which meets it in that plane. ‘ 3. A plane is perpendicular to a plane, when every right line in the one, which is perpendicular to their com- “mon feGion, is perpendicular to the other. 4. The inclination of a right line to a plane, is the angle it makes with another line, drawn from the point ' of fe&ion, to that point in the plane, which is cut by a perpendicular falling from any part of the former. - g. The inclination of a plane to a plane, is the angle contained by two right lines, drawn from any point in the common fe&ion, at right angles to that fection ; one ~ in one plane, and the other in the other. 6. Parallel planes, are fuch as being produced ever fo far both ways will never meet. - 7. A plane is faid to be extended by, er to pafs through a right line, when every part of that line lies in the plane. ¥ PROP S SOOK OCTHE "SSEVEN T H. . 197 PROP. I Tuiokem - The common fe&;on of any two planes 18 a rzght line. e Let AB, cDbe two planes, whofe common fe&xon is £F ; then will £F be aright line. | For if not, let FGE be a right line, drawn in the plane AB; and FKE another right line, drawn in the plane cp. Then, fince the lines FGE, FKE are in d:ﬂ’erent planes, they muft fall wholly without each other. But the line FGE, having the fame extremities with the line FkE, will coincide with it: whence they coincide and fall wholly without each other, at the fame time, which 1s abfurd. s oy The lines FGE, FKE cannot, therefore, be right lines ; and confequently the line £F, which lies in each of the planes, muft be a right line, as was.to be fhewn. Scuorrum. One part of a right line cannot be in a plane, and another part out of it. = For fince the line can be produced in that plane, the part out of the plane, and the part produced would have different direions, which is abfurd. O PROP § N 798 ELEMENTS OF GEOMETRY. P RO P-1l. " "FEEOREM. Any three right lines which mutually in- terfect each other, are all in the fame plane. E (\/ D i B\ . " Lot AB, Bc, ca be three right lines, which mterfeé‘t ~ each other in the points A, B, C; then will thofe lines be in the fame plane. | ' | TR Iet any plane AD pafs through the points a, B, and be turned round that line, as an axis, till it pafs through| the point c.. Then, becaufe the points A, c are in the plane AD, the whole line ac muft alfo be in it; or othe:wifé‘ its parts would not lie in the fame direQion. | | And, becaufe the points B, C are alfo in this plane, the whole hne BC muft likewife be in it ; for the fame reafon. ' But the line AB is in the plane Ap, by hypothefis ; whence the three lines AB, BC, CA are all in the fame plane, as was to be fhewn. CoRr. Any two right lines which interfect each other, are both in the fame plane ; and through any three pomts 2 plane may be extended. | PR QYT . g BOOK R E B SENENT 1. . Tyl PROW.HE S Tutoren If a right line be perpendicular to two other right lines, at their point of interfec~ tion, it will alfo be perpendicular to the plane which paffes through thofe lines. !t.i , N \ B C Let the right line AB be perpendicular to each of the two right lines Bc, BD, at their point of interfection B ; then will it al(o be perpendicular to the plane which paﬁ'es through thofe lines. For make Bp equal to Bc; and, in the plane which pafles through thofe lines, draw any right line ge; and join the points €D, AD, AE and Ac: \ Then becaufe the fide sc is equal to the fide BD (by Con/l.), and the perpendicular AB is common to each of the triangles ABc, AeD, the fide ap will alfo be equal to the fide-ac (I. 1.) | And fince the triangles cap, cBp are ifofceles, the rectangle of cE, ED, together with the fquare of Eg, is equal to the {quare of pr; and the re€angle of cE, ED together with the {quare of EA, is equal to the fquare of ap (ll. 20.) From each of thefe equals, take away the reQangle of CE, Ep which is common, and the difference of the O 4 fquares 200 ELEMENTS OF GEOMETRY. fquares of EB, EA will be equal to the difference of the fquares of pB, AD. But the difference of the fquares of DB, AD is equal to the fquare of A (IL. 14. Cor.) ; whence the difference of the fquares of £8, EA will alfo be equal to the fquare of AB; and confequently A® is perpendlcular to BE, as was to be thewn. COROLL If a right line be perpendicular to three other right lines, at their point of interfection, thofe lines wﬂl be all in the fame plane. For if either of them, as BE, were above or below the plane which pafles through the other two, the angle ABF would be lefs or greater than a right angle ‘ PVR.O P. \IV. THEOREM. If two ught lines be perpendicular to the {ame plane, they W111 be parallel to each -»othex. F E Let the right lines AB, cD be each of them perpendi- cular to the plane FG, then w111 thofe lines be parallel to each other. - For join the points D, B3 and, in the plane Fe, make pE perpendicular to DB, and equal to AB (I 1T 3 ); and join the pomts AE, AD. i Then, 0 BOOK THE SEVENTH. 201 Then, fince the right lines aB, cD are perpendicular to the plane ¥G (4y Hyp.), the angles ABD, ABE, CDB and cpE will be right angles (VII. Def. 2.) And becaufe the fide aAB, is equal to the fide ED (by Conf?. ), the fide DB common to each of the triangles BAD, BED, and the angles ABD, BDE right angles (4y Hyp. and Conji.), the fide ap will alfo be equal to the fide EB (1. 4.} Again, fince the fides AD, DE are equal to the fides | EB, BA, and the fide AE is common to each of the tri- angles EBA, EDA, the angle ApE will alfo be equal to the angle aBe (I. 7.), and is, therefore, a right angle. And, becaufe the line Ep is at right angles with each of the three lines DA, DB, DC, thofe lines, together with the line aB, will be all in the fame plane (VIL. 3. Cor.) Since, therefore, the lines AB, BD, DC are all in the fame plane, and the angles ABD, CDB are each of | them a right angle, the line A will be parallel to the line cp (L. 23.), as was to be fhewn. Cor. Any two parallel right lines AE, ¢D, are in the fame plane ; and any right line DA, which interfeéls *hofe parallels, is in the fame plane with them. PR OP, 202 ELEMENTS OF GEOMETRY, PR OP: V. THEOREM. If two right lines be parallel to each other, and one of them be perpendicular to a plane ; the other will alfo be perpendicular to that plane, 1 AN, J F 3 Let aB, cp be two parallel right lines, one of which, 'AB, is perpendicular to the plane FG ; then will the other cb be alfo perpendicular to that plane. For join the points D, B; and, in the plane rG, make DE perpendicular to DB, and equal to Ba (I.11.3.); and ~join AE, AD and EB: ” Then, becaufe AB is perpendicular to the plane G (4y Hyp.) the angles aBp, ABE will be right angles (VII. Def. 2.} And, fince the fide AB is equal to the fide D (5y Hyp.), the fide pB common to each of ‘the triangles BAD, BED, and thejangles ABD, BDE right angles (by Con/t. and Hyp.), the fide Ap will be equal to the fide B (I. 4.) ‘Again, fince the fides Ap, DE are equal to the fides EB, BA, and the fide AE is common to each of the tii- angles;E}AD, EBD, the angle ADE will be equal to the angle aBe (I. %.), and is, therefore, a right angle. And, becaufe the right lines AB, cD are parallel to each other (by Hyp.), and the line AD interfeéts them, they will be all in the fame plane (V1L 4. Cor.); and the angle BOOK YEHE SEVENTH. 203 angle ABD being a right angle, the ang‘le cpr will alfo be a right angle (I.25.) But fince ED is at right angles to DB, DA, itis alfo at right angles to the plane which pafles through them (VH ~ 3.) 5 and confequently to pc (VIL. Def. 2.) , The line Dc is, therefore, perpendicular to each of the lines DE, DB; whence it is alfo perpendicular to the plane rG (VIL. 3.), as was to be fhewn. PROP. VI. THEOREM. If two right lines be parallel to the fame right line, though not in the fame plane with it, they will be parallel to each other. Let the right lines AB, ¢D be each of them parallel to - the right line EF, though not in the fame plane with it ; then will AB be parallel to cp, For take any point G in the line £F, and draw the right lines ‘GH, Gk, each perpendicular to eF (L. 11.}, in the planes AF, ED of the propofed parallels : Then fince the right line EF is perpendicular to the two right lines GH, GK, at their point of interfection G, it will alfo be perpendicular to the plane aGK which paffes through thofe lines (VIIL. 3.) And becaufe the lines AB, EF are parallel to each other (by Hyp.), and one of them, EF, is perpendicular to the plane HGK, the other, aB, will alfo be perpendicular to that plane (VIL 5.) §4 1 And, 204 ' ELEMENTS OF GEOMETRY. And, in like manner, it may be proved that the line - ¢p is alfo perpendiculaf to the plane HGk. s But when two right lines are perpendicular ‘to the ‘fame plane, they are parallel to each other (VII. 4.); whence the line aB is parallel to the lme CD, as was to | be fhewn. | ” PR OP, VII. Tueorgnm. If two right lines that meet each other, be parallel to two other right lines that meect each other, though not in the fame plane with them, the angles contained by thofe lines will be equal. b : ‘ E D E B A C Let the two right lines AB, Bc, which meet each other in the point B, be parallel to the two right lines pE, &F - which meet each other in the point E 3 then will the angle ABC be equal to the angle DEF. For make BA, BC, ED, EF all equal to each other (4¢3 35 and Jom AD, CF, BE, AC and DF. "T'hen, becaufe BA is equal and parallel to Ep (by Hyp. ), AD will be equal and parallel to e (1. 29.) ’ And, for the fame reafon, CF will alfo be equal and pa- :Uls..,] {0 BE. ; ¥ But lines Wthh are equal and parallel to the fame line, though not m the fame plane with it, are equal and | : parallel O B - - - 49 BOOK T HE SEVENTHN. 205 parallel to each other (I. 26. and VII. 6.); whence AD is equal and parallel to cr. | “And fince lines which join the correfpondmg extremes of two equal and parallel lines are alfo equal and parallel (L. 29.), ac will be equal and parallel to DR | The three fides of the triangle ABc are, therefore, equal to the three fides of the triangle prF, each to each; whence the angle ABc is equal to the angle per (L. 7.), ~as was to be thewn. @ PROP VI Prosngeu To draw a right line perpendicular to a given plane, from a given point in the plane. O \H N /: B 4 A Fl f’;( 5 re Let A be the giv‘en point, and Bc the given plane; it is required to draw a right line from the point a that fhall be perpendicular to the plane Bc. | Take any point E above the plane Bc, and join EA and throughf A draw AF, in the plane Bc, at right angles with ga (I. 11.); then if EA be alfo at right angles with any other line which meets it in that plane ; the _thing required is done. | But if not, in the plane Bc, draw AG at right angles W‘ to ar (I. 11.); and in the plane kg, Wthh pafies | through the points E, A, G, make AH perpendicular to AG, 206 ELEMENTS OF GEOMETRY. aG (L. 11.), and it will alfo be perpendicular to the plane BC, as was required. /7 For, fince the right line FA is perpendlcular to each of the right lines AE, AG (by Confl.), it will alfo be perpen- dicular to the plane EG which pafles through thofe lines Vit g3:) : _‘ And becaufe a right line which is perpendxcu]ar to a plane is perpendxcrular to every right line which meets it \ _ in that plane (Def. 2.), Fa will be perpendicular to AH. But Ac is alfo perpendicular to an (4y Con/l.) ; whence Ad, being perpendicular to each of the right lines FA, AG, ‘it will alfo be perpendicular to the plane sc (VIL. 3), as was to be thewn. PROP. IX. ProBrLEM. To draw a right line perpendicular to a given plane, from a given point above it. B D C Let a be the given point, and BG the given plane ; it ‘is required to draw a right line from the point A that fhall be perpendicular to the plane BG. Take any right line Bc, in the plane BG, and draw ap perpendicular to B¢ (I, 17.) ; then if it be alfo perpen- dicular to the plane BG, the thing required is done. But if not, draw DE, in the plane BG, at right angles ~ tosc (L.11), and make AF perpendicular to pE (L. 12.) ; ' then BOOK ' THE SEVENTH, 207 then will A¥ be perpendicular to the plane BG, as was required. For, through the point ¥, draw the line rc parallel to the line Bc (L. 27.) | | Then fince the right line Bc is perpendicular to each of the right lines pa, DE, it will alfo be perpendicular to the plane which pafles through thofe lines (VII. 3.) And becaufe the lines Bc, HG are parallel to each other, and one of them, Bc, is perpendicular to the plane ADF, the other, G, will alfo be perpendlcular to that plane (VIL 5.) | : But if a line be perpendicular to a plane it will be per- pendicular to all the lines which meet it in that plane (VIL. Def. 2.) 