Demonstration. Because the right angle BDA is equal to the angle ECA in a semicircle (III. 31), and the angle ABD equal to the angle AEC in the same segment (III. 21); the triangles ABD, AEC are equiangular; therefore E 1. As BA to AD, so is EA to AC (VI. 4); and consequently 2. The rectangle BA, AC is equal to the rectangle EA, AD (VI. 16). If therefore from any angle, &c. Q.E.D. PROPOSITION D.-Theorem. The rectangle contained by the diagonals of a quadrilateral figure inscribed in a circle, is equal to both the rectangles contained by its opposite sides. Let ABCD be any quadrilateral figure inscribed in a circle, and join AC, BD. Then the rectangle contained by AC, BD shall be equal to the two rectangles contained by AB, CD, and by AD, BC. B C A Construction. Make the angle ABE equal to the angle DBC (I. 23). Demonstration. Because the angle ABE is equal to the angle DBC, add to each of these equals the common angle EBD, then 1. The angle ABD is equal to the angle EBC; and the angle BDA is equal to the angle BCE, because they are in the same segment (III. 21); therefore 2. The triangle ABD is equiangular to the triangle BCE; wherefore, as BC is to CE, so is BD to DA (VI. 4); and consequently 3. The rectangle BC, AD is equal to the rectangle BD, CE (VI. 16). Again, because the angle ABE is equal to the angle DBC, and the angle BAE to the angle BDC (III. 21), 4. The triangle ABE is equiangular to the triangle BCD; therefore as BA to AE, so is BD to DC; wherefore 5. The rectangle BA, DC is equal to the rectangle BD, AE; but the rectangle BC, AD has been shown to be equal to the rectangle BD, CE; therefore 6. The whole rectangle AC, BD is equal to the rectangle AB, DC, together with the rectangle AD, BC (II. 1). Therefore the rectangle, &c. Q.E.D. PROPOSITION E.-Theorem. If a segment of a circle be bisected, and from the extremities of the base of the segment, 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 base, will have to the line draw from the point of bisection the same ratio which the base of the segment has to the base of half the segment. ៤ C Let ABD be a circle, of which AB is a segment bisected in C; and from A, C, and B to D, any point whatever in the circumference, let AD, CD, BD be drawn. Then the sum of the two lines AD and DB has to DC the same ratio that AB has to AC. B C Demonstration. Because ACBD is a quadrilateral inscribed in a circle, of which the diagonals are AB and CD, therefore 1. The rectangles AD, CB, and DB, AC, are together equal to the rectangle AB, CD (VI. D). But the rectangle AD, BC is equal to the rectangle AD, AC; because BC is equal to AC; and therefore 2. The two rectangles AD, AC, and BD, AC, are equal to the rectangle AB, CD. But the two rectangles AD, AC, and BD, AC, are the rectangle contained by AC, and the sum of the lines AD and DB (II. 1); wherefore 3. The rectangle contained by AC and the sum of the lines AD, DB, is equal to the rectangle AB, CD; and because the sides of equal rectangles are reciprocally proportional (VI. 14); therefore 4. The sum of AD and DB is to DC as AB to AC. Wherefore, if a segment, &c. Q.E.D. PROPOSITION F.-Theorem. 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 on the semidiameter; and if from these points two straight lines be inflected to any point whatsoever in the circumference of the circle, the ratio of the lines inflected will be the same with the ratio of the segments intercepted between the two firstmentioned 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 on AD; from E and F to any point B in the circumference, let EB, FB, be drawn. Then FB is to BE as FA is to AE. Construction. Join BD. A Demonstration. Because the rectangle FD, DE, is equal to the square on AD, that is, on DB, therefore 1. FD is to DB as DB is to DE (VI. 17). The two triangles FDB, BDE, have therefore the sides proportional that are about the common angle D; therefore 2. The triangles FDB, BDE, are equiangular (VI. 6). the angle DEB being equal to the angle DBF, and DBE to DFB. Now since the sides about these equal angles are also proportional (VI. 4); therefore 3. FB is to BD as BE is to ED; and alternately, 4. FB is to BE as BD is to ED (V. 16), or as FB is to BE so is AD to DE. But because FD is to DA as DA is to DE, by division (V. 17), 5. FA is to DA as AE is to ED; and alternately, 6. FA is to AE as DA is to ED. Now it has been shown that FB is to BE as AD is to DE, therefore 7. FB is to BE as FA is to AE (V. 11). Wherefore, if two points, &c. Q.E.D. Cor. If AB be drawn, because FB is to BE as FA is to AE, therefore 1. The angle FBE is bisected by AB (VI. 3). A Again, since FD is to DC as DC is to DE, by composition (V. 18), 2. FC is to DC as CE is to ED; and since it has been shown that FA is to AD (DC) as AE is to ED; therefore, by equality, 3. FA is to AE as FC is to CE. But FB is to BE as FA is to AE; therefore 4. FB is to BE as FC is to CE; so that if FB be produced to G, and if BC be drawn, 5. The angle EBG is bisected by the line BC (VI. A). PROPOSITION G.-Theorem. If, from one extremity of the diameter of a circle, a straight line be drawn, and a perpendicular be drawn to the diameter, so as to cut it and the straight line either internally or externally, the rectangle contained by the diameter and its segment reckoned from that extremity, is equal to the rectangle contained by the straight line and its corresponding segment. Let ABC be a circle, of which AC is a diameter, let DE be perpendicular to the diameter AC, and let AB meet DÉ in F. Then the rectangle BA, AF, is equal to the rectangle CA, AD. |