TO cut a given straight line in extreme and mean ratio. Let AB be the given straight line; it is required to cut AB in extreme and mean ratio. Upon AB describes the square BC, and produce CA to D, a 46. 1. so that the rectangle CD.DA may be equal to the square CBb; take AE equal to AD, and complete the rectangle DF under DC and DA. Because the rectangle DF is equal to the square CB, take away the common part CE from each, and the remainder FB is equal to the remainder DE. But FB is the rect D A E B b 29. 6. angle contained by AB and BE, and DE is the square upon AE; therefore AE is a mean proportional be tween AB and BEC, or AB is to AE as AE to EB. But AB is greater than AE, wherefore AE is greater than EBd. Therefore d 14. 5. the straight line AB is cut in extreme and mean ratio in E. e Def. 3. 6. Which was to be done. Book VI. g 11. 2. Otherwise. Let AB be the given straight line; it is required to cut it in extreme and mean ratio. Divide AB in the point C, so that the rectangle contained by AB, BC may be equal to the square of AC. Then, because the rectangle C B as BA to AC, so is AC to CBC. was to be done. PROP. XXXI. THEOR. a 8. 6. b 4. 6. c 2. Cor. 20.6 IN right angled triangles the rectilineal figure described upon the side opposite to the right angle is equal to the similar and similarly described figures upon the sides containing the right angle. Let ABC be a right angled triangle, having the right angle BAC; the rectilineal figure described upon BC is equal to the similar and similarly described figures upon BA, AC. Draw the perpendicular AD, then the triangles ABD, ADC are similar to the whole triangle ABC, and therefore, inverselyd, as DB to BC, so is the figure upon BA Book VI. to that upon BC. For the same reason, as DC to CB, so is the figure upon CA to that upon CB. Wherefore, as BD d B. 5. and DC together to BC, so are the figures upon BA and AC together to the figure upon BC; therefore the figures on e 24. 5. BA and AC are together equalf to that on BC; and they are f 9.5. similar figures. Wherefore, in right angled triangles, &c. Q. E. D. PROP. XXXII. THEOR. IF two triangles, which have two sides of one proportional to two sides of the other, be joined at one angle, so as to have their homologous sides parallel to each other, the remaining sides will be in a straight line. Let ABC, DCE be two triangles, which have two sides BA, AC proportional to the two CD, DE, BA to AC, as CD to DE; and let AB be parallel to DC, and AC to DE; BC and CE are in a straight line. Because AB is parallel to DC, and the straight line AC meets them, the alternate angles BAC, ACD are equala. For a 29. 1. the same reason the angle CDE is equal to the angle A BA to AC, as CD to DE, they are equiangular; therefore the angle ABC is equal to b 6. 6. the angle DCE. Wherefore, because the angle BAC is equal to ACD, the whole angle ACE is equal to the two Book VI. angles ABC, BAC. Add the angle ACB to both, then the angles ACE, ACB are equal to the angles ABC, BAC, ACB. But ABC, ВАС, ACB are equal to two right angles; therefore also the angles ACE, ACB are equal to two right angles; therefore BC and CE are in a straight lined. Wherefore, if two triangles, &c. Q. E. D. с 32. 1. d 14. 1. a 27.3. PROP. XXXIII. THEOR. IN equal circles angles, whether at the centres or circumferences, have the same ratio which the arches on which they stand have to one another; and so also have the sectors. Let ABC, DEF be equal circles, and let BGC, EHF be angles at their centres, and BAC, EDF angles at their circumferences; as the arch BC to the arch EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF; and also the sector BGC to the sector EHF. Take any number of arches CK, KL, each equal to BC, and any number whatever FM, MN, each equal to EF; and join GK, GL, HM, HN. Because the arches BC, CK, KL are equal, the angles BGC, CGK, KGL are also equala; therefore whatever multiple the arch BL is of the arch BC, Book VI. the same multiple is the angle BGL of the angle BGC. For the same reason whatever multiple the arch EN is of the arch EF, the same multiple is the angle EHN of the angle EHF. But if the arch BL be equal to the arch EN, the angle BGL is also equala to the angle EHN; or if the arch BL be greater than EN, the angle BGL is greater than EHN; and if less, less. Therefore, as the arch BC is to the arch EF, sob is the angle BGC to the angle EHF. But as b Def. 5. 5. the angle BGC is to the angle EHF, so the angle BAC to c 15.5. the angle EDF, for each is double of eachd Therefore, as d 20. 3. the arch BC is to EF, so is the angle BGCEHF, and the angle BAC to EDF. Again, as the arch BC is to EF, so is the sector BGC to the sector EHF. Join BC, CK, and in the arches BC, CK take any points X, O, and join BX, XC, CO, OK. Because in the triangles GBC, GCK, the two sides BG, GC are equal to the two CG, GK, and also contain equal angles, the base BC is equale to the base CK, and the triangle GBC to the e 4. triangle GCK. Because the arch BC is equal to the arch CK, the remaining part of the whole circumference of the circle ABC is equal to the remaining part of the whole cir cumference of the same circle; wherefore the angle BXC is equal to the angle COK; therefore the segment BXC is similar to the segment COKf; and they are upon equal f Def. 9. 3. straight lines BC, CK; therefore the segment BXC is equal to the segment COKs. Now the triangle BGC is equal to g 24. 3. the triangle CGK; therefore the sector BGC is equal to the sector CGK. For the same reason the sector KGL is equal to each of the sectors BGC, CGK. In the same manner the |