3 whence the line HG is perpendicular to AF. | , e And fince the line AF is perpendicular to each of the lines HG, ED, at their point of interfetion F, it will alfo be perpendicular to the plane 8G (V1I. 3), as was to be fhewn. FRRGFE X Intorim Planes to which the fame rigﬁt line is perpendicular, are parallel to each other. D Let the right line AB be perpendicular to each of the planes ¢cp, EF; then will thofe planes be parallel to each ~ pther. (3 | For 208 ELEMENTS OF GEOMETRY. For if they be not, let them be produced till they meet each other ; and in the line cn, which is their common feCtion, take any point K ; and join KA, KE : Then, becaule the line AB is perpendicular to the plane gr (by Hyp.), it will alfo be perpendicular to the line Bk, which lies in that plane (VIL. Def. 2.) ; and the angle - aBk will be a right angle. And, for the fame reafon, the line AB, which is perpen- dicular to the plane pc (by Hyp.), will be perpendicular to the line AK ; and the angle Bax will alfo be a right angle. The angles ABK, BAK are, therefore, equal to two right angles, which is abfurd (I. 28. ); and confequently the planes can never meet, but muft be parallel to each other (VIL. Def. 6.), as was to be thewn. PROP. XI. THEOREM. If two r1ght lmes which meet each other, be parallel to two other right lines which meet each other, though not in the fame plane with them, the planes which pafs through thofe lines will be parallel. A\ Let the right }ines AB, BC, which meet each other 1n g, be parallel to the right lines DE, EF, which meet each other in E, though not in the fame plane with them ; then will the plane aBc be parallel to the plane DEF. 4 , Lty For POoOR THE SEVERTH. - 208 For through the point 8 draw BG perpendicular to the plane pFE (VIl. ¢.) ;5 and make cu parallel to DE, and 6K to eF (. 27.) Then becaufe G is at right angles with the plane DFE, it will alfo be at right angles with each of the lines cH, GK which meet it in that plane (Def. 2.) And fince gu is parallel to DE or AB (by Confil. and + VIL 6.), and BG interfeéts them, the angles BGH, GBA ~are, together, equal to two right angles (L. 23.) But the angle BGH has been thewn to be a right angle 5 whence the angle GBa is alfo a right angle ; and confe- quently ¢B is perpendicular to BA, And, in the fame manner, it may be fhewn, that GB 1s perpendxcular to BC. - The right line GB, therefore, being perpendicular to each of the right lines BA, BC, will alfo be perpendicu- lar to the plane acB through which they pafs (VIL. 3.) But planes to which the fame right line is perpendicu- lar are parallel to each other (VII. 10.); whence the plane acB is parallel to the plancf DFE, O ED, p s Ny 210 ELEMENTS OF GEOMETRY, PR O P XIL. THEOREM. If any two mmllei pl nes be cut by ano- ther plane, their (‘ommon f@&mm W1ll be Tet the tvon pn,ra}lel ;Japes AB, ¢p be cut by the 3 EGHT ; then will their common fg&xons EF, GH be pa- ¥ailc] to eachinther~ iy - | ; For if EF, GH be not pa;allel t}"*ey may be pro.. duced till ‘they meet, . either on the fide I«H, or the fide EG. Let them be pmduced on the fide FH, and meet each other in the point K. Then, fince the whole line EFk is in the plane as, or the plane produced, the point X muft be in that plane. And becaufe the whole line GHK is in the plane cp, or the plane produced the pomt K muft alfo be in that plane. Since, therefore, the point K is in each of the planes AB, CD, thofe planes, if produced wxll meet in that _ point. ¢ » , ; But the two planes are parallel to each other, by hypo- thefis ; whence they meet, and are parallel at the fame time, Wthh 1$ a abﬁlfd The TSHOORIEHTE SEVREN T, 211 - 'The lines £F, cH, therefore, do not meet on the fide FH ; and, in the fame manner, it may be proved, that ’ they do not meet on the fide EG 3 confequently they are parallel to each other. ‘ Q | 5 E p R OP. XHI.- THEOREM.- If a right line be perpendic&lartp a plane, every plane which paffes through it will alfo be perpendicular to that plane. | DoG'A R ¥ YERR AN b il > ARl Let the right line as be perpendicular to the plane ¢k 3 then will every plane which paﬁes through that line be alfo perpendicular to cx. For let Ep be any plane which pafles by the lme AB; and in this plane draw any right line 6¥ perpendicular to the common feétion ce (I. 11.) .. Then, becaufe the line AB 1s perpéndicular to the plane cx (by Hyp.), it will alfo be perpendicular to the line cE ; and the angle ABF will be a right angle (VH Df 2. And fince the angles ARF, GFB are each of them a right angle, and the lines AB, GF are in the fame plane, they will be parallel to each other (VII. 4.) Since, therefore, thefe lines are parallel to each other, and one of them, aB, is perpendicular to the plane ¢k, Py the 2¥2° " ELEMENTS OF GEOMETRY. the other, GF, will alfo be perpendicular to that plane V1L 5.) g e | Fg But one plane is perpendicular to another, when any _yight line that can be drawn in it, at right angles to the ~ common fection, is alfo at right angles to the other plane (VIL. Def. 3.) ; whence the plane Ep is perpendicular to the plane ¢k, as was to be thewn. el PRO P. XIV. THEOREM. If two planes. which cut each other, be each of them perpendicular to a third plane, their common {ection will alfo be perpendi- cular to that plane. "y /\ S i A c Let the two planes AB, cB be each of them perpendi- cular to the plane Acp; then will their common fection sD be alfo perpendicular to Acp. | | For if not, let pE be drawn in the plane aB, at right angles to the common feftion ADp ; and DF in the plane cp at right angles to the common feétion »c (I. 11.) Then becaufe the plane AB is perpendicular to the plane acp (by Hyp.), the line pE will alfo be perpendi- cular to that plane (VII. 3.) And - 4 FEOR "THE SEVENTN. 2313 And fince the plane cB-is perpendicular to the plane Acp (by Hyp.), the line oF will alfo be perpendicular to that plane (VII. 3.) But lines which are perpendicular to the fame plane are parallel to each other (VIL. 4.); whence the lines DE, DF meet, and are parallel at the fame time, which is abfurd. | , Thefe lines, therefore, are not perpendicular to the plane AcD ; and the fame may be thewn of any other line but pB; whence DB is perpendicular to AcD, as was te be fthewn. | ! P 3 BOQK 214 ELEMENTS.OF GEOMETRY. B O 0K vIm D P NTEI N . 1. A folid angle is ‘that which is made by three or more plane angles, wm,ch meet each other in the fame point. - \ 2. Similar folids, contamed by plane ﬁcrures, are fuch as have all their folid angles equal, each to each, and are bounded by the fame number of fimilar planes. a2, A prlﬁn is a folid whofe ends are parallel, equa{, and like plane figures, and its fides parallelograms, 4. A parallelepipedon is a prifm contained by fix pa- rallelograms, every oppofite two of which are equal, alike, and parallel. : 5. A refangular parallelepipedon is that whofe bound- ing planes are all re&angles, which are perpendicular to each other. : ; 6. A cube is 2 prifm, contamed by fix equal fquare fides, or faces. . | 7 A pyramid is a folid whofe bafe is any right lined plane figure, and its fides trlangles, which meet each other in a point above the bafe, called the veitex. 8. A cylinder is a folid generated by the revolution of a right line about the circumferences of two equal and parallel circles, which remain fixed. 9. The axis of a cylinder is the right line joining the centres of the two parallel circles, about which the figure is defcribed, ro. The y BOOK-THE EiIGHTIR 205 10. A cone'is a folid gencrated by the revolution of a right line about the circumference of a circle, one end of which is fixed at a point above 'the plane of that circle. 11. The axis of a cone is the right line joining the vertex, or fixed point, and the centre of the circle about which the figure is defcribed. 12. Slmllar cones and Cylmders are fuch as have their altitudes and the diameters of their bafes proportional, 13. A fphere is a folid generated by the revolution of a femi-circle about its diameter, which remains fixed. ‘14. The axis of a fphere is the right line about ‘which the femi-circle revolves ; and the centre is the fame as that of the femi-circle, | 15. The diameter of a fphere is any right line pafling ‘through the centre, and terminated both ways by the {urface. PROP 7 L pgMMaA; If from the greater of two magnitudes, there be taken more than its half; and from the remainder, InOre than its half; and fo on : there will at length remain a magm-— | tude lefs than the leaft of the propofed mag- nitudes. | U errc it o “— B D e B i Let ag and ¢ be any two magnitudes, of which B is the greater ; then; if from AB there be taken more than P 4 1ts 216 ELEMENTS OF . GEOMETRY. its half ; and from the remainder more than its half ; and fo on: there will at length remain a magnitude lefs than c. , e, For fince AB and c are each finite magnitudes, it is evident that ¢ may be taken fuch a number of times as at length to become greater than as. Let, therefore, DE be fuch a multiple of c as is greater than A8, and divide it into the parts DF, FG, GE, each equal to c. Alfo from AB take BH greater than its half; and from the remainder AH, take HEK greater than its half, and fo on, till there be as many dxvxﬁons in AB as there are in DE. ' Then becaufe DE is greater than AB, and BH; taken from AB, is greater than its half, but G, taken from DE, is not greater than its half; the remainder gp will be greater than the remainder HA. ‘And, again, becaufe Gp is greater than Ha, and HX, taken from mA, is greater than its half, but GF, taken from GD, is not greater than its half; the remainder rp will be greater than the remainder AK. But Fp is equal to ¢ by conftru&ion, whence c is greater than AK ; or, which is the fame thing, AK islefs than c, as was to be {fhewn, RR O P. o % e i G| | F. 3 :r; 7 | | AGOK THE EIGHTH. . s PR OP. IL T\H‘EORE M. Similar polygons infcribed in circles, are to each other as the fquares of the diameters of thofe circles. Let ABCDE, FGHKL be two fimilar polygons, infcribed in the circles ABD, FGK : then will the polygon ABCDE be to the polygon FGHKL as the {quare of the diameter BM is to the {quare of the diameter GN, For join the points B, E and A, M, G, Land F, N Then, becaule the polygon ABCDE is fimilar te the polygon FGHKL (&y Hyp.), the angle BAE is equal to.the angle GFL, and BA is to AE, as GF is to FL (VL Def 1.) And, fince the angle BAE, of the triangle ABE, is equal to the angle GFL, of the triangle FGL, and the fides about thofe angles are proportional, the angle aAes will alfo be equal to the angle FLg (V1. 5.) | But the angle AEB is equal to the angle AMB, and the angle FLG to the angle FNG (III. 15.), confequently the angle AMB is alfo equal to the angle FnG. | And fince thefe angles are equal to each other, and the. . angles BAM, GFN are each of them rightangles (111. 16.), the angle mBA will alfo be equal to the angle NGF (1. 