b 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 ACE is equal to the two angles ABC, BAC. add the common angle ACB, then the angles ACE, AUB are equal to the angles ABC, BAC, ACB. but ABC, BAC, ACB are equal to two right angles; therefore alío 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 oppofite fides of it, make the adjacent angles ACE, ACB equal to two right angles; therefore BC and CE are d. 14. I. in a straight line. Wherefore if two triangles, &c. Q. E. D. equal circles, angles whether at the centers or cir. See N. cumferences have the fame ratio which the circumferences on which they ftand have to one another. fo 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 circumferences. as the circumference BC to the circumference EF, fo is the angle BGC to the angle EHF, and the angle BAC to the angle EDF; and also the sector BGC to the fector 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 all equal, the angles BGC, CGK, KGL are alfo all equal. there- a. 27. 3a fore what multiple foever the circumference BL is of the circumfe rence BC, the fame multiple is the angle BGL of the angle BGC. for the fame reafon, whatever multiple the circumference EN is of 2. 27.3. Book VI. the circumference EF, the fame multiple is the angle EHN of the angle FHF. and if the circumference BL be equal to the circumfe rence EN, the angle BGL is alfo equal to the angle EHN; and if the circumference BL be greater than EN, likewise the angle BGL is greater than EHN; and if lefs, lefs. 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 C. 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 lefs, lefs. as therefore the circumference BC b. 5. Def. 5. to the circumference EF, fob is 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 the angle EDF, for each is double of each . therefore as the circumference BC is to EF, fo is the angle BGC to the angle EHF, and the angle BAC to the angle EDF. d. 20. 3. e. 4. I. Alfo, as the circumference BC to EF, fo is the sector BGC to the sector EHF. Join BC, CK, and in the circumferences BC, CK take any points X, O, and join BX, XC, CO, OK. then because in the triangles GBC, GCK the two fides BG, GC are equal to the two CG, GK, and that they contain equal angles; the bafe 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 fame circle. wherefore the angle BXC is equal to the angle COK ; and. 11.Def. 3. the segment BXC is therefore fimilar to the segment COK £; and they are upon equal straight lines BC, CK. but fimilar segments of 8. 24. 3. circles upon equal straight lines, are equal to one another. there fore the fegment BXC is equal to the fegment COK. and the tri- Book VI. angle BGC is equal to the triangle CGK; therefore the whole, the fector BGC is equal to the whole, the fector CGK. for the fame reason the sector KGL is equal to each of the fectors BGC, CGK. in the fame manner the fectors EHF, FHM, MHN may be proved equal to one another. therefore what multiple soever the circumference BL is of the circumference BC, the fame multiple is the sector BGL of the sector BGC. for the fame reason, whatever multiple the circumference EN is of EF, the fame 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 fector BGL is greater than the fector EHN; and if lefs, lefs. fince then there are four magnitudes, the two circumferences BC, EF, and the two fectors 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 fector EHF, the circumference EN and 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 as the circumference BC is to the circumference EF, fo b. 5. Def. 5. is the sector BGC to the fector EHF. Wherefore in equal circles, &c. Q. E. D. b Book VL PROP. B. THEOR. See N. 5.4. F an angle of a triangle be bisected by a straight line, which likewife cuts the bafe; the rectangle contained by the fides of the triangle is equal to the rectangle contained by the fegments of the bafe, together with the square of the straight line bifecting the angle. Let ABC be a triangle, and let the angle BAC be bifected by by the straight line AD; the rectangle BA, AC is equal to the rectangle BD, DC together with the fquare of AD. A 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 21. 3. angle ABD to the angle ↳ AEC, for they are in the fame segment; the triangles ABD, AEC are equiangular to one another. therefore as BA to AD, fo is EA to AC, and confequently the rectangle BA, AC d. 16. 6. is equal d to the rectangle EA, AD, that is to the rectangle ED, DA together with the fquare of c. 4. 6. e. 3. 2. f. 35. 3. B D AD. but the rectangle ED, DA is equal to the rectangle f BD, See N. I to F from an angle of a triangle a ftraight line be drawn perpendicular to the base; the rectangle contained by the fides of the triangle is equal to the rectangle contained by the perpendicular and the diameter of the circle defcribed 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 defcribed about the triangle. Defcribe the circle ACB about the triangle, and draw its diameter b AE, and join EC. because the right angle BDA is equal to the angle ECA in a femicircle, and the angle ABD to the angle AEC in the fame fegment; the triangles ABD, AEC are equiangular. therefore as a BA to AD, fo is EA to AC, and confequently the rectangle BA, AC is e d equal to the rectangle EA, AD. If therefore from an angle, &c. e. 16. 6. Q. E. D. THE HE rectangle contained by the diagonals of a qua- the rectangles contained by its oppofite fides. 44 Let ABCD be any quadrilateral inscribed in a circle, and join AC, BD; the rectangle contained by AC, 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 fame segment; therefore the triangle ABD is equi E A b. 4. 6. C c. 16. 6. BAE to the angle BDC, the triangle * This is a Lemina of Cl. Ptolemaeus in page 9. of his μsydan oÚrTağı, |