Book VI. sectors EHF, FHM, MHN may be proved equal to one another. Therefore, whatever multiple 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. Therefore, as the arch BC is to the arch EF, so is the sector BGC to the sector EHFb. Wherefore, in equal circles, &c. Q. E. D. PROP. B. THEOR. T 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. a 5. 4. Describe the circlea ACB a A C D CAE, and the angle ABD to the b 21. 3. angleb AEC, the triangles ABD, AEC are equiangular to each other; therefore BA : AD: : EA € 4. 6. : ACC, consequently BA.AC=d E d 16. 6. AD.AE=ED.DA+DA. But € 3.2. f 35.3. ED.DA = BD.DC; therefore BA.AC= BD.DC + DA Wherefore, if an angle, &c. Q. E. D. PROP. C. THEOR. 1 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. Book VI. 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 a the circle ACB a bout the triangle, and draw its diameter AE, and join EC. Because the right angle BDA is equal to the angle ECA in a semicircle, and the angle ABD to the angle AEC in the same segment, the triangles ABD, AEC are equiangular; therefore, as BA to AD, so is EA to ACd; consequently the rectangle BA.AC is equale to the rectangle EA.AD. Therefore, if from any angle, &c. Q. E. D. c 21.3. d 4. 6. e 16. 6. PROP. D. PROB. THE rectangle contained by the diagonals of a See N. quadrilateral inscribed in a circle is equal to both the rectangles contained by its opposite sides. Book VI. a 21. 3. 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. 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. But the angle BDA is equala to the angle BCE; therefore the d 1. 2. C because the angle ABE is equal therefore BA.DC=BD.AE. But BC.DA=BD.CE; wherefore BC.DA + BA.DC = BD.CE + BD.AE = BD.AC d. 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 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. AD.AC + DB.AC = b AD+DB.AC = AB.CD; therefore b 1. 2. AD+DB: DC::AB: ACC. Wherefore, if an arch of a c 16.6. circle, &c. Q. E. D. PROP. F. THEOR. 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 whatever 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. Book VI. Join BD; then because the rectangle FD.DE is equal to the square of AD, or DB, FD : DB : : DB : DE. There b 6. 6. с 4. 6. d 16. 5. fore the two triangles FDB, BDE have the sides proportional about the common angle D; therefore they are equiangular, the angle DEB being equal to the angle DBF, and DRE to DFB; wherefore, FB: BD :: BE : ED, alternatelyd, FB : BE :: BD or AD : ED. e 17. 5. Because FD: DA : : DA: DE, by divisione, FA: DA :: AE : ED, alternately, FA: AE :: DA : ED. f 11. 5. But FB : BE : : AD : DE, thereforef FB : BE :: FA: AE. g 3. 6. Therefore, if two points, &c. Q. E. D. Cor. Because FB: BE:: FA: AE, the angle FBE is bisected by ABg. PROP. G. THEOR. IF from the extremity of the diameter of a circle a straight line be drawn in the circle, and if it meet a line perpendicular to the same diameter, either within the circle or, produced, without it, 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. |