28. Gor.), and BM will be to N as Ba is te g2 (VI §) But 218, ELEMENTS OF GEOMETRY. But the polygén ABCDE is to the polygon FGHKL as the fquare of BA is to the fquare of G (VI. 17.), there- fore the polygon ABCDE is alfo to the polygon FGHKL as the {quare of BM is to the {quare of Gn. Q. E. D. P RO P U A ko A polygon may be infcribedﬁ in a Circle that fhall differ from it by lefs than any afligned magnitude whatever‘ iy Let aBcp bé a c1rcle, and. s any given ‘magnitude whatever ; then may a polygon be inferibed in the cxrcle' Aecp that fhall dlﬁ'er from it by lefs than the magm-‘ tude s. For, let ac, EG be two, fquares, the one defcribed in the circle ABcp, and the other about it (IV. 6, 7.); and bife¢t the arcs aB, BC, CDs DA, in the points m, 7, r and s (III 3. ), and join Am, B, Bny, nC, Cr, rD, XD and AL Then fince the {quare Ac is half the fqﬁare ECLE 359y and the fquare £G is greater than the circle Arcp, the fquare Ac will be greater than half the circle ascp. In like manner, if tangents be drawn to the circle through the points m, #, r, 5, and parallelograms be de- {cribed BOOK THE EIGHTH. 219 fcribed upon the right lines AB, BC, cD, DA, the trian- gles AmB, BnC, CrD, DsA will each of them be half the parallelogram in which it ftands (I. g} But every fegment is lefs than the paramlscrram which A circum{cribes it ; and therefore each of the triangles AmBy BnC, CrD, DsA is greater than half the fegment of ;hg circle which contains it. " And, if each of the arcs am, mB, &c. be again dzv1ded into two equal parts, and right lines be drawn to the points of bifeCtion, the triangles thus formed, may in like manner, be fhewn to be greater than half the fegments which contain them ; and {o on continually.- Since, therefore, the circle ABcD is greater than the fpace s, and from the former there has been taken ‘morg than its half, and from the remainder more than its half; &e. there will at length remain fegments which, takeq together, fhall'be lefs than the excefs of the circle ARcp above the fpace s (VILI. 1.), as was to be {hewn. PROP. IV. THEOREM. A polygon may be circumfcribed about & circle that thall differ from it by lefs than any affigned magnitude whatever, Let ABcp be the circle, and s any given magnitude whatever ; then may a polygon be circumfcribed about 4 the 220 ELEMENTS OF GEOMETRY. the circle aBcp, that fhall differ from it by lefs than the magnitude s, _ For let the circle AECD be circumfcribed by the fquare EFGH (IV 7.}, and bifect the arcs AB, BC, €¢B, DA with the lines OE, OF; 0G and oH ; and to the points of bifeGion draw the tangents £/, mn, pr, st (IIL. 10.) Then fince #/is a tangent to the circle, and ok is drawn from the centre through the peint of contad, the angle Exf is a right angle (IIl. 12.), and Eé will be greater than kx (L. 17.) orits equal fa. But triangles of the fame altitude are to each other as their bafes (VI. 1.) ; whence the bafe EZ being greater than the bafe £a, the triangle Exé will alfo be greater than the triangle ZxAa. o And becaufe the triangle Exf | is greater than the triane gle kxa, it Wﬂl alfo be greater than half the curvelineal fpace ExA: and the fame may be thewn of any other tri- angle and the curvelineal fpace to which it belongs. In like manner, if the arcs Ax, ¥B, &c. be agéin bi- fefted, and tangents be drawn to the points of bifeétion, the triangles thus formed will be greater than half the curvelineal {paces to which they belong. Since, therefore, the excefs of the {fquare - above the circle is greater than the magnitude s, and from the former there has been taken more than its half, and from the remainder more than its half, and fo on, there will at length remain fpaces, which, taken together, fhall be lefs than the magnitude s (VILL, 1.), as was to be fhewn. 2R OP, @ BOOK THE EIGHT H. 221 £ £ RO Praan —TH'EORE‘MS. 3 Circles are to each other as the fquares of theu dxameters. Let ABcp, EFGH be two circles, and Bp, Fi’gf%heir diameters : then will the fquare of BD be to the fquare of ¥H as the circle ABcD s to the circle EFGH. i S For, if they have not this ratio, the fquare of BD will be to the fquare of rH, as the circle aBcp is to fome fpace either lefs or greater than the circle Ergum. e Firft, let it be to a fpace sT lefs than the circle Erou; and infcribe the two fimilar polygons AROPQ, ERLMN fo that the circle ErFcH may exceed the latter by lefs than it exceeds the fpace st (VIIIL. 3.) o Then,ﬁnce the circle EFGH exceeds the polygohEKLMN by lefs than it exceeds the fpace sT, the polygon exLmN will be greater than the fpace sT. | | And, becaufe fimilar polygons, inferibed in circles, are to each other as the fquares of their diameters (VIII. 5, ), the fquare of 8D will be to the fquare of Fu as the polygon AROPQ _is to the polygon EXLMN. : But the fquare of 8D 15 alfo to the fquare of Fu as the circle ABcD is to the fpace sT (&y Confl.) ; whence the circle apep will .be. to the fpace st, as the polygon AROPQ_Is to the polygon EXLMN. | Pl | \The 222 ELEMENTS OF GEOMETRY. The circle aBcD, therefore, beihg greater than the polygon AROPQ, which is contained in it; the fpace sT will alfo be greater than the polygon EKLMN. ¢ isy therefore; lefs and greater at the fame time, which is 1mpoﬁib1 e; confequently the fquare of BD is not to the fquare of FH as the circle ABcD is to any fpace lefs than the circle EFGH: ' And; in the famé rhannet, it may be demonftrated, thit the fquare of #¥ is not to the fquare of BD as the €ifele £FGH is to any fpace lefs than the circle aBco. Nor, is the fquare of BD to the {quare of FH as the €ircle ABCD is to a fpace greater than the circle EFGH, ?or, if it be pofﬁble, let it be fo to the fpace s:x, which s gxeater than the circle EFGH. Then, ﬁnce the fquare of BD is to the {quare of FH as the circle aBcD is to the fpace sx, therefore, alfo, in- veifely, the fquare of FH is to the fquare of BD as Lhe fpace §X is to the cxrcle ABED (V. #.) But the fpace sx is greater than the circle ereu (& Hyp. ) whence. the fpace sx is to the circle ABcD as the eircle EFGH is to fomc fpace lefs than the circle ABCD (V. 14.) The fquare of FH is, therefore, to the fquare of BD as the circle EFcH is to a fpace lefs than the cxrch, ABCD (V. 11.); which has been fhewn to be lmpoﬁibl bmce, therefore, the fquare.of BD is not to the fquare 6f 1 as the circle ABcD is to any fpace either lefs or preater than the circle EFGH, the fquare of ep muft be to the fquare of ¥ as the circle ABcD is to the circle EFGH: 7 ‘ ety B 1) CoRr. 1 ercles are’ to ‘each other as the fquaies of their radii thefe being half the diameters. COR'Q ~B00K THE ELGHTH. 223 ~CoRr. 2. If the radii or diameters of three circles be re- fpe&ively equal to the ‘thhﬂree fides of a. right angled trian- ~ gie, that whofe radius or diameter is the hypothenufe will be equal to the other two taken together (II 14.) PROP. V. THEOREM. Every circle is equal to the re®angle of its radius, and a right line equal to half its circumference. s g ‘b'J/'Vz Let Zmps be a circlé, and ov 2 rectangle contained under the radius o and a right line ow equal to half the circumference ; then will the circle kmps be equcu to the recta ncrle ov. | For if it be not, it muft be elther oreater or lefs. Let it be greater ; and let the rectangle oz be equal to the cxrcle kmps; and infcribe 2 polygon [z 7¢ in the circle kmps that fhall diffcr from it by lefs than the magnitude wz (VIL.3.) Then ﬁnce the triangle Z£ot is equal to half a reGtangle under the bafe ¢ and the pcrpcndxcular ox (L. 22:)y the whole polygon will be equal to half a re@angle under its perimeter and the perpendicular ox. ' And becaufe ow is greater than half the perimeter of -any polygon that can be infcribed m the circle dmps (&y 222 FLEMENTS OF GEOMETRY. (%y f{yp Y, and of is greater than ox (I. 17.), the re@- angle ov will alfo be greater than the polygon @5 S ‘But the polygon differs from the citcle, or from the reCtangle 0z, by lefs than the maghitude wz (& Conft.), and ov differs from oz by wz; confequently the poxyocn is greater than the rectangle ov. It is, therefore, both greater and lefs at the fame time, ~which is abfurd ; whence the circle /émps is not greater than the retangle ov. ' | “Again, let it be lefs than ov, by the re&anvle Wy and let BDFH be a polygon circumicribed about the circle, that fhall differ from it by lefs than the magnitude wy (VIII. 4.) | | Then fince the triangle BOA is equal to half a re€tangle ‘under the bafe BA and the perpendicular o (1. 32.), the whole polygon will be equal to half a rectangle under its perimeter and perpendicular o#. And becaufe ow is lefs than half the penmeter of any polygon that can be circumfcribed about the circle (dy Hyp.), and ok is common, the reltangle ov will alfo be lefs than the polygon BDFH. But the polygon differs from the circle, or from oy, by lefs than the miagnitude wy (&y Hyp.), and ov differs from oy by wy ; confequently the re&angle ov will be ‘vreate‘r than the polygon BDFH, which is abfurd. Since, therefore, the rectangle ov is neither greater rior lefs than the circle 2m ps, it muft be equal to it, as - was to be thewn., PROPR . BOOK THE EIGHTH. 238 PROP; VH. THEQR.&:M: The circumferences of circles are in pro= portion to each other as their diameters. o N B 1 L R A B‘V\O ; g FQ Hl " ‘? SR @ : Let aArcp, EFcH be any two circles, whofe diameters I€ BD, FH ; then will the circumference aBcp be to the circumference EFGH as the diameter BD is to the diame= ter FH. : | For let om, sp be two right lines equal to the femi- circumferences DAB, HEF, and on the radj; OA, SE make the {quares ok, st (IL. 1.), and complete the rectangles ON, SR : | - Then fince the re&angles on, sr are equal to the cir- cles arcp, EFGH (VIIL 6.), and the circles are to each other as the fquares of their radii (VIIL. 5. Gor.) the rectangle on will alfo be to the fquare ok as the re@an~ gle sRr is to the fquare sL (V. g.) But ox is to ox as omto op (VI 1.), and sk to SL as sP to sH (VI 1.); therefore, by equality, om will be to op as sp is to su (V. i1 And becaufe any equimultiples of four proportional quantities, are alfo proportional (V. 13.), twice om will be to twice oD as twice sp is to twice sH ; or, by alter- nation, twice oM is to twice sp as twice ob is to twice SH. : , S . But 226 ELEMENTS OF GEOMETRY. ~ But twice om and twice sp are equal to the circum=- ferences ABCD, EFGH (by Confl.) ; and twice oD and twice sH are equal to the diameters BD, FH ; whence the circum- ference ABCD is to the circumference EFGH as the diame- ter BD is to the diameter FH. Q. 5, b, PROP VL. TnrEorREM. If a prifm be cut by a plane parallel to its bafe, the fection will be equal and like the bafe. | | A B Let A be a prifm, and KLMN a plane parallel to the bafe ABcD ; then will kLMN be equal and like ABCD. ~ For join the points NL, and DB : | Then fince kM, Ac are parallel planes (2y Hyp.), and the plane AN cuts them, the fection KN Willbe parallel to the fetion ap (VIL 12.) | And fince AK is alfo parallel to pn (VIIL Déf. 3.), the figure AN is 2 parallelogram ; and confequently KN is ~equal to ap (L 30.) | [ . In like manner it may alfo be fhewn, that KL is equal to AB, LM t0 BC, and MN to CD. ‘And fince kXN, KL in the plane KM, are parallel to AD, AB in the plane AC, the angle Nk1 will be equal to the angle DAB (Vil. 7.) The o D Y BOOK: - T"HRE EIGHNTI: 24% The two fides kN, XL of the triangle kLN, being, therefore, equal to the two fides Ap, A of the triangle aBD, and the angle NK 1 to the angle paB, the triangle KLN will be equal and like the triangle asp (1. 