angle BGC, have been taken any equimultiples whatever, viz. the arch BL, and the angle BGL; and of the arch EF, and of the angle EHF, and equimultiples whatever, viz. the arch EN, and the angle EHN : And it has been proved, that if the arch BL be greater than EN, the angle BGL is greater than EHN; and if equal, equal; and if less, less As therefore, the arch BC to the arch EF, so (def. 5. 5.) is the angle BGC to the angle EHF: But as the angle BGC is to the angle EHF, so is (15. 5.) the angle BAC to the angle EDF, for each is double of each (20. 3.): Therefore, as the circumference BC is to EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF. : Also, as the arch BC 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: Then, 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 equal (4. 1.) to the base CK, and the triangle GBC to the triangle GCK: And 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 circumference of the same circle: Wherefore the angle BXC is equal to the angle COK (27. 3.); and the segment BXC is therefore similar to the segment COK (def, 9. 3.); and they are upon equal straight lines BC, CK But similar segments of circles upon equal straight lines are equal (24. 3.) to one another: Therefore the segment BXC is equal to the segment COK: And the triangle 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; and in the same manner, the sectors EHF, FHM, MHN may be proved equal to one another Therefore, what multiple soever the arch BL is of the arch BC, the same multiple is the sector BGL of the sector BGC. For the same reason, whatever multiple the arch EN is of EF, the same multiple is the sector EHN of the sector EHF: Now if the ; arch BL be equal to EN, the sector BGL is equal to the sector EHN and if the arch 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 arches BC, EF, and the two sectors BGC, EHF, and of the arch BC, and sector BGC, the arch BL and the sector BGL are any equimultiples whatever; and of the arch EF, and sector EHF, the arch EN and sector EHN, are any equimultiples whatever; and it has been proved, that if the arch BL be greater than EN, the sector BGL is greater than the sector EHN; if equal, equal; and if less, less; therefore (def. 5. 5.), as the arch BC is to the arch EF, so is the sector BGC to the sector EHF. Wherefore, in equal circles, &c. Q. E. D. PROP. B. THEOR. If an angle of a triangle be bisected by a straight line, which likewise cuts the base; the rectangle contained by the sides 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 the straight line AD; the rectangle BA.AC is equal to the rectangle BD.DC, together with the square of AD. B Describe the circle (5. 4.) 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 angle ABD to the angle (21. 3.) AEC, for they are in the same segment; the triangles ABD, AEC are equiangular to one another: Therefore BA: AD: EA: (4. 6.) AC, and consequently, BA.AC (16. 6.) AD.AE = ED.DA (3. 2.) + DA3. But ED.DA = BD.DC, therefore BA.AC BD.DC + DA. Wherefore, if an angle, &c. Q. E. D. = U E A PROP. C. THEOR. If from any angle of a triangle a straight line be 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 diameter 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. Describe (5. 4.) the circle ACB about the triangle, and draw its diameter AE, and join EC: Because the right angle BDA is equal (3. 3.) to the angle ECA in a semicircle, and the angle ABD to the angle AEC, in the same segment (21. 3.); the triangles ABD, AEC are equiangular Therefore, as (4. 6.) BA to AD, so is EA to AC: and consequently the rectangle BA.AC is equal (16. 6.) to the rectangle EA.AD. ̄If, therefore, from an angle, &c. Q. E. D. B PROP. D. THEOR. E D The rectangle contained by the diagonals of a quadrilateral 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 let AC, BD be drawn; the rectangle AC. BD is equal to the two rectangles AB.CD, and AD.BC. B 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 (21. 3.) the angle BCE, because they are in the same segment therefore the triangle ABD is equiangular to the triangle BCE. Wherefore (4. 6.), BC: CE :: BD: DA, and consequently (16. 6.) BC.DA BD.CE. Again, because the angle ABE is equal to the angle DBC,and the angle (21.3.) BAE to the angleBDC, the triangleABÉ is equiangular to the triangle BCD; therefore BA : AE: : BD: DC, and BA.DCBD.AE: But it was shewn that BC.DA=BD.CE; wherefore BC. A DA+BA.DC=BD.CE + BD.AE = BD.AC (1. 2.). That is the rectangle contained by BD and AC, is equal to the rectangles contained by AB, CD, and AD, BC. Therefore the rectangle, &c. Q. E. D. PROP. E. THEOR. If an arch of a circle be bisected, and from the extremities of the arch, and from the point of bisection, straight lines be drawn to any point in the circumference, the sum of the two lines drawn from the extremities of the arch will have to the line drawn from the point of bisection, the same ratio which the straight line subtending the arch has to the straight line subtending half the arch. Let ABD be a circle, of which AB is an arch bisected in C, and from A, C, and B to D, any point whatever in the circumference, let AD, CD, BD be drawn; the sum of the two lines AD and DB has to DC the same ratio that BA has to AC. For since ACBD is a quadrilateral inscribed in a circle, of which the diagonals are AB and CD, AD.CB + DB.AC (D. 6.) = AB.CD: but AD. CB+DB.ACAD.AC+DB.AC, because CB=AC. Therefore AD.AC+ DB.AC, that is (1. 2.), (AD + DB) AC AB.CD. And because the sides of equal rectangles are reciprocally proportional (14. 6.), AD+DB: DC AB : AC. Wherefore, &c. Q. E. D. PROP. F. THEOR. D B C If two points be taken in the diameter of a circle, such that the rectangle contained by the segments intercepted between them and the centre of the circle be equal to the square of the radius: and if from these points two straight lines be drawn to any point whatsoever in the circumference of the circle, the ratio of these lines will be the same with the ratio of the segments intercepted between the two first mentioned points and the circumference of the circle. Let ABC be a circle, of which the centre is D, and in DA produced, let the points E and F be such that the rectangle ED, DF is equal to the square of AD; from E and F to any point B in the circumference, let EB, FB be drawn; FB: BE :: FA : AE. Join BD, and because the rectangle FD, DE is equal to the square of AD, that is, of DB, FD : DB :: DB : DE (17. 6). The two triangles, FDB, BDE have therefore the sides proportional that are about the common angle D; therefore they are equiangular (6. 6.), the angle DEB being equal to the angle DBF, and DBE to DFB. Now since the sides about these equal angles are also propor tional (4. 6.), FB : BD : : BE : ED, and alternately (16. 5.), FB : BE :: BD: ED, or FB : BE:: AD: DE. But because FD: DA :: DA: DE, by division (17. 5.), FA : DA :: AE : ED, and alternately (11. 5.), FA: AE :: DA : ÉD. Now it has been shewn that FB: BE:: AD: DE, therefore FB : BE :: FA : AE. Therefore, &c. Q. E. D. COR. If AB be drawn, because FB : BE: : FA: AE, the angle FBE is bisected (3. 6.) by AB. Also, since FD: DC :: DC : DE, by composition (18. 5.), FC: DC :: CE: ED, and since it has been shewn that FÀ: AD (DC) : : AE : ED, therefore, ex æquo, FA : AE: FC: CE. But FB: BE:: FA: AE, therefore, FB : BE:: FC: CE (11. 5.); so that if FB be produced to G, and if BC be drawn, the angle EBG is bisected by the line BC (A 6.). PROP. G. THEOR. If from the extremity of the diameter of a circle a straight line be drawn in the circle, and if either within the circle or produced without it, it meet a line perpendicular to the same diameter, the rectangle contained by the straight line drawn in the circle, and the segment of it, intercepted between the extremity of the diameter and the perpendicular, is equal to the rectangle-contained by the diameter, and the segment of it cut off by the perpendicular. Let ABC be a circle, of which AC is a diameter, let DE be per |