Demonstration. Because EA is equal to EB, therefore the angle EBA is equal to the angle EAB (I. 5); therefore 1. The angles EAB, EBA are double of the angle EAB; but the angle BEF is equal to the angles EAB, EBA (I. 32); therefore also 2. The angle BEF is double of the angle EAB; for the same reason, 3. The angle FEC is double of the angle EAC; therefore 4. The whole angle BEC is double of the whole angle BAC. Secondly. Let the centre of the circle be without the angle BAC. Construction. Join AE, and produce it to meet the circumference in F. Demonstration. It may be demonstrated, as in the first case, that 1. The angle FEC is double of the angle FAC, and that 2. FEB a part of FEC; is double of FAB, a part of FAC; therefore 3. The remaining angle BEC is double of the remaining angle BAC. Therefore the angle at the centre, &c. Q.E.D. H PROPOSITION 21.-Theorem. The angles in the same segment of a circle are equal to one another. Let ABCD be a circle, and BAD, BED angles in the same segment BAED. Then the angles BAD, BED shall be equal to one another. Construction. Take F, the centre of the circle ABCD (III. 1), and join BF, FD. Demonstration. Because the angle BFD is at the centre, and the angle BAD at the circumference, and that they have the same part of the circumference, viz., the arc BCD for their base ; therefore 1. The angle BFD is double of the angle BAD (III. 20); for the same reason, 2. The angle BFD is double of the angle BED; therefore 3. The angle BAD is equal to the angle BED (Ax. 7). Next. Let the segment BA ED be not greater than a semicircle. Construction. Draw AF to the centre, and produce it to C, and join CE. Demonstration. Then the segment BADC is greater than a semicircle, therefore 1. The angles BAC, BEC in the segment BADC are equal by the first case. For the same reason, because CBED is greater than a semicircle, 2. The angles CAD, CED are equal ; therefore 3. The whole angle BAD is equal to the whole angle BED (Ax. 2). Wherefore, the angles in the same segment, &c. Q.E.D. PROPOSITION 22.-Theorem. The opposite angles of any quadrilateral figure inscribed in a circle, are together equal to two right angles. Let ABCD be a quadrilateral figure in the circle ABCD. Then any two of its opposite angles shall together be equal to two right angles. Construction. Join AC, BD. Demonstration. And because the three angles of every triangle are equal to two right angles (I. 32), the three angles of the triangle CAB, viz., 1. The angles CAB, ABC, BCA are equal to two right angles ; but 2. The angle CAB is equal to the angle CDB (III. 21), because they are in the same segment CDAB; and 3. The angle ACB is equal to the angle ADB; because they are in the same segment ADCB; therefore 4. The two angles CAB, ACB are together equal to the whole angle ADC (Ax. 2); to each of these equals add the angle ABC; therefore 5. The three angles ABC, CAB, BCA are equal to the two angles ABC, ADC (Ax. 2); but ABC, BAC, BCA are equal to two right angles ; therefore also 6. The angles ABC, ADC are equal to two right angles. In the same manner it may be shown, that 7. The angles BAD, DCB are equal to two right angles. Therefore, the opposite angles, &c. Q.E.D. PROPOSITION 23.-Theorem. Upon the same straight line, and upon the same side of it, there cannot be two similar segments of circles, not coinciding with one another. If it be possible, upon the same straight line AB, and upon the same side of it, let there be two similar segments of circles, ACB, ADB, not coinciding with one another. Construction. Then, because the circumference ACB cuts the circumference ADB in the two points A, B, they cannot cut one another in any other point (III. 10); therefore 1. One of the segments ACB, ADB, must fall within the other ; let ACB fall within ADB; draw the straight line BCD, and join CADA. Demonstration. Because the segment ACB is similar to the segment ADB (hyp.), and that similar segments of circles contain equal angles (III. Def. 11); therefore 2. The angle ACB is equal to the angle ADB; the exterior angle to the interior, which is impossible (I. 16). Therefore, there cannot be two similar segments of circles upon the same side of the same line, which do not coincide. Q.E.D. PROPOSITION 24.-Theorem. Similar segments of circles upon equal straight lines, are equal to one another. Let AEB, CFD be similar segments of circles upon the equal straight lines AB, CD. Then the segment AEB shall be equal to the segment CFD. Demonstration. For if the segment AEB be applied to the segment CFD, so that the point A may be on C, and the straight line AB upon CD, then 1. The point B shall coincide with the point D, because AB is equal to CD; therefore, the straight line AB coinciding with CD, 2. The segment AEB must coincide with the segment CFD (III. 23); and therefore 3. The segment AEB is equal to the segment CFD (I. Ax. 8). Wherefore, similar segments, &c. Q.E.D. PROPOSITION 25.- Problem. A segment of a circle being given, to describe the circle of which it is the segment. |