4.) And in the fame manner it may be thewn, that the triangle LMN is equal and like to the triangle BcD. But the triangles kLN, LMN are, together, equal to the fection KLMN; and the triangles ABD, BcD to the fection ABcp ; whence the fe@tien kLMN is equal and like to the fection aBCD. ‘ Q;, E. DO FPROP IX, TuHroREM Prifms of equal bafes and altitudes are equal to each other. N T ; 7 M T/\/S Let am, Es be any two prifms, ftanding upon the equal bafes ABCD, EFGH, and having equal altitudes; then will Am be equal to Es. . For parallel to the bafes, and at equal dxf’cances from them, draw the planes mp and vw. Then, by the laft propoﬁtlon, the fection mnps will be equal to the bafe ABcD, and the fection wowr to the bafe EFGH, Q.2 _ But 208 ELEMENTS OF GEOMETRY. But the bafe ABcb is equal to the bafe EFGH by hypo- thefis ; whence the feGtion mmps is, alfo, equal to the feCtion vowr. o And in the fame manner it may be fhewn, that any other fections, at equal diftances from the bafes, are equal to each other. | Since therefore every fe&xon in the prifm AM is equal to its correfpondmo fection in the prifm £s, the prifms - themfelves, which are compofed of thofe feGtions, muft alfo be equal. Q. ks B Cor. Every prifm is equal to a re&angular parallele- pipedon of an equal bafe and altitude. PR OP X, THEDREM, Re&angular parallelepipedons, of equal al- titudes, are to each other as their bafes. Let ac, MP be two reGtangular parallelepipedons, hav- ing the equal altitudes ED, QR; then will Ac be to MmP as the bafe BE is to the bafe NQ. For in AB, produced take any number of nght lines AF, FL each equal to AB; and in MN, produced, take any number of right lines NT, TX each equal to MN. Complete the parallelograms rE, FK, MV, TZ, and make the upright folids AG, FH, NW, TYX of equal alti- tudes with Ac or mp. 4 ‘ Then, BOOK THE FIcHEH. 229 Then, becaufe AF, F1 are each equal to AB, and NT, TX are each equal to MN (4y Ca;yi )» the parallelograms FE, FK will be each equal to BE, and the parallelograms NV, 77 to no (1l 2.)- And, fince the folids AG, Fu have equal bafes and al- titudes with the folid ac, they will be each equal to Ac (VIIL q.) 5 and, for the fame reafon, the folids Nw, 1Y, will be each equal to Nr. Whatever multiple, therefore, the bafe Bx is of the bafe BE, the fame multiple will the folid Bu be of the folid Ac; and, for the fame reafon, whatever multiple the bafe Mz is of the bafe nq, the fame multiple will the folid MY be of the folid Nx. If, therefore, the bafe Bk be equal to the bafe MZ, the folid B will be equal to the folid My : ; and if greater, greater 5 and if lefs, lefs ; whence the bafe BE is to the bafe nQ, as the folid Ac is to the folid N® (V. Defs 50) $ E L. Cor. From this demonftration, and the Cor. to the laft Prop. it appears that all prifms of equal altitudes, are to each other as their bafes ; ; every prifm being equal to a reCtangular parallelepipedon of an equal bafe and alti- tude, Q3 e B R P, 230 ELEMENTS OF GEOMETRY, PROP. XI. THEOREM. Reftangular parallelepipedons of equal bafes are to each other as their altitudes. ' K | } M/~ | b 'Vfl—— » ) S 2 A 2 i i # Jiis H & 1} e —— C Gl 7 A & A X5 I3 Let Ak, EP be two re&tangular parallelepipedons {tand- ing on the equal bafes AC, EG; then will Ak be to EP as the altitude AM is'to the altitude Es. For let Aw be a reCtangular paralleleplpedon on the bafe Ac, whofe altitude Av is equal to the altitude Es of the parallelepipedon EP : ' Then, fince the bafe Ac is equal to the bafe EG (by Hyp.), and the altitude Av 1s equal to the altitude Es (&y Confl.), the folid Aw will be equal to the folid ep (VIIL. g.) And if AL, Ay be confidered as bafes, the folid AK will be to the folid aw as the bafe AL 1s to the bafe ay (VIII. 10.) But the bafe AL is to the bafe Ay as the fide Am is to the fide Av '(VI. 1‘.'); whence by‘ equality the folid AK will be to the folid Aw as the altitude AM is to the alti- tude av (V. 11.) o ‘ Smce, therefore, the folid Aw is equal to the folid EP, and the altitude Av to the altitude Es, the folid AK w111 alfo be to the folid Ep as am is to Es (V. g.) o SOk, BOOKlTHE EIGHT H. 291 CoRr. From the reafon given in the Cor. to the laft Prop. it follows, that all prifms of equal bafes, are to cach other as their altitudes. PROP, XII. THEOREM, The bafes and altitudes of equal rectangu- lar parallelepipedons are reciprocally propor- tional ; and if the bafes and altitudes be re- ciprocally proportional, the paralleleplpedons \ will be equal. ] I Let the reCtangular parallelepipedon AR be equal to the reCtangular parallelepipedon v ; then will the bafe AC be to the bafe EG, as the altitude Eo is to the alti- tude Aw. | | For let ar be a reQangular parallelepipedon on the bafe AcC, whofe altitude AP is equal to Eo, the altitude of the parallelepipedon Ev. ' Then fince the altitudes AP, Eo are equal to each other (%y Gonft.), the folid AL will be to the folid Ey as the bafe Ac is to the bafe ec (VIII. 10.) And becaufe the folid AR is equal to the fohd EY (¢y Hyp.), the folid AL will be to the folid AR as ac is to EG (V 9-) Q4 But 232 ELEMENTS OF GEOMETRY. But the folid AL is to the folid AR as AP is to Aw (VIII. 11.) ; whence, alfo, Acisto EG as AP is t0 AW MV . 11.), or ACto BG B BO 1O AW. _, Again, let Ac beto EG as Eo is to Aw ; then will AR be equal to EY. ' For, fince AL is to EY as ac to G (VIIL 10. ), and AC to EG as EO to AW (& Hyp ), AL will be to EY as £o0 to Aw (V. 11, ) But Eo, or ap, isto Aw as AL is to AR (VIIL 11.); therefore AL will be to EY as AL is to AR (V. 11.) And fince the antecedents are equal, the confequents will alfo be equal ; whence the folid AR is equal to the folid ey, as was to be fthewn. Cor. The fame proportion will hold of prlfms in gene- ral ; thefe being equal to rectangular parallelepipedons of ~ equal bafes and altitudes, : i P ROP. XIII THEOREM. Similar 1eé’cangular paralleleplped@ns are to cach other as the cubes of their like fides. G 13 L4 P . T s e 1 R by é 2 By V V- Y X , Z. D (o N 7%1 . e ¥ SR B B 4 1 1 Let ar, x? be two fimilar re&angular parallelepipe- dons, whofe like fides are ABy KL then w1ll AF be to KP as the cube of AB is to the cube of kL. ‘For let AT, kw be two cubes &andmg on AX, KZ, the fquares of the {ides AB, KL. Then )y ROQQK THE EIGHTH, 273 Then fince parallelepipedons on the fame bafe are to ~ each other as their altitudes (VIII. 11.), aF will be to An as AH to Av, or AB; and KP to Ks-as KR to KY, or KL. But the planes ABEH, KLOR being fimilar (VIII. Def. 2.)s AH will be to a8 as kR is to KL (VL. Defi1.) 5 whence AF is to azas KP to Ks (V. 11.); or AF to KP as Az to Ks (V. 15.) Again, fince parallelepipedons of the fame altitude are to each other as their bafes (VIIL 10.), AT will be to An as AX to AC; and KW to Ksas KZ to KM. And becaufe Ax, or the {quare of AB, is to Ac, as Kz, or the fquare of k1, is to km (VI, 17. ); ar will beto Azas kwisto ks (V.11.); ofr AT to Kw as A?t to ks (Vors:) But AF has been fhewn to be to kP as Az isto Ks3 therefore, alfo, AF is to KP as AT to KW (V.11.) W Ea D. Cor. 1. Similar reGtangular parallelepipedons are to each other as the cubes of their altitudes; thefe being confidered as like fides of the folids, Cor. 2. Every prifm being equal to a parallelepipedon of an equal bafe and altitude (VIII. g. Cor.), all fimilar prifms will be to each other as the cubes of their altitudes, or like fides. . PR OP. 234 ELEMENTS OF GEOMETRY. PROP. XIV. TuErorEM. If a pyramid be cut by a plane parallel to its bafe, the feGion will be to the bafe as the fquares of their diftances from the vericx., ; Let EpABC be a pyramid, and 0 a feGion parallel to the bafe Ac; then will 70 be to Ac as the fquares of their diftances from the vertex. For draw Es perpendicular to the plane of the bafe Ac (VII. g.) ; and join Ds and pr. Then, fince mp, mn are parallel to an, ag (VII. 12), the angle pmn will be equal to the angle pas (VIL. 7. ), and pm will be to DA as Em to EA, or as mn to AB (V1. 3.) For a like reafon each of the angles in the fection mo are equal to their correfponding angles in the bafe ac, and the fides about them are proportional ; whence mo is fimilar to ac (VI. Def. 1.) And becaufe pm is parallel to pa, and pr to ps (VIL, I;.), pm will be to DA as Ep to ED, or as Er to ES (VL. 3) _ @ e ‘ - The it b oS £ i i e R | G P = s g 1 & - =z 2 i & 3 i v BOOK THE EIGHTH. 2135 The lines pm, DA, Er and Es being, therefore, pro~ portional, the fquare of pm will be to the {quare of DA, as the fquare of Er is to the fquare of Es (V1. 19. Cor.) But the fquare of pm is to the {quare of DA as mo 1s to Ac (V1. 17.) 3 whence the fquare of Er is to the fquare of Es as mo is to ac (V. 11.) L g Cor. If a pyramid be cut by a plane parallel to its bafe, the feGtion will be fimilar to the bafe. | P RO P XY, 1 HrORE M. Pyramids of equal bafes and altitudes are equal to each other. //f | \‘\\\ / 4 \'\\\ LTS Gk N i ///Z/i/ﬁ y /’ﬁ“r ? / \ B o B R T o \ / A KL\:_.?___%) \\u \ / Yo & g A B ¥ G Let EDABC, LKFGH be any two pyramids, of which the bafe ac is equal to the bafe ¥H, and the altitude s to the altitude LP; then will EDABC be equal to LKFGH. | For make Er equal to Lo ; and draw the {ections mn, vw, pafallel to the bafes Ac, FH. | Then, by the laft propofition, the {quare of Er is to the fquare of Es as mn isto AC ; and the {quare of Lo to the fquare of Lp as vw is to FH. And fince the fquare of £ is equal to the {quare of Lo (Conf. and 11 2.) ; and the fquarc of Es to the fquare , i 2356 = ELEMENTS OF GEOMETRY. of Lp (Hyp. and 11, 2.), mn will be to ac as vw is to ¥ (V.9.) 0 But Ac is equal to FH, by h)potheﬁ s whence mn is, alfo, equal to vw (V. 10.) | And, in the fame manner, it may be {hewn, that any other feGtions, at equal diftances from the vertices, are equal to each other. | Since, therefore, every fection in the pyramid EpaBC is equal to its correfponding fection in the pyramid LK FGH, the pyramids themfelves, which are compofed of thofe feCtions, muft alfo be equal Q. E. D, YR U R Tk Every pyramid of a triangular bafe, is the third part of a prifm of the fame bafe and, altitude. A St \T:a B Let pasc be a pyramid, and FDABE a prifm, {tanding upon the fame bafe ABC, and having the fame altitude ; then will paBc be a third of FDABE. | For in the planes of the three fides of the prifm, draw the diagonals nB, pc and CE: Then Lo s dtan LR Gl S gt B PP SRS i S I LR T A S s N A .?2@%‘9:‘%"«"" s e B b A B o R BOOK THE EIGHTH. 239 - Then becaufe pz divides the parallelogram AE into two equal parts, the pyramid whofe bafe is ABD, and vertex c, is equal to the pyramid whofe bafe is BED and ver- tex ¢ (Vilhis) | And fince the oppofite ends of the prlfm are €qu'ﬁ to each other (VIII. Def. 3.), the pyramid whofe bafe is ABc and vertex D, is equal to the pyramid whofe bafe is DEF and vertex ¢ (VIIIL 15.) | But the pyramid whofe bafe is Arc and vertex b, is equal to the pyramid whofe bafe is ABD and vertex ¢, be- ing both contained by the fame planes. ' The three pyramids pABc, CBED and CEFD are, there- fore, all equal to each other ; and confequently the prifm FDABE, which is compofed of them, is triple the py- ramid DABC, as was to be thewn. Cor. Every pyramid is the third part of a prifm of the fame bafe and altitude ; fince the bafe of the prifm,- what- ever be its figure, may be divided into triangles, and the whole folid into triangular prifms, and pyramids. ~ ScHoLiuM. Whatever has been demonftrated of the proportionality of prifms, holds equally true of pyramids; the former being always triple the latter. PROF. 238 ELEMENTS OF GEOMETRY. Pl 0D M0l Haronsn If a cylinder be cut by a plane parallel to its bafe, the fection will be a circle, equal to the bafe. | \ B Let AF be a cylinder, and GHK a fettion parallel to its bafe ABc ; then will cHK be a circle, equal to azc. For let the planes NE, NF pafs through the axis of the ~cylinder LN, and mect the feCtion GHK in M, H and K. Then, fince the circle DEF is equal and parallel to the circle asc (VIIL. Def. 8.), the radii Lr, LE will be equal and parallel to the radii nc, ~z (1L 5. and VIL 12.) A And becaufe lines which join the correfponding ex- tremes of equal and 'parallel lines are themfelves parallel (I.zg.), Fc, EB will be parallel to LN ; or KC, HB to MN. o In like manner, fince the circle uk is parallel to the circle ABc (&y Hyp.), Mk, MH will be parallel to NC, NB. " Pl And, becaufe the oppefite fides of parallelograms are equaf (I. 30.), Mk will be equal to Nc, and MH to NB. But S S v o & ¢ BOOK THE EIGHTH., 239 But Nc, NB are equal to each other, being