DA it is required to draw a straight line from A which shall touch Book IIT. the circle. Find a the centre E of the circle, and join AE: and from a l. 3. the centre E, at the distance EA, describe the circle AFG: from the point D drawb DF at right angles to EA, and join b 11. 1. EBF, AB: AB touches the circle BCD. Because E is the centre of the circles BCD, AFG, EA is equal to EF: And ED to EB; therefore, the two sides A E, EB are equal to the two FE, ED, and they contain the angle at E, common to the two triangles A EB, FED; therefore, the base DF is equal to the base AB, and the triangle FED to the triangle A EB, and the other angles to the other angles : c Therefore, c 4. 1. the angle EBA is equal to the angle EDF: But EDF is a right angle, wherefore EBA is a right angle: And EB is drawn from the centre: But a straight line drawn from the extremity of a diameter, at right angles to it, touches the circle: d Therefore AB touches the circle; and it is drawn d Cor. 16.3. from the given point A. Which was to be done. But if the given point be in the circumference of the circle, as the point D, draw DE to the centre E, and DF at right angles to to DE; DF touches the circled. PROP. XVIII. THEOR. If a straight line touches a circle, the straight line drawn from the centre to the point of contact is perpendicular to the line touching the circle. Let the straight line DE touch the circle ABC in the point C; take 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, GCF is a an acute a 17. 1. angle; and to the greater angle the greater side is opposite : b 19. 1. If a straight line touches a circle, and from the point of contact a straight line is drawn at right angles to the touching line, the centre of the circle is in that line. a 18. 3. Let the straight line DE touch the circle ABC in C, and from C let 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, FC is perpendiculara to DE; therefore FCE is a right angle: But ACE is also a right angle; therefore, the angle B FCE is equal to the angle ACE, 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, 7 that no other point which is not in CA, is the centre; that is, the centre is in CA. Therefore, “ if a straight line,” &c. Q. E. D. PROP. XX. THEOR. See N. The angle at the centre of a circle is double 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, Book III. and BAC an angle at the circumference, which have the m same circumference BC for their base ; the angle BEC is double the angle BAC. First, Let E, the centre of the circle, be within the angle BAC, and join A E, and produce it to F: Because EA is equal to EB, the angle EAB is equal a to the angle EBĂ; therefore a 5. 1. the angles EAB, EBA together, are double the angle EAB; but the angle BEF is equal to the angles EAB, b 32. 1. EBA; therefore also the angle BEF is double the angle EAB: For the same reason, the angle FEC is double the angle EAC: Therefore the whole angle BEC is double the whole angle BAC. Again, Let E, the centre of the circle, be without the angle BDC, and join DE, and produce it to G. It may be demonstrated, as in the first case, that the angle GEC is double the angle GDC, and that GEB, a part of the first, is double GDB, a part of the other; therefore, the remaining angle BEC is double the remaining angle BDC. Therefore, " the angle at the 66 centre," &c. R. E. D. А PROP. XXI. THEOR. The angles in the same segment of a circle are See N. 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. Take F, the centre of the circle ABCD: And, first, let the segment BAED be greater than a semicircle, and join BF, FD: And because the B angle BFD is at the centre, and the angle BAD at the circumference, both having the same part of the cir Book III. cumference, viz. BCD for their base; therefore the angle m BFD is double a the angle BAD: For the same reason, the a 20. 3. angle BFD is double the angle BED: Therefore the angle BAD is equal to the angle BED. But, if the segment BXED be not greater than a semicircle, let BAD, BED be angles in it; these also are equal to one an A E other: Draw AF to the centre, and produce it to C, and join CE: Therefore, the segment BADC is greater than a semicircle; and the angles in it BAC, BEC are equal, by the first case : 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. Wherefore, " the angles in the same segment,” &c. Q. E. D. PROP. XXII. THEOR. a 32. 1. b 21. 3. The opposite angles of any quadrilateral figure described 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. Join AC, BD; and because the three angles of every triangle are equala to two right angles, the three angles of the triangle CAB, viz. the angle CXB, ABC, BCA, are equal to two right angles : But the angle CAB is equal to the angle CDB, because they are in the same segment BADC, 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 CĂB, ACB: To each of these equals add the angle ABC; therefore, the angles ABC, CAB, BCA, are equal to the angles ABC, ADC: But ABC, CĂB, BCA are equal to two right angles; therefore also the angles ABC, ADC are equal to Book III, two right angles. In the same manner, the angles DAB, DCB, may be shown to be equal to two right angles. Therefore, 6 the opposite angles,” &c. Q. E. D. PROP. XXIII. THEOR. Upon the same straight line, and upon the same See N. 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. a 10. 3. other in any other pointa. One of the segments must therefore fall within the other: let ACB fall within ADB, draw the straight line BCD, and join CA, DA: And because the segment ACB is similar to the segment ADB, A and that similar segments of circles contain equal angles; the angle ACB is equal to the angle bll. def. 3. ADB, the exterior to the interior opposite, which is impossible c. Therefore, " there cannot be two similar segments of c 16. 1.. circles upon the same side of the same line, which do not coincide." Q. E. D. PROP. XXIV. THEOR. Similar segments of circles upon equal straight lines, See N. 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 as the point A be on C, and the straight line AB B C upon CD, the point B will coincide with the point D, because AB is equal |