DEMONSTRATION And because FGC is assumed to be a right angle, therefore GCF must be an acute angle; (1. 17) and to the greater angle the greater side is opposite; (1. 19) therefore FC must be greater than FG, but FC is equal to FB, (1. def. 15) therefore FB must be greater than FG, the less than the greater, which is impossible; wherefore FG is not perpendicular to DE. In the same manner it may be shown, that no other is perpendicular to it besides FC; that is, FC is perpendicular to DE. Therefore, if a straight line, &c. Q. E. D. PROP. XIX.-THEOREM. If a straight line touch a circle, and from the point of contact a straight line be drawn at right angles to the touching line, the centre of the circle shall be in that line. (References - Prop. II. 18.) Let the straight line DE touch the circle ABC in C; and from C let CA be drawn at right angles to DE. Then the centre of the circle shall be in CA. For, if not, let F be the centre, if possible, and join CF. DEMONSTRATION Because DE touches the circle ABC, and FC is drawn from the assumed centre to the point of contact, therefore FC must be perpendicular to DE; (m. 18) and therefore FCE must be a right angle; but ACE is a right angle, (hyp.) therefore the angle FCE must be equal to the angle ACE, (ax. 1) the less to the greater, which is impossible. wherefore F is not the centre of the circle ABC. In the same manner, it may be shown that no other point which is not in CA is the centre; that is, the centre of the circle is in CA. Therefore, if a straight line, &c. Q. E. D. PROP. XX.-THEOREM. The angle at the centre of a circle is double of the angle at the circumference upon the same base, that is, upon the same part of the circumference. (References-Prop. 1. 5, 32.) Let ABC be a circle, and BEC an angle at the centre, and BAC an angle at the circumference, which have the same circumference BC for their base. Then the angle BEC shall be double of the angle BAC. First, let the centre E of the circle be within the angle BAC. CONSTRUCTION Join AE, and produce it to the circumference in F. Because EA is equal to EB, DEMONSTRATION the angle EAB is equal to the angle EBA, (1. 5) therefore the angles EAB, EBA are double of the angle EAB; but the angle BEF is equal to the angles EAB, EBA; (1. 32) therefore also the angle BEF is double of the angle EAB; for the same reason, the angle FEC is double of the angle EAC; therefore the whole angle BEC is double of the whole angle BAC. Secondly, let the centre E of the circle be without the angle BAC. F B E CONSTRUCTION Join AE, and produce it to meet the circumference in F. DEMONSTRATION It may be demonstrated, as in the first case, that the angle FEC is double of the angle FAC, and that FEB, a part of FEC, is double of FAD, a part of FAC; therefore the remaining angle BEC is double of the remaining angle BAC. Therefore, the angie at the centre, &c. Q. E. D. PROP. XXI.—THEOREM. The angles in the same segment of a circle are equal to one another. (References-Prop. I. 1, 20.) 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. First, let the segment BAED be greater than a semicircle CONSTRUCTION Take F the centre of the circle ABCD, (III. 1) and join BF, FD. DEMONSTRATION And, 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. BCD for their base; therefore the angle BFD is double of the angle BAD; (III. 20) for the same reason, the angle BFD is double of the angle BED; therefore the angle BAD is equal to the angle BED. (ax. 7.) Next, let the segment BAED be not greater than a semicircle. E B D F 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 by the first case, the angles BAC, BEC, in it, are equal; for the same reason, because CBED is greater than a semicircle, the angles CAD, CED, are equal; therefore the whole angle BAD is equal to the whole angle BED. (ax. 2.) Wherefore, the angles in the same segment, &c. Q. E. D. PROP. XXII.-THEOREM. The opposite angles of any quadrilateral figure inscribed in a circle, are together equal to two right angles. (References-Prop. I. 32; 111. 21.) Let ABCD be a quadrilateral figure in the circle ABCD. Then any two of its opposite angles shall be together equal to two right angles. Because the three angles of every triangle are equal to two right angles, (1. 32) the three angles of the triangle CAB, viz. the angles CAB, ABC, BCA, are equal to two right angles; but the angle CAB is equal to the angle CDB, (III. 21) because they are in the same segment CDAB; and the angle ACB is equal to the angle ADB, because they are in the same segment ADCB; therefore the whole angle ADC is equal to the angles CAB, ACB; to each of these equals add the angle ABC; therefore the three angles ABC, CAB, BCA, are equal to the two angles ABC, ADC; (ax. 2) but ABC, CAB, BCA, are equal to two right angles; therefore also the angles ABC, ADC, are equal to two right angles. In the same manner, it may be shown that the angles BAD, DCB, are equal to two right angles. Therefore, the opposite angles, &c. Q. E. D. PROP. XXIII.- 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. |