radii of - the fame circle ; whence Mk, My are alfo equal to each other. And the fame may be fhewn of any other hnes, drawn from the point M, to the circumference of the fection GHK ; canfequently GHK is a circle, and equal to ABC, as was to be thewn, PROP. XVII. THEOREM. Every cylinder is equai to a prifm. of an equal bafe and altitude. kg SRR ke U e A ‘KZ £ Let An be a cylinder, and pM a prifm, ftanding upon equal bafes AcB, DEF, and having equal altitudes ; then will AH be equal to pm. For parallel to the bafes, and at equal diftances from them, draw the planes onm, and vrw. Then, by the laft Prop and Prop. 8, the fection onm is equal to the bafe acs, and the fection vrw to xhc bafe DEF. | But the bafe AcB is equal to the bafe DEF, by hypo- thefis ; whence the fection onm is alfo equal to the fec-: tion vrw, | And, 240 TLEMENTS OF GEOMETRY. And, in the fame manner, it may be fhewn, that any other fections, at equal diftances from the bafe, are equal to each other. ‘ | | Since, therefore, every feCtion of the cylinder is equal to its correfpondent feGion in the prifm, the folids them. felves, which are compofed of thofe fe@ions, muft alfo be equal. ‘ _ ‘ 1 D, ScuoriuM. Whatever has been demontftrated of the proportionality of prifms, holds equally true of cylinders ; the former being equal to the latter. PR O P, XIX THEbR'EM. If a cone be cut by a plane parallel to its bafe, the feCtion will be to‘the bafe as the {quares of their diftances from the vertex, Let pasc be a cone, and #mp a feQtion parallel to the bafe aBc; then will nmp be to ABC as the fquares of their diftances from the vertex. For draw the perpendicular pr; and let the planes cDP, BDP pafs through the axis of the cone, and meet the fetion in o, p, and m, | Then BOOK THE EIGHTH, 241 Then fince the fe&ion #mp is parallel to the bafe apc (by Hyp.), and the planes Bo, co cut them, op will be parallel to pc,y and om to PB (VII, 12.) _ And becaufe the triangles formed by thefe lines are equiangular, om will be to PB as Do to DP, or as op to Pe-(Vi g ‘ But pr is equal to pc, being radii of the fame circle; wherefore om will alfo be equal to op (V. 10.) ~ And the fame may be thewn of any other lines drawn from the point o to the circumference of the fe@ion nmp ; whence nmp is a circle. _ Again, by fimilar triangles, Ds is to Dr as Do to DP, or as om to PB; whence the fquare of s is to the fquare of pr as the fquare of om is to the fquare of g (VI. 19.) But the fquare of om is to the fquare of PB as the circle nmp is to the circle asc (VIIIL. 5.); therefore the fquare of s is to the fquare of D7 as the circle nmp is to the circle asc (V.11.) kD Cor. If a cone be cut by a plane parallel to its bafe the fection will be a circle. R PR QPR ; 242 ELEMENTS OF GEOMETRY. CRERIDY P, XX, T'HEOREM. Every cone is equal to a pyramid of an equal bafe and altitude. 1et pAnc bé a cone, and KEFGH a pytamid; ftanding upon equal bafes ABC, EFGH, and having equal altitudes pP, ks; then will paBC be equal to KEFGH. For parallel to the bafes; and at ¢qual diftances Doy k# from the vertices, draw the planes zmp and vw. Then, by the laft Prop. and Prop. 13; the' fquare of Do is to the fquare of D as #mp is to ABC 5 and the {quare of xr to the fquare of ks as vw to EG. ' And fince the fquares of Do, DP are equal to ) the fqumes of kr, ks (Confl. and 1. 2.); nmp is to ABC as vw is k0 tc (Vi) But ABC is equal to EG, by hypothefis ; wherefore nmp 1S, alfo, eqml to vw (V. 10.) And, in the fame manner, it may be fhewn, that any other fe&ions, at equal diftances from the vertices, are equal to each other. Since, therefore, every {eQion in the cone is equal to b 3 1 its correfponding feftion in the pyramid, the folids pABC, KEFGH of which they are compofed, muft be equal. . D, PR OP, bl e & = A 4 # & tBOOK "PHD BEICQOHTH. 243 PR OP XXL. TuaurorneEm Every cone is thc thlrd part of a cylmdet of the fame bafe and altitude. D BT Let £ar be a cone, and DABc a cylinder, of the fame bafe and altitude ; then will EaB be a third of DazC. For let kxFG, KFGH be a pyramid and prifm, having an equal bafe and altitude with the cone and cylinder. Then fince cylinders and prifms of equal bafes and al- titudes are equal to each other (VIII. 18.), the cylinder pagc will be equal to the prifm Kk FGH, And, becaufe cones and pyramids of equ: J bafes and altitudes are equal to each other (VIIL 20.), the cone eAB will be equal to the pyramid xkrG. - But the pyramid xFg is a third part of the prifm K:.FGH (VIIL 16.), wherefore the cone EAB is, alfo, a third part of the cylinder bABc. | Q. E. D. - Scrorrum 1. Whatever has been demonftrated of the proportibnality of pyramids, prifms, or cylinders, holds equally true of cones, thefe being a third of the latter, 2. 1t is alfo to be obferved, that fimilar cones and cy- linders are to each other as the cubes of their altitudeés, R 2 or 244 ELEMENTS OF GEOMETRY. or the diameters of their bales ; the term like fides being here inapplicable. PROP. XXII. TrroREM. If a {phere be cut by a plane the fection erl be a circle. Let the fphere EBD be cut by the plane Bsp; then will BsD be a circle. | | For let the planes ABC, Asc pafs through the axis of the fphere Ec, and be perpendicular to the plane BsD. Alfo draw the line 8p; and join the points A, D and Fg 8 Then fince each of thefe planes are perpendicular to the plane BsD, theit common fetion Ar will allo be per- pendicular to that plane (VII. 14.) And, becaufe the fides aB, Ar, of the triangle Amr, are equal to the fides A5, Ar of the triangle Asr, and the ‘angles ArB, Ars are right angles, the fide »2 will be equal to the fide s (1. 4.) | In like manner, the fides AD, ar, of the triangle ADr, being equal to the fides As, ar, of the triangle Asr, and the angles ArD, ars right angles, the fide rp will alfo be equal to the fide s (1. 4.) The lines #B, b and rs, are, therefore, all equal ; and the fame may be thewn of any other lines, drawn from the point Le R AT T S I e S R SRR RS SR S e e w,’?‘-’fx,,ﬁ;_?vv‘{} el e R e e N T e BOOK 'THE FiOUER. - 248 point # to the circumference of the fection ; whence Bsp is a circle, as was to be thewn. . Cor. The centre of every fection of a {phere is always in 2 diameter of the fphere, ' ER O P XXIEl TREOREM Every {phere is two thirds of its circum- {cribing cylinder. Let 7Esm be a {phere, and DABC“itS circumf{cribing cylinder ; then will 7Esm be two thirds of pasc. For let Ac be a fection of the fphere through its centre F ; and parallel to bc, or AB, the bafe of the cylinder, draw the plane LH, cutting the former in z and 7 ; and join FE, Fn, FD and Fr, Then, if the {quare Er be conceived to revolve round the fixed axis ¥r, it will generate the cylinder Ec ; the quadrant FEr w111 alfo generate the hemifphere EmrE; and the triangle FDr the cone FDC. And fince FHz is a right angled triangle, and Fu is equal to Hm, the {quares of FH, Hn, or of Hm, Hu, are equal to the fquare of Fuz. : But Fz is alfo equal to FE or HL ; . whence the fquares f of Hm, H#n are equal to the {quare of HL : or the circular R 3 fections 246 FLEMENTS OF GEGMETRY. feQions whofe radii are Bm, Hz are equal to the circular fe@ion whofe radius is mr (VIII. 5. Cor.) And as this is always the cafe, in every parallel pofition of HL, the cone FDC and cylinder Ec, which are com- pofed of the former of thefe fe&ions; are equal to the hemifphere EMrE, which is compofed of the latter. , ~ But the cone Fnc is a third part of the cylinder Ec (VIIL. 21.); whence the hemifphere EmrE is equal to ‘the remaining two thirds; or the whole {phere 7Esm to two thirds of the whole cylinder nABC, as was to be fhewn. | ' i Cor. 1. A cone, hemifphere, and cylinder, of the fame bafe and altitude, are to each other as the num- bers 1, 2 and 3. - - Cor. 2. All {pheres are to each other as the cubes of | e e Coar ] their diameters ; thefe being like parts of their circum- fcribing cylinders, e e S B R Lw 1 | NOTES NOTES asxp OBSERVATIONS PDEF. 1.. Boord: THE definition of a {olid, contrary to the ufual me- thod, is here made the firft of the firft Book; as thofe of a point, line and fuperficies are 21l derived from it, and cannot be underftood without it. + ucLip feems to have placed it in the eleventh book of the Elements, for the fake of uniformity; but arrangements of this kind, which are merely arbitrary, are but of little confequence, and fhould therefore always bc made to give place to .- perfpicuity and the natural order of things. | D EF. 2, 3 4 Booxk L Thefe definitions are. now, by means of the former, rendered perfe@ly clear and intelligible, fo that any far- there lucidation of them is altogether unneceflary. Dg, o1MsoN has endeavoured to fhew, by a formal proof, drawn from the confideration of a folid, thata point, ac- cording to EucrLip’s definition, is without parts, a line “without breadth, and a furface without thicknefs ; but this, and all.other demonftrations of the fame kind, are unfcientific and fuperfluous ; for thefe properties are fo Ouv'ouﬂy eflential to the things defined, that they cannot, even in idea, be feparated from them. If a point had parts, it would be a line; if a line had breadth it would be afuperficies ; and if a fuperficies had thicknefs, 1t would "be a folid ; which are all manifeft contradi¢tions. It is, belides, a {ure fign that a definition 1s badly exprefled, ' R 4 when 248 | )JO’TFZS AND when it requires a number of prohx arguments to efta- blifh its truth and propriety. - DEr.g Boox I Evucrip’s definition of a right line is not exprefled in fo accurate and fcientific 2 manner as could be wifhed ; the lying evenly between its extreme points, 1S too vague and indefinite a term to be ufed in a fcience fo much celebrated for its ftri&nefs and fimplicity as Geometry. ARCHIMEDES defines it to be the fhorteft diftance between any two points ; but this is equally exceptionable, on ac- count of the uncertain fignification of the word diftance, which, in common lan%age, admits of various mean- ings. That which 1s here given, is, perhaps, not much preferable to either of thefe. The term right, or ftraight line, is, indeed, fo common and fimple, that it feems to convey its own meaning, in a more clear and fatisfactory manner, than any explanation which can be given of it. ‘Dr. AusTin, in his' Examination of the firft fix books of the Elements, propofes a fingular emendation of this de- finition, which includes the confideration of right lines, inftead of a right line, as the cale manifeftly requires. Der. 6. Boox L Some call a plane fuperficies that which is the leaft of all thofe having the fame bounds ; and others, that which is generated by the motion of a right line, not moving " in the direéion of itfelf; but thefe definitions are too complex and obfcure to anfwer the purpofe required. EvcLip defines it to be that which /lies evenly between its lines ; which is liable to the fame exceptions as that given of a right line : nor is the one which has been fubftituted in OBSERVATIONS, 249 in the place of this, by Dr. Simson, and other Editors, fo fimple and perfpicuous as could be wifhed. Nothing is gained by the explanation of a term, if the words in which it is exprefled are equally, or more, ambiguous, than the term itfelf : for this reafon, that which is here given, has been preferred to either of thofe abovemen- ‘ tioned ; though, perhaps, it may not be equally com- modious 1n certain cafes. It is alfo to be remarked, that Eucrip never defines one thing by the intervention of another, as is the cafe in Dr. Simson’s emendation; fo that if this method had occured to him, he would certainly have rejected it. ~ Der.7. Boox i The general definition of an angle in EucLip, has been properly objeéted to, by feveral of the modern Editors, as being unneceflary, and conveying no diftinct meaning ; and in Dr. SiMmson’s emendation of the ninth, there feems to be fill a fuperfluous condition. He defines a recti- lineal angle, to be ¢ the inclination of two ftraight lines to one another, which meet together, but are not in the fame ﬁraight line .”