PROP. XVIII.-THEOREM. If a straight line touches a circle, the straight line drawn from the centre to the point of contact, shall be perpendicular to the line touching the circle. Let the straight line DE touch the circle ABC in the point C; take (III. 1.) the centre F, and draw the straight line FC; FC is perpendicular to DE. For, if it be not, from the point F draw FBG perpendicular to DE; and because FGC is a right angle, (I 17.) 1. GCF is an acute angle; and to the greater angle the greater side (I. 19.) is opposite; therefore 2. FC is greater than FG; but FC is equal (I. Def. 15.) to FB; therefore the less than the greater, which is impossible; wherefore In the same manner it may be shown, that no other is perpendicular to it besides FC; that is 5. 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. Let the straight line DE touch the circle ABC in C; and from Clet CA be drawn at right angles to DE; the centre of the circle is in CA. For, if not, let F be the centre, if possible, and join CF. Because DE touches the circle ABC, and FC is drawn from the centre to the point of contact, (III. 18.) 1. FC is perpendicular to DE; therefore 2. FCE is a right angle: but ACE is also a right angle (Hyp.); therefore 3. The angle FCE is equal to the angle ACE, the less to the greater, which is impossible. Wherefore 4. 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, 5. The centre of the circle is in CA. Therefore, if a straight line, &c. Q.E.D. PROP. XX.-THEOREM. The angie 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. 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; the angle BEC is double of the angle BAC. First let E the centre of the circle be within the angle BAC, and join AE, and produce it to F. Because EA is equal to EB, the angle EAB is equal (I. 5.) to the angle EBA; therefore 1. The angles EAB, EBA, are double of the angle EAB; but the angle BEF is equal (I. 32.) to the angles EAB, EBA; therefore also 4. The whole angle BEC is double of the whole angle BAC. Again, let E the centre of the circle be without the angle BAC. B It may be demonstrated, as in the first case, that 1. and that 2. therefore 3. The angle FEC is double of the angle FAC, FEB, a part of FEC, is double of FAB, a part of FAC; The remaining angle BEC is double of the remaining angle BAC. Therefore, the angle at the centre, &c. . Q.E.D. PROP. XXI.—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: the angles BAD, BED, are equal to one another. C Take First, let the segment BAED be greater than a semicircle. (III. 1.) F the centre of the circle ABCD, and join BF, FD. 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 (III. 20.) 1. The angle BFD is double of the angle BAD: for the same reason, 2. The angle BFD is double of the angle BED: therefore (Ax. 7.) 3. The angle BAD is equal to the angle BED. Next, let the segment BAED be not greater than a semicircle. Draw AF to the centre, and produce it to C, and join CE: therefore the segment BADC is greater than a semicircle, and 1. The angles BAC, BEC, 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 (Ax. 2.) 3. The whole angle BAD is equal to the whole angle BED. 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. Let ABCD be a quadrilateral figure in the circle ABCD; any two of its opposite angles are together equal to two right angles. D B Join AC, BD; and because the three angles of every triangle are equal (I. 32.) to two right angles, the three angles of the triangle CAB, viz. 1. The angles CAB, ABC, BCA, are equal to two right angles: but (III. 21.) 2. The angle CAB is equal to the angle CDB, because they are in the same segment BADC; and 3. The angle ACB is equal to the angle ADB, because they are in the same segment ADCB: therefore 4. The whole angle ADC is equal to the angles CAB, ACB: to each of these equals add the angle ABC; therefore (Ax. 2.) The angles ABC, CAB, BCA, are equal to the angles 5. ABC, ADC: but ABC, CAB, 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. 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. If it be possible, let the two similar segments of circles, viz. ACB, ADB, be upon the same side of the same straight line AB, not coinciding with one another. Then, because the circle ACB cuts the circle ADB in the two points A, B, they cannot cut one another in any other point (III. 10.): therefore other: 1. One of the segments ACB, ADB, must fall within the let ABC fall within ADB, and draw the straight line BCD, and join CA, DA. And because the segment ACB is similar (Hyp.) to the segment ADB, and that similar segments of circles contain`(IÏÏ. Def. 11.) equal angles; 2. The angle ACB is equal to the angle ADB, the exterior 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. 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; the segment AEB is equal to the segment CFD. For if the segment AEB be applied to the segment CFD, so that the point 4 may be on C, and the straight line AB on CD, 1. The point B shall coincide with the point D, because AB is equal to CD: therefore the straight line AB coinciding with CD, (III. 23.) 2. The segment AEB must coincide with the segment CFD, and therefore (Ax. 8.) 3. The segment AEB is equal to the segment CFD. Wherefore similar segments, &c. Q.E.D. |