timilar and alike situate to Do, and equal to the excess by which Book V1. EF exceeds Cb; then EF is equal to C, and KLMN together; W therefore KLMN is less than EF; and, because they are similar, the a 18. fide GF is greater than LM, and GE than LK; make GO equal b 25. to LM, and GX to LK; and compleat the parallelogram GP, which will be similar to, and about the fame diameter with EF"; let this diameter GB be drawn, and produce XP to R, and © 26. OP to S; then TS will be equal to C, and wanting in figure by SR, which is similar to Dd: Which was to be done. d 27. To apply a parallelogram upon a given right line, equal to a gi ven right lined figure, exceeding by a parallelogram similar to another given parallelogram. Upon the given right line AB, it is required to apply a parallelogram equal to the given right lined figure C, exceeding by a parallelogram similar to the given parallelogram D. Bisect AB in E ; upon EB describe the parallelogram EL, fimilar and alike situate to Da; and the parallelogram GH equal a 18. to C and EL together b, and similar and alike situate to Da; b 25& let KH be a fide homologous to FL, and KG to FE; then, because the parallelogram GH is greater than the parallelogram EL, the right line KH is greater than FL, and KG than FE : Produce FL and FE to M and N, so that FM be equal to KH, and FN to KG; compleat the parallelogram NM ; then MN is equal and similar to GH ; but GH is similar to EL; therefore MN is similar to EL"; therefore EL is about the same diameter c 21. with MN ; let FX, their diameter, be drawn, and describe the d 26. figure : Then, since GH is equal to EL and C together, as also to MN ; therefore MN is equal to EL and C together. Take KL, which is common, from both; then the gnomon MPE is equal to C; and, because the parallelograms AN, NB, are equal, AN is equal to LOf; and, if BX be added, AX is equal to the gnomon MPE; therefore AX is equal to C. Where- Ax. 1. J. fore, &c. f 43. and Το ει cut a given right line into extreme and mean ratio, b 29. Book VI. It is required to cut the given right line AB into extreme and mean ratio. a 40. I. Upon AB describe the square BC, and to AC apply the parallelogram CD, equal to the square BC, exceeding by the figure AD 6, similar to BC; but BC is a square ; therefore AD is also a square. From the equal parallelograms BC, CD, take away the common parallelogram CE, then the remainder BF will be equal to AD; but BF is equal to AD; therefore FE is to ED as AE is to EB®; that is, AB is to AE as AE is to EB: Or, let AB be cut in E, so that the rectangle under AB, BE, be equal to the square of AEd. Wherefore, &c. C 14. PRO P. XXXI. THE O R. IN every right angled triangle, any figure described upon the side subtending the right angle, is equal to the two similar figures defcribed upon the sides containing the right angle. 8. a Let ABC be the right angled triangle, the figure described on BC, fubtending the right angle, is equal to the two similar figures described on BA, AC; for, from the point A, let fall the perpendicular AD; then the triangle ABC is divided into the two fimilar triangles ADB, ADC; then, because the triangle ABC is fimilar to the triangle ABD, CB is to BA as BA is to BD, and CB is to BD as the figure described on CB is to b Cor. 20. the similar figure described on BA b. For the same reason, as BC is to CD, so is the figure described on BC to the similar one ther, so is the figure described on BC to the two similar figures © 24. 5. described on BA, AC“, together ; but BC is equal to BD, and DC together; therefore the figure described on BC is equal 1 PRO P. XXXII. THE O R. TF two triangles having two sides proportional to two sides, be so compounded or set together at one angle, 'that their homologous sides be parallel; then the other sides of these triangles will be in one right line. If the triangles ABC, DCE be fo placed at the point C, that the fide DE be parallel to AC, and DC to AB; then BCË will be a right line. For, because the homologous aides AB, DC, are parallel, and Book VI. AC falls upon them, the alternate angles BAC, ACD, are equal“; for the fame reason, CDE is equal to ACD; then, a 29. 1. since the two triangles BAC, CDE, have the angles at A and D equal, and the sides about them proportional, viz. BA to AC as CD to DE, the triangles are equiangular b, viz. theb 6. angle ABC equal to DCE, and ACB to DEC; but the angle AČD is proved equal to BAC; therefore, the whole angle ACE is equal to the two angles ABC, BAC. Add the common angle ACB to both, then the two angles ACE, ACB, are equal to the three angles ABC, ACB, BAC; that is, equal to two right angles"; therefore BCE is one right line, Wherefore,c 32. I. &c. d 14. I. PRO P. XXXIII. THE O R. IN equal circles, the angles are in the same proportion to one an other as the circumferences on which they stand, whether the angles be at the centres or the circumference ; so likewise are sectors, as being at the centres. Let ABC, DEF, be equal circles, and the angles BGC, EHF, at the centres G, H; and BAC, EDF, angles at their circumferences, then the angle BGC will be to the angle EHF as the circumference BC is to the circumference EF; and like, wise the angle BAC to the angle EDF, and the fector BGC to the sector EHF, as the circumference BC to EF. For, take any number of circumferences, as CK, KL, each equal to BC; and any number of circumferences, as FM, MN, each equal to EF; join GK, GL, HM, HN, then, because the circumferences BC, CK, KL, are equal, the angles BGC, CGK, KGL, are likewife equal; therefore, BL is the same multiplea 27. 3. of BC, that the angle BGL is of the angle BGC; for the same reason, EN is the same multiple of EF, that EHN is of EHF; therefore, if the circumference BL be equal to the circumference EN, the angle BGL is equal to the angle EHN, if greater greater, and if lefs less ; therefore, as BC is to EF, fo is BGC to EHFb; and so is BAC to EDFC. Again, as the b def. 5. 5. circumference BC is to EF, so is the sector BGC to the sector 15. 5. and EHF; for, join BC, CK, EF, FM, and assume the points X, 0, in the circumference BC, CK, and join BX, XC, CO, OK ; then, because BG, GC, are equal to CG, GK, and contain equal angles, the base BC is equal to the base CK 4,4 and the triangles equal ; and, because the right line BC is e qual 20. 3. e 28, 3 Book VI. qual to the right line CK, the circumference BXC is equal to the circumference COK : therefore the angle BXC is equal to the angle COK: Therefore, the segments BXC, COK, are equal f 24. 3. and and similar f; but the triangles BGC, CGK, are equal; there. def. II. 3. fore the whole sector BGCX is equal to the whole sector CGKO; in the same manner the sectors EHF, FHM, are proved equal ; therefore BK is the fame multiple of BC, that BGK is of BGCX; and EM the same multiple of EF that the sector EHMF is of the sector EHF; therefore, if the circumference BK is equal to the circumference EM, the sector BGK is equal to the sector EHM, if greater, greater, and if less, less; thereb def. 5. 5. fore, as BC is to EF, so is the sector BGCX to the sector EHFb. Wherefore, &c. Cor. I. An angle at the centre of, a circle is to four right angles, as the arch on which it stands is to the whole circumference; for, as the angle BAC is to a right angle, so is the arch BC to a quadrant, the consequents quadrupled ; then BAC is to four right angles as BC is to the whole circumference. Cor. II. The arches IL, BC, of unequal circles, which subtend equal angles, whether at the centres or circumferences, are similar : For IL is to the whole circumference ILE as the angle IAL, or BAC, is to four right angles; and so is the arch BC to the whole circumference BCF; therefore the arches IL, BC, are similar. COR. III. Two femidiameters AB, AC, cut off similar arches IL, BC, from concentric circumferences. THE |