> Now their not being in the fame ftraight line, is a neceflary confequence, obvioufly in- cluded in their having an inclination to each other ; and, therefore, to make this an eflential part of the defini- tion, is certainly improper, and unfcientific. S EF..8, 0o . BOO R, Evcurip includes a right angle and a perpendicular in the fame definition, which appears to be immethodical, and contrary to his ufual cuftom. ‘T'hey are certainly diftin&t things, though dependent upon each other, and have 3 250 N0 7 ES AND have as much claim to be feparatley deﬁned, as 2 circle and its diameter. , " Der.13. Beox 1. The _definition of a circle from its generation, has been thought by Dr. BARROW and others, to be preferable to Eucrip’s, or the one here given; as it is {uppofed to furnifh its propertics more readily, a’}d to have tje {till farther advantage of fhewing the aClual exiftence of fuch a figure, mdepende it of any hypothefis, but that of granting the pofiibility of motion. = But the requifition of this poftulatum, appears to be a fufficient reafon why EvcLip rejetted {uch a definition. The principles of pure Geometry, have no dependence upon motion, and it is, therefore, never ufed in the ’Elcrﬁents, but in two or three places of the eleventh book, where it could not, without much obfcurity and circumlocution, have been eafily avoided. It is, befides, neither fo {imple, nor con+ venient to refer to, as Eucrin’s; which, in thele refpecls, is as commodious as could be withed. ' 1JEF. 208. Book 1l ~ Dr.Barrow, and other writers of confiderable emis nence, have cenfured Evcrip for defining parallel lines, from the negative property of their never meeting each A e i S e S e i A A other ; and to this they attribute all the perplexity and confufion, which has hitherto attended this delicate fub- ject: affirming it as an utter impoffibility, that any of the properties of thefe lines, can be derived from a defini- tion which contains only a fimple negation. But thefe aflertions appear to be groundlefs ; for the definition is founded on one of the moft familiar, fimple and obvious propertics OBSERVATIONS. 251 properties of parallel lines, which either reafon or fcience can difcover: ‘and, on this account, it is certainly pre- ferable to any other that could have been formed from more abftrufe and complicated affeGtions of thofe lines, . how ready and ufeful foever fuch a definition might have been found in its application. The affertion, likewife, that none of the other pro- perties of parallel lines can be derived from this definition, has been unadvifedly made; for the 27th Prop. of the firft Element, which is the fame as the 22d of the prefent performance, is fairly and eleor:mtly de- monfirated by it; and by means fomething fimilar to thofe made ufe of bV Dgr. Stmsow, in his Notes upon the 29th Prop. it would not be difficult to fhew that all the other properties of thofe lines may be derived from this definition, without the afliftance of the 12th axiom, or any other of the fame kind, . Dr. Simsow, indeed, in his attempt to demonftrate this axiom; has made feveral paralogifms which render his reafonings altogether in- valid, and nugatory. Paffing by others, of lefs confe- quence, it will be {ufficient to obferve, that in his fifth Prop. he takes it for granted, that a line, which is per. - pendicular to one of two parallel lines, may be pro- duced till*it meets the other: now this is a particular cafe of the very thing he is endeavourmw to prove, which 1s fo ftrange an overfight, that;l.t is remarkable how it could efcape his obfervation. i This, however, is not the only inftance of an unfuc- cefsful attempt to prove the truth of the 12th axiom; for Cravius and others have committed {imilar mlf.. takes, and Dr. AusTiN, who has endeavoured to de- monftrate it by means of a new definition of parallel lines, 252 N Q T R'5 AND lines, has made ufe of an affumption equally unwar- rantable with that mentioned above. That the theory of parallel lines, as it is given in the Elements, is very imperfeét, cannot be denied; but no one has yet been fubftituted in its place which is not equally defective ; and in fome inftances ftill more exceptionable: par- ticularly as they are founded on a definition which is de- rived from 'an adventitious property of thofe lines, inftead of one which is inherent and neceffary, as the nature of - the {fubject requires. Whether Eucrip was the author of this axiom cannot perhaps, at this time, be eafily determined ; butit is cer- tainly a difgrace to the Elements. The truth of the pro- perty here affumed as a thing to be granted, is fo far from being obvious, that it requires demonftration as much as any Prop’. in the Elements ; and it is always obferved that learners, inftead of giving that ready aflent to it which an axiomatical principle requires, receive it with doubt and hefitation,” and are fcarcely able to comprehend the mean- ing of it. The one which is here made the 4th poftu- late, though nearly the fame thing in effedt, is much | ‘more clear and intelligible. \, " Prorp. 1. Book L : In the demonftration of this propofition, by Evcrip, that part which relates to the interfection of the circles is, very improperly, omitted; for in a work of this kind, nothing, however evident, ought to be taken for grant- ed ; and particularly at the firft outfet, where a flrictnefs of elucidation is peculiarly neceflary. The pafling of the circles through each other’s centres is, indeed, a {ufficient reafon why they muft cut each other; but this thould cer- tainly have been mentioned in the demonftration. Prop, / § OBSERVATIONS. 253 Propr. 2. Book I.:- Procrus, and other writers, have obferved, that this problem admits of feveral cafes, according to the fituation of the point A ; but there is only one of them that can properly be called a feparate cafe, which is when the point A is at either of the extremities of the given line: and in this cafe, if a circle be defcribed from the given point, at the diftance cB, any of the radii of that circle will be the line required. In all other fituations of the point A, whether in the line AB, or out of it, the conftru&ion and demonfiration will be the fame as that given in the text; which differs from Evcrip’s only in the producing of the line DA, after the circle FHG is defcribed 5 this ‘being thought more conformable to the terms of the propo- {ition. TROP. 70 BUOK L In the conftruclion of this problem the line Ap may fall upon the line a3, and then the thing required is done. The given lines ¢ and AB may alfo meet each other, at the point A ; and then a circle defcribed from that point, with the radius ¢, will cut off from AB the part required. This cafe occurs in the conftruction of the sth prepo- fition following, and in feveral other parts of the Ele-. ments, and, for that reafon, ought to have been men- “tioned. In all other pofitions of the two given lines Ev- crip’s conftruétion and demonftration are general. Pror. 4. Book L The demonftration of this propofition has been fre- quently objected againft, as being too mechanical. But this complaint is frivolous and iil founded ; for the ope- ration 284 NOT S AnND ration of placing one triangle upon the other, is 2 mental - oney and what is to be confidered as poffible to be effe&- ed, rather than altnally done. There is, befides, no other way in which the equality of thefe figures can be eftablifhed, fo that any cavils about its merits or defects are entirely precluded. | Prop. 5. Book I v Evctip, in his demonftration of this propofitiofi, has fhewn that the angles below the bafe are, alfo, equal to each other. But as this property is never referred to ~ throughout the Elements, except in the demonfiration of the 2d cafe of the 7th propofition following, the whole of which is both aukward and unneceffary, it would have been better to have omitted it, and confined the demon- fration, in the prefent inftance, to the equality of the ~angles above the bafe, which is a property much more generally ufeful. e PRrop. 6. BOOK‘.L The demonfiration of this propoﬁtzon, in Evciip, is” smmethodical; and defeéive. It is not fufficient to thew fhat one fide is not greater than the other, but it ought; alfo, to be thewn that it is not lefs, before their equality can be fairly inferred. - It is true, indeed, that either of _ the fides may be taken at pleafure, and the fame thing will follow : but this obfervation fhould have been made; and then the premifes, which they do not at prefent, would have authorized the deduétion required. The fame objection may be made to feveral other propofitions in the Elements, Prorp. Pror. 7. Book L - This is the fame as Evevrin’s 8th propofition; bu¢ de= monftrated in a different manner, in order that the pres ceding one, which is altogether ufelefs, might be omit- ted. Procrusdemonflrates itin nearly the fame manner; but he makes three cafes of it, when it may be done gene- rally in one; for if the longeft fides, or rather thofe which are not fhorter than any other, be applied together, there can be ho ambiguity in the fpecies of the triangles. PR o p. 8. BOOKL - This propofition is made an axiom by Evcrin ; but it is certainly not a truth of that kind ¢ for when two right angles are found in feparate and diftinét figures, there is nothing in the definitions or poftulates, from which their equality to each other can be fairly inferred. Prob 12 'Bberx L It is not thewn by EucLip, in his demonfiration of this problem, that the circle made ufe .of in the conftruétion; will cut the given line in two points; which as much re- quires proof as Prop. 2. Beok I1l., which is fiearly its converfe. For this reafon the conftru&ion given in the text has been preferred ; but in a2 work where the utmoft {cientific rigour is required; it would be better to con. ftruét the problem without the intervention of the circle, by means of right lines only, which may eafily be done, Fror. 132 Boog I In the enunciation of this Prop. by Eucrip, the angles are faid to be cither equal to two right angles, or toge- ‘ X ‘ ther ouh op ! N B8 AND ther equal to two right angles; but the former part of this feems to be unneceflary; for in all cafes, whether the - angles be each of them a right angle, or not, they are together equal to two ﬁght angles. Paov 16 Boow I, . As the outward angle of a triangle is afterwards fhewn to be equal to the two inward oppofite angles, it is much to be wifhed that the prefent Prop. which is only a par- tial cafe of the former, could be removed from the Ele- ments : but this cannot eafily be done; for the following propofition, and the firft relating to parallel lines, are not otherwife to be demonftrated. The next Prop. in EvucLip, is, however, quite unneceflary, as the firft place in which it occurs is Prop. 18. B. 3, where a reference . may be as readily made to the general propofition. Propr. 19 Boox1. The demonftration of this propofition, as it is given by EucLip, is extremely defective ; for the whole defign of the problem is to fhew that of three right lines, under certain fpecified reftri¢tions, a triangle may be formed; and as no ufe whatever 15 made of thefe reftrictions, ei- ~ther in the conftruction or demontftration, both the ar- gume‘nts adduced, and the conclufions derived from them, are entirely nugatory. 'This defect was obferved by MR. Simpson, between whom and his antagonift Dr. Sim- soN, it occafioned fome controverfy, which drew from the latter fome very hafty unfcientific expreflions, not much comporting with the character of fo {trict and ac- curate a (Geometrician. ’ TR OB, bt i O'BEER YV A TP O NS, 257 Provr. 