upon ca, BA. But bc is equal to BD, DC; therefore also the figure upon Bc is equal to the similar and similarly described figures upon BA, AC. Therefore in right-angled triangles, &c. 'Q. E. D. PROPOSITION XXXII. THEOREM. D * 29. 1. If two triangles having two sides of the one proportional to two sides of the other, be joined at one angle, so that their homologous sides be parallel ; the remaining sides of those triangles shall be in one right line. Let ABC, CDE, be two triangles, having the two sides BA, AC, proportional to the two sides CD, DE, viz. as AB to ac so is DC to DE, also AB parallel to Dc, and ac to DE; then BC, CE, are in one right line. For because AB is parallel to Dc, and ac falls upon them, the alternate angles, BAC, ACD, are equal to one another. For the same reason, also, CDE is equal to ACD; wherefore also BAC is equal to CDE. And because there are two triangles ABC, DCE, having the angle at A equal to the angle at D, and the sides about the equal angles proportionals, viz. as BA to Ac so is CD to De. Therefore the triangle ABC is equiangular to the triangle CDE;b hence the angle Abc is equal to DCE. But it has been shown that Acd is equal to BAC: hence the whole ace is equal to the two ABC, BAC; add the common angle ACB; therefore the angles BAC, ABC, BCA, are equal to two right angles, and ACE, ACB. To any“ 32. 1. right line Ac, and at any point c, two right lines be, CE, not placed at the same parts, make the adjacent angles ACE, ACB, equal to two right angles; hence Bc is in the same right line with ce.d B С E b 6. 6. a 14. 1. N D N M K к E F a 27. 3. PROPOSITION XXXIII. THEOREM. Let ABC, DEF, be equal circles, and BGC, EHF, at Take any number of circumferences ck, KL, each CK, KL, For the same therefore the circumference Bl be equal to the circum27. 3. ference En, the angle bou is also equal to EHN; and and if less, less; therefore it is as the circumference • 5 Def. 5. BC to Ef so is the angle bgc to EHF. But as the angle bgc to ehf so is BAC to Edf, for each is double A D OK. G H L BY Again, as the circumference bc is to the circumference EF so is the sector 6 Bc to the sector HEF. For join BC, CK, and take any points x, o, in the circumference, BC, CK, and join BX, xc, co, And because the two BG, GC, are equal to the two CG GK, and M they comprehend equal do K angles, the base BC is also equal to ck; therefore the triangle bgc is equal to the triangle GCK. And because the circumference BC is equal to the circumference ck, also the remaining circumference of the whole circle is equal to the remaining circumference of the whole circle; wherefore also the angle bxc is equal to the angle cok; hence the segment sxc is similar to the segment cok; and they are upon equal right lines BC, CK. But similar segments of circles upon equal right lines are equal to one another; hence the segment bxc is equal to the segment cok. But the triangle bgc is equal to the triangle GCK; and therefore the whole sector Gbc is equal to the whole sector GCK. For the same reason the sector GKL is equal to each of them GKC, GCB; hence the three sectors GBC, GCK, GKL, are equal to one another. For the same reason also the sectors HEF, HFM, HMN, are equal to one another; therefore whatsoever multiple the circumference bl is of the circumference BC, the same multiple is the sector GBL of the sector GBC. For the same reason also whatsoever multiple the circumference En is of the circumference EF, the same multiple is the sector hen of the sector HEF; if therefore the circumference bl is equal to the circumference En, the sector BGL is also equal to the sector HEN; and if the circumference BL exceed the circumference EN, the sector GBL will also exceed the sector HEŅ; and if less, less. Hence there being four magnitudes, the two circumferences BC, EF, also the two sectors GBC, HEF, and any equimultiples of the circumference bc and of the sector GBC, are taken, viz. the circumference BL and the sector Gel; also any equimultiples of the circumference EF and sector HEF, viz. the circumference EN, and sector HEN. And it has been shown that if the circumference BL exceed the circumference En, the sector Gel will also exceed the sector hen; if equal, equal; and if less, less; therefore as the circumference BC is to EF so is the sector GBC to the sector HEF. Therefore in equal circles, &c. Q. E. D.* Deduction. To trisect a given circle, by dividing it into three equal sections. * The latter part of this proposition was added by Theon, as he informs us in his Commentaries on Ptolemy's Almagest. He says, έτι δε και οι Touets and moreover also the sectors ; which are the only words of Theon added to Euclid's proposition, for what is subjoined atET POOTOīs nevt pois OUVIOTAPEVOI, when constituted at the centres, must be some marginal note very absurdly put in, as supposing there were another kind of sectors, besides what are stated at the centre of the circle, according to def. 10th, lib. 3d. Indeed the figures at the circumference are not, as their angles are, in the same ratio with the arcs on which they insist; but these figures are not called sectors, neither have they any note or name in Geometry to give occasion for such a needless caution-an oversight too great for Theon to be guilty of. |