b. 6. 6. the triangle ABC is equiangular to DCE. therefore the angle Book VI. ABC is equal to the angle DCE. and the angle BAC was proved to be equal to ACD. therefore the whole angle Ac£ is equal to the two angles ABC, BAC. add the common angle ACB, then the angles ACE, ACB are equal to the angles ABC, BAC, ACB. but ABC, BAC, ACB are equal to two right angles“; therefore allo C. 32. I. the angles ACE, ACB are equal to two right angles. and fince at the point C in the straight line AC, the two straight lines BC, CE, which are on the opposite sides of it, make the adjacent angles ACE, ACB equal to two right angles; therefore d BC and CE are d. 14. I. in a straight line. Wherefore if two triangles, &c. Q. E. D. IN equal circles, angles whether at the centers or cir. Sec N. cumferences have the same ratio which the circumferences on which they stand have to one another. so also have the sectors. Let ABC, DEF be equal circles; and at their centers the angles BGC, EHF, and the angles BAC, EDF at their circumfejences. as the circumference BC to the circumference EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF; and also the fector BGC to the sector EHF. Take any number of circumferences CK, KL each equal to BC, and any number whatever FM, MN each equal to EF; and join GK, GL, HM, HN. Because the circumferences BC, CK, KL are K M E F all equal, the angles BGC, CGK, KGL are also all equalthere4. 27. 3. fore what multiple foever the circumference BL is of the circumfe. rence BC, the same multiple is the angle BGL of the angle BGC. for the same reason, whatever multiple the circumference EN is of Book VI. the circumference EF, the same multiple is the angle EHN of the angle FHF. and if the circumference BL be equal to the circumfe. &. 37. 3. rence EN, the angle BGL is also equal a to the angle EHN; and if the circumference BL be greater than EN, likewise the angle BGL is greater than EHN; and if less, less. there being then four magnitudes, the two circumferences BC, EF, and the two angles BGC, EHF; of the circumference BC, and of the angle BGC, have been taken any equimultiples whatever, viz. the circumference BL, and, the angle BGL; and of the circumference EF, and of the angle A LD EHF, any equimultiples whatever, viz. the circumference EN, and the angle EHN. and it has been proved that if the circumference BL be greater than EN, the angle BGL is greater than EHN; and if equal, equal; and if less, less. as therefore the circumference BC b. 5. Def. 5. to the circumference EF, so bis the angle BGC to the angle EHF. C. 15. 5. but as the angle BGC is to the angle EHF, fo is the angle BAC to d. 20. 3. the angle EDF, for each is double of each d. therefore as the cir cumference BC is to EF, fo is the angle BGC to the angle EHF, and the angle BAC to the angle EDF. Also, as the circumference BC to EF, so is the sector BGC to the sector EHF. Join BC, CK, and in the circumferences BC, CK take any points X, 0, and join BX, XC, CO, OK. then because in the triangles GBC, GCK the two sides BG, GC are equal to the two e. 4. I. CG, GK, and that they contain equal angles; the base BC is equal to the base CK, and the triangle GBC to the triangle GCK. and because the circumference BC is equal to the circumference CK, the remaining part of the whole circumference of the circle ABC, is equal to the remaining part of the whole circumference of the same circle. wherefore the angle BXC is equal to the angle COK *; and 4.11. Def. 3. the fegment BXC is therefore similar to the segment COK '; and they are upon equal straight lines BC; CK. but similar segments of 6. 24. 