28. Book L . R In moft editions of EuvcLip, two corollaries are affixed to this propofition, which are equally or more intricate than the propofition itfelf. DRr.AusTIN has endeavoured to prove that thefe, and moft of the other corollaries, to be found in the Elements s, were not introduced by Evu- cLID, but by fome of his commentators, or interpreters ; and there are many reafons for believing that this opinion is not ill founded. It is generally allowed, that EvcLip wrote a book entitled Corollaries, which were a colle@ion of confequences deducible from his Elements ; and, there- fore, it is not to be imagined that they were originally inferted in that work ; as in that cafe it would have been quite unneceflary to have publifhed them in a feparate performance. Befides this, the chain of reafoning is complete without them, as is evident from their being feldom referred to in any propofition. In all cafes, how- ~ever, where a ufeful truth of this kind can be readily deduced from a preceding demonftration, there appears to be no impropriety in making it a corollary. Pror 21+ Book 1. ‘This propofition, as it ftands in moft of the early editions, has three diftin& cafes, which all require to be {eparately demonftrated. - Dr. Simson, by changing the mode of demonftration, has reduced it to two: but by an obvious alteration in the enunciation of the I, 26. Evc. which is the fame as our 21ft, the propofition, both for parallelograms and triangles, might have been demon- ftrated generally, in one cafe only ; which, when it can S be 258 NOTES aND be done, is always to be preferred. In this part of the work, alfo, feveral other alterations have been made, the reafon for which will be given in the notes to the fecond book. “ Pror. 33. Booxk L ~ This propofition is fubftituted in the place of Prop. 42, 44, 2nd 45 of Euc. B. L as being lefs intricate, and equally ufeful in its application. One of the principal defigns of thefe propofitions, is to fhew, that a parallelogram; un- der certain conditions, can be formed; and as this can be more readily effeted by other methods, the preference has been given to that which appears the moft fimple. It may alfo be obferved, that the 44th propofition of Eu- cL1Dp is not legally demontftrated ; for the parallelogram BF, which makes a part of the conftrution, cannot be formed from Prop. 42, as is direGed, being entirely 4 different cafe : and as the 45th is derived from the 44th, it muft ~ alfo be liable to the fame objection. PrROP. 34, 35, Booxk L Thefe propofitions are delivered by EucLip in a dif- ferent form, and not given till the 6th book ; but as they are extremely eafy, and of frequent ufe in their applica= tion to other propofitions, in the preceding books, they have been here introduced as early as poflible, and de- monfirated independently of the dotrine of proportion which, it is imagined, beginners will confider as an ad- vantage, as they feldom arrive to fuch a proficiency, in a knowledge of the Elements, as to obtain clear and fatis- falory ideas of that intricate fubject. PRroc>. B S OBSERVATIONS Pror. 1. Book 15 In EvcLip’s demonftration of this problem, it ought to have been proved that the lines which are direéted to be drawn parallel to A, Ap, will meet each other; or otherwife it is not certain that the fquare required can be formed. On this account, 2 mode of conftru&ion has been here obferved, which is not liable to that obje&ion. b} Fropmogua, o Beo g il Thefe propofitions, which are not in EvcLip, may, by fome, be thought unneceflary; but they muft either be demonftrated, or aflumed; as the firft, in particular, is wanted in almoft every propofition of the fecond book ; and the others are frequently required in feveral parts of the Elements. Why they were omitted by Evcrip does not appear ; they are certainly not axiomatical, nor more evident in themfelves than many others which he has fcrupuloufly demonftrated. Pro?. 5. Book Ik This propofition is placed in the fecond book, for the purpofe of demonflrating it in 2 more general manner than has been done by EucLip ; and in order that fome others, of little importance, might be more eafily omitted. The demonftration depends principally upon the firft propofition, mentioned above; and this, among other inftances, is {ufficient to fthew the utility of that theorem, and the neceflity of its being introduced into the Elements, | S 2 Prop. 260 st N OTES adD Pror. 7.' Booxk II. ‘ This theorem, which is not in EvcLip, is given chiefly on account of its application to fome of the follow~ ing propofitions, the demonftrations of which are; by this means; rendered more concife and elegant. Pror. 13. Boox 1L All the theorems in Evcrip’s fecond book, which re= late to the divifion of a line into more than two parts, are here omitted, as they are commonly found tedious and embarrafling to beginners, and are not of any very exa tenfive ufe. The prefent propofition, which is not in EucLip, is much more generally applicable; and this, together with the preceding ones, will be found fufficient for moft geometrical purpofes. Prop. 16, 19, 20 and 21. Boox IL Thefe propofitions, though not in EvcLrip, are fre- quently wanted, particularly the 1fl, 2d, and 3d, which are, alfo, equally remarkable both for their elegance and utility. \ 4 » " _y Prov 1. Book 11 It is properly obferved by Dr. Simson, in his notes “upon this book, that the objections which have been “ufually made againft the indire&t method of proof, ufed in this and feveral other propofitions in the Ele- ‘ments, are injudicious and ill founded ; as it js obvious to every one, who has duly confidered the fubject; that there are many things which cannot be proved in any other e e A e OBSERVATBEONS . 6k other way. There is, however, a real defe& in the de- monftration of this propofition, that efcaped his notice ; which is, that the fictitious centre, or point G, may be taken in the line Ec ; and in this cafe the demonftration given by Eucrip will not hold, ' ProPa4e Boog lli. This propofition is the fame as the gth of Evucrip, Book III. but demonftrated in a manner which it is ima- gined will appear fomething more clear and fatisfactory, “at leaft to beginners. According to his method the pro- pofition admits of feveral cafes ; and in that which he has chofen as a general one, the fictitious centre, or point E, is {o taken, that the proof would be exacily the fame for two equal right lines as for three, which is a manifeft imperfection, "Pror. s Boox III. Eucrip has given this theorem in his 3d Book, in the form of a definition; which is the more remarkable, as he appears, in feveral parts of the Elements, to be well aware, that the equality of no two figures can be admitted but from the teft which he has laid dowh in the 8th axiom. Pror. 6, 7. Boox IIL Thefe theorems are, in fubftance, the fame 2s Ey- cLID’s, but differently enunciated, in order to accom- modate beginners, who are generally embarrafled with the aukwardnefs of the figures, and the two fi&titious centres in the laft propofition ; the latter of which are D 3 ~ here 262 ' NOTES AND here avoided. It may alfo be obferved that the demen« ftrations of thefe propofitions are not ftrittly fcientific ; fince, for aught that appears to the contrary, the circles may touch each other in more points than one, in which cafe the proof he has given would be nugatory. To avoid this, the fucceeding propofition fhould have been placed prior in order to the prefent ones, and demon- {trated independently of them ; which, hoWever, cannot eafily be done. For this reafon, and to avoid as much as poflible all theorems which are otherwife of little impor- tance, the touching of the circles in one point only, has been here inferred from the definition. A fimilar objzc- tion may, likewife, be made againft the sth, 6th, and Ioth theorems of Evcrip, Book III. ; the{la'{’c of which fhould have been demonflrated firft; for, as they now ftand, feveral things which require proof as much as the propog {itions themfelves, are taken for granted, Propr. 10. Booxk IIL In this propofition, no mention is made of the corni- cular angle, or that which is fuppofed to be formed by the cxrcumference and tangent, at the point of contact; as it is of no ufe whatever in Geometry, and ought never to have been admitted into the Elements. Dr. Simsown {ufpects, with VIETA, that it is an mterpolatlon, and on that account has properly reje&ed it; but there are fill fome partlculars, in his enunciation Of this propofition, which appear to be equally umeceﬂ’am’{ The theorem is, therefore, here propofed in as ﬁmple a manner as pofiible, and reftricted to that cafe which rnoﬂ: frequently Occurs. PROE» OBSERVATION S. 263 Prop. 14, Boox HE oX In Eucrip’s demonftration of the fecond cafe of this theorem, the following propofition has been taken for granted, viz. ¢ If one magnitude be double of another, and a part taken from the firft, be double of a part taken from the fecond, the remainder of the firft will be double the remainder of the fecond.”” But as this aflumption, which has hitherto been tacitly acquiefced in, is not de- rived from the axioms, or any thing which has been pre- vioufly demonflrated, it is certainly improper, and un- juftifiable. In order, therefore, to render the demon- firation of this cafe more ftrict and fcientific, it is here given in a different manner, which is equally eafy with the former, and not liable to the fame objection, Pror. 15. Book 1. Dr. AusTiN in his examination of the firft {ix books of the Elements, is of opinion that the fecond cafe of this propofition, which has been added by Dr. Simsow, and other modern Editors, is unneceflary. ¢ The former propofition, he obferves, is general; and, therefore, itis immaterial, whether the part of the circumference upon which the angles at the centre and circumference *and, be greater or lefs than a femicircle.”” But this obferva- tion is foreign to the purpofe; for as the arc which fub- tends an angle at the centre, muft always be lefs than a femicircle, no fuch angle can be introduced into the con-. ftruction of this cafe ; and therefore the demonttration of it muft neceflarily be obtained in fome way different from the former, ' g4 Prop. 264 NOTES 4Axp Pror. 18, 19. Boox III. '_ The firft of thefe propoﬁtiéhs is the vfame in effet ag the 25th of Eucrip, Book III, but fomething more fimple, being demonftrated general!y in one cafe. The other is the converfe of our 17th, or Evcrin’s zlzd which he haa omitted, but for what reafon does not apm pcar, as it 1s of frpquent ufe in its apphcatxon. ProP.20. Book il This propofition differs from the 24th of EucLip, Book I1I. only in the enunciation, which was done to avmd the neceflity of defining fimilar fegments of circles. For as this definition, whxch is nothing more than Ev- cL1p’s 21t propofition, in another form, cannot poffibly be underitood, tillit is fhewn that all angles in the fame fegment are equal to each other, it is altorrether ufelefs, ~and contrary to the nature of a definition, which requires that it thould be exprefled in fuch terms, and derived from fuch properties as are fimple and obvious. Proe: 57, 28, 29; Boor HI The demonftrations of thefe propofitions are confined to one cafe, Which? though it does not include every pofli- ~ ble pofition of the lines, will, itis conceived, be thought {ufficiently general. EucCLID, in this inftance, is much more particular§ having {crupuleufly demonftrated cafes of lefs moment ;th_an many others in the Elements which are taken for grantéd. ~ And the fame want of unifo'rmity may be obferved in feveral other propofitions, with refpect to the ftriGtnefs or laxity of their demonftrations, Prorpr OBSERVATIONS: 265 Prori 3 Boox IV, Dr. S1mson in his note upon Evcrip, Prop. a4y -obferves ¢ that the demonftration of this problem, has been fpoiled by fome unfkilful hand. For he does not demonftrate, as is neceffary, that the two ftraight lines, which bife& the fides of the mazmle, at right angks, muft meet one another.” After which, it appears fome thing fingular, that this able Geometrician fhould not per- ceive that a fimilar omiffion had been made in the de- monftration of the 3d, and feveral other propofitions of this book ; in which the neceflity of provim that certain lines will meet each other is equally obvious. In all thefe cafes, therefore, that. part of the demonftration is now fupplied, and the different felutzons, by that means, ren- de*‘ed more complete, ' Frov 16, Baok V- This propofition has been purpofely altered from Eu- CLID, in order to render the conftruction of the following one more practical and fimple. It is now, alfo, properly limited, which EvucrLip’s is not; for according to his enunciation of the problem, an infinite number of trian- gles may be formed, which will anfwer the conditions required. ‘T'he fame objection is likewife applicable to {everal other propofitions in the Elements; and though it may, to fome, appear trifling, it is certainly a de- parture from that ftriétnefs and prcczﬁon which, in a work of this nature, are generally confidered as eflential requifites. An inftance of this kind occurs even in the 2d Prop. of B. 1. which, however, is not fo caﬁyre- medied., Prop, 266 NOTES anp Pror 1:. Boox 1V. ' Some of the Commentators have obferved, that Evu- crip’s method of infcribing a pentagon in a circle, is much lefs fimple and elegant than that of ProLEmy in the 1ft Book of his Almageft ; and for that reafon think it ought to have been given in the Elements. Whilft others maintain that the demonftration of ProrEmY’s ~conftruction depends wholly on the 13th Book, and that, therefore, if Eucrip had known it, he could not have inferted it in the prefent Book ; the materials for it not being yet prepared. This, however, is not true; for it has been clearly fhewn by the author, in a periodical publication for the year 1786, that the truth of this con- firu&tion may be proved by means of the firft three Books of the Elements only ; but the reafon why it'has not been given in the text is, that the demonftration, being fome- thing more intricate, might not have been fo readily com- -prehended by beginners, for whofe ufe this work is prin- cipally defigned. Der. 5. Book V. T'his definition, which is the fame in effet as that given by Evcrip, has been the occafion of much con- troverfy and difpute among Mathematicians ; many of them thinking it foreign to the purpofe, and others too difficult and obfcure to be made the leading principle of a doctrine fo ufeful and neceflary as that of proportion, But, from a mature confideration of the {ubje, it is {uffi- ciently evident that no other definition, equally applicable and general, could have been given ; and, therefore, the 5 _ neceflity OBSERVATIONS, 267 neceflity of the cafe required that the prefent one fhould be adopted in preference to all others. That it is not fo fimple and evident as that which may be given of numbers, or commenfurable magnitudes, cannot be denied ; but this arifes from the nature of the fubje& and is not to be avoided. There are fome pro- portional magnitudes, fuch, for inftance, as the fide of a fquare and its diagonal, which have no common meafure, and confequently cannot be defined by that means. Some other definition was, therefore, to be found, which would hold in this, and all other cafes, without exception; and as the one in queftion anfwers thefe conditions, and is, at the fame time, equally commodxous in practice, nothing farther can be expe&ted. In order, however, to accommodate learners, who are feldom able to comprehend Evcrin’s sth Book, fuch alterations have been made in this part of the work, as it is prefumed will render it much more clear and intel- ligible. Among others, the definition above-mentioned is more concifely enunciated ;1 by which it is made to ap- pear lefs intricate and involved ; and confequently may be more eafily remembered and applied. Every thing which relates to greater and lefs ratios is alfo rejeted, as being obfcure and unneceflary. And as brevity was here thought particularly reqmﬁte, fuch propofitions only have been introduced, as are obvmuﬂy ufeful ; the reft being con- fidered as impediments in the way of the learner, and, for that reafon, unfit for an elementary performance, whofe principal aim fhould be clearnefs and perfpicuity. For a farther account of the do&rine of ratios, as de- Ewered by EUCLID, the reader is referred to the 7th and 8th 268 NOTES ann 8th of Dr. Barrow’s Mathematical Le&tures, where the ufual objeCtions which are made to this method are fully refuted, Propr. 2. Booxk V. This Prop. is the fame as the Cor. to Eucrip’s 2d Prop. of Book V. which Dr. Simson has marked with inverted commas, as being unneceflary. - But, whoever confiders this Book with attention, will obferve, that the coroﬂary is much more ufeful and general than the pro- pofition. . It is, indeed, ftrictly fpeaking, no corollary to the propofition in queftion, and for that reafon is properly enough difcarded ; but it would have been much better to have firuck out the propofition, and fubftituted the corol- lary in its place. ‘ Priomrigv B oo k V. Thls propofition, which is required in the demonftra- tion of fome of the following ones, is not exprefly enun- ciated by Eucrip, being introduced into the demon- ftration of his 8th propofition, without any farther ho- tice. But as it is a diftin& theorem, the truth of which is much lefs obvious than that oif {feveral others in this Book, it ought to have, bee wratclv demonftrated ; and particularly as it is of cmﬁ\ig ufe in its apph«« catio ' Prop 6517, Boor Vs It has been properly obferved, by Mnr. SimPpson, that the manner in which the compofition and divifion of ratios is treated of by Evciip, is defeciive, as not being [ufficiently general. 1t 1s alfo commonly found very | ' abitrufe OBSERVATIO N s . o abftrufe and embarraffing to beginners, on account of the complicated terms in which it is enunciated, and the number of cafes to be feparately demonftrated. For thefe reafons, it was deemed neceflary to give the propofitions a more {imple and general form, and to render the de- monftrations of them as concife and petfpicuous as pofli- -ble. The 17th, in friGtnefs, has two diftin&t cafes; but as the fecond differs from the firft only by inverting the terms, it was judged fufficient to mention it in a Scholium, | Der. 1, Booxr VI Dr. AusTIN obje&ts to this definition, becaufe it does not yet appear that any rectilineal figures can have their angles equal, each to each, and the fides about them pro- portional. He would therefore define fimilar triangles firft, which may be done only from the equality of their - angles ; and after this to cull thofe fimilar reéilineal figures, of more than three f{ides, which confift of an equal number of fimilar triangles, fimilarly fituated. This is no doubt an alteration which might prove advantageous to learners'; but as it appears illogical, and contrary to the method made ufe of in other cafes of a like nature, the original definition was thought preferable. Pro® 12y 13 16417 Book VL Several alterations have been made in the demonfira- tions of thefe propofitions, as they were given by Evu- crip, with the view of rendering them more fimple and eafy ; butasthey are in general fuch as may be readily dif= covered by infpection, or by comparing the theorems with thofe in the Elements, it will be unneceflary to point them out to the reader, . Proep. F 4 270 : NOT ES anp 3 Pror. 19, 20, 22, 23. Booxk VL. "Thefe propofitions are not in EucLip, though their utility and elegance certainly entitled them to a place in the Elements. 'The latter, in particular, may fupply the place of the 27th, 28th, and 2gth of Evcrip; which are certainly very aukward propofitions, and are feldom well underftood by beginners. The ufe which was made ‘of them by fome of the ancient Geometers, has been urged as a {ufficient reafon why they ought to be retained in the Elements ; but, upen this principle, numberlefs other pro- pofitions might be inferted, which would fwell this com- pendious and beautiful fyftem into a large volume; and make it appear more like 2 common place book than a fimple regular performance, judicioufly arranged in all its parts, and difplaying only the firft and moft important principles of the fcience. Propr. 25, 26, 27. Boox VI Thefe elegant and ufeful theorems were not originally in EvucLip; but have been inferted in fome of the late editions, by Dr. S1imson and other writers. MR. Simp- son has given them in the third book of his Elements, and demonftrated them independently of proportion ; which he conceives to be an alteration much for the better: but ' this will fcarcely be allowed, when it is confidered that he was obliged to introduce a new theorem for this pur- ‘pofe, manifeftly derived from the principles laid down in the sth Book; and that the advantages attending this method, are entirely deftroyed by a forced and unnatural arrangement, Prop, i OBSERVATIONS. 271 Pror. 2. Booxk VIL | - Dr. SiMson, in his notes upon this propofition, ob- ferves, that the enunciation of it, in moft editions, has been changed and vitiated, by its being propofed to fhew that every part of a triangle is in the fame plane, inftead of the way in which it now ftands ; as the property here alluded to is faid to be contained in the definition of a tri- angle, and is conftantly taken for granted in every part of the firft fix books. But he did not perceive that a fimilar objection might be made to the preceding propofition ; in which it is proved that one part of a right line cannot be in a plane, and another part above it ; this being, in like manner, a neceflary confequence of the definition he has given of a plane, or rather the very property from which it is derived. ' ) "‘Prop. 3. Book Vil The demontftration of this theorem, which is new, and more concife than that given by EucLip, is derived from a property of the ifofceles triangle, not mentioned in the Elements, but which is found of confiderable ufe in its application to other propofiticns. Several other altera- tions have alfo been made in different parts of this book, of which, as they will be readily difcovered by thofe who are verfed in the fubjet, any farther account is un- neceflary, Boox VII. T'he method here followed, of treating the folids, is not materially different from that which has been employed by b Ard L BTPRO A e by other writers upon the like occafion; but the propo- fitions, ‘it ‘is prefumed, will be found much’ better ar- -ranged, and the fubje& rendered more eafy and familiar ‘than has hitherto been done. The author however, i3 ~ well aware that it is liable to critical objections ; and had it not been incompatible with the plan of the performance,j'he certainly would have given it a more fcientific forin ; which even from the fimple prin- ciples here employed, might eafily have been done. But - as the advantages attending’the prefent mode of demon- ftration, efpecially to learners, appeared fuperior to every - other confideration, it was adapted in preference to that of Eucrip ; which, though more accurate, is frequently found to be tedious and obfcure. ' b T~ 1 N1 5, Publifhed by the fame AUTHOR. | 1, The ScHOLAR’s GUIDE to ARITHMETIC, 5th Ed. Pr. as, 2. An INTRODUCTION to MENSURATION and PRACTI- cAL GEOMETRY, 2d Ed. Pr. gs. | 3. AnINTRODUCTION to ALGEBRA, 2d Ed. Pr. gs. 4. An INTRODUCTION to ASTRONOMY, od Ed. Pr. 8s, Printed for J. JouNsoN, No 72, St. Paul’s Church-Yard. o o i A g E LB @ E NT S &t oM ET R Y CONTAINING 'THS PRINCIPAL 'PROPOSITIONSI 1% * o FIRST 81X, AND THE ELEVENTH AN] TWELFTH BOOKS s WiTH NOTES CRITICAL AND EXPLANATORYJ By JOHN BONNYCASTLE, OF THE ROYAL MILITARY ACADEMY, VVOC}LWICH,' s S I 5 I 0 N B O N PRINTED FOR J. JOHNSONy N© 72, 8Ts PAUL'S CHURCH-YARDs | o M:DCCLXXXIX, (=) - VierFarbSelector Standard * - Euroskala Offset Copyright 4/1999 YxyMaster GmbH www.yxymaster.com