3. circles upon equal straight lines, are equal & to one another. there. fore the segment BXC is equal to the segment COK. and the tri- Book VI. angle BGC is equal to the triangle CGK; therefore the whole, the sector BGC is equal to the whole, the sector CGK. for the same reason the sector KGL is equal to each of the sectors BGC, CGK. in the fame manner the sectors EHF, FHM, MHN may be proved equal to one another. therefore what multiple soever the circumference BL is of the circumference BC, the same multiple is the sector BGL of the sector BGC. for the same reason, whatever multiple the circumference EN is of EF, the same multiple is the sector EHN of the sector EHF. and if the circumference BL be equal to EN, the sector BGL is equal to the sector EHN; and if the circumfe rence BL be greater than EN, the sector BGL is greater than the sector EHN; and if less, less. since then there are four magnitudes, the two circumferences BC, EF, and the two sectors BGC, EHF, and of the circumference BC and sector BGC, the circumference BL and sector BGL are any equimultiples whatever; and of the circumference EF and sector EHF, the circumference ENand sector EHN are any equimultiples whatever; and that it has been proved if the circumference BL be greater than EN, the sector BGL is greater than the sector EHN; and if equal, equal; and if less, less. Therefore b as the circumference BC is to the circumference EF, so b. 5. Def. 5. is the sector BGC to the sector EHF. Wherefore in equal circles, &c. Q. E. D. Book VL PROP. B. THEOR. See N. IF* (F an angle of a triangle be bisected by a straight line, which likewise cuts the base; the rectangle con. tained by the fides of the triangle is equal to the rectangle contained by the segments of the base, together with the square of the straight line bisecting the angle. Let ABC be a triangle, and let the angle BAC be bisected by by the straight line AD, the rectangle BA, AC is equal to the rectangle BD, DC together with the square of AD. 2. S. 4o Describe the circle a ACB about the triangle, and produce AD to the circumference in E, and join EC. then because the angle BAD is equal to the angle CAE, and the А. they are in the same segment; the gular to one another. therefore as B C. 4. 6. BA to AD, so is EA to AC, and D AD, that is e to the rectangle ED, e. 3. 2. f. 35. 3. Sec N. IF drawn perpendicular to the base; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the perpendicular and the dia. meter of the circle described about the triangle. Let ABC be a triangle, and AD the perpendicular from the angle A to the base BC; the rectangle BA, AC is equal to the rectangle contained by AD and the diameter of the circle described about the triangle. b. 31. 3. Describe a the circle ACB about Book VI. A the triangle, and draw its diameter 2. 5.'4. AE, and join EC. because the right angle BDA is equal o to the angle ECA in a semicircle, and the angle B C D ABD to the angle AEC in the same segment"; the triangles ABD, AEC c. 21. 3. are equiangular. therefore as d BA d. 4. 6. to AD, so is EA to AC, and conse E quently the rectangle BA, AC is equal to the rectangle EA, AD. If therefore from an angle, &c. e. 16. 6. Q. E. D. E PROP. D. THEOR. TH HE rectangle contained by the diagonals of a qua. drilateral inscribed in a circle, is equal to both the rectangles contained by its opposite sides. Let ABCD be any quadrilateral inscribed in a circle, and join AC, BD; the rectangle contained by AÇ, BD is equal to the two rectangles contained by AB, CD and by AD, BC* Make the angle ABE equal to the angle DBC; add to each of these the common angle EBD, then the angle ABD is equal to the angle EBC. and the angle BDA is equal“to the angle BCE, because a. 21. 3. they are in the same segment; therefore the triangle ABD is equiangular to the triangle BCE. whereforebas BC is to CE, fo is BD to DA, B b. 4. 6. and consequently the rectangle BC, C AD is equal to the rectangle BD, CE. again, because the angle ABE is equal to the angle DBC, and the angle • BAE to the angle BDC, the triangle ABE is equiangular to the triangle BCD, as therefore BA to AE, fo is A BD to DC; wherefore the rectangle BA, DC is equal to the rectangle BD, AE. but the rectangle BC, AD has been thewn equal to the rectangle BD, CE; therefore the whole rectangle AC, BD is equal to the rectangle AB, DC together with the rectangle AD, BC. Therefore the rectangle, &c. Q. E. D. * This is a Lemina of Cl. Ptolemaeus in page g. of his peydan cúrtatis, C. 16. 6. |