BCD; it is required to draw a straight line from A which Book 111. shall touch the circle. Find a the centre E of the circle, and join AE; and from * 1. 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 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 AE, EB are equal to the two FE, ED, and they contain the angle at E common to the two triangles AEB, FED; therefore the base DF is equal to the base AB; and the tri angle FED to the triangle AEB, and the other angles to the other angles: Therefore the angle EBA is equal to 4. 1. 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 circled: Therefore AB Cor. 16.3. touches the circle; and it is drawn 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 DE; DF touches the circle. d PROP. XVIII. THEOR. 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 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 isa a 17. 1. an acute angle; and to the greater angle the greatest side ↳ 19. 1. b Book III. is opposite; Therefore FC is greater than FG; but FC is equal to FB; therefore FB is 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. F B D GE PROP. XIX. THEOR. IF a 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 B ABC, and FC is drawn from the centre to the point of contact, 18. 3. FC is perpendiculara to DE; therefore FCE is a right angle But ACE is also a right angle; therefore the angle FCE is equal to the angle ACE, the less to the greater, which is impossible: Wherefore Fis 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 is in CA. Therefore, if a straight line, &c. Q. E. D. D C E PROP. XX. THEOR. See N. 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. Let ABC be a circle, and BEC an angle at the centre, Boox III. and BAC an angle at the circumference, which have them same circumference BC for their base; the angle BEC is double of the angle BAC. 5. 1. E b52.1. 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 a to the angle EBA; therefore the angles EAB, EBA are double of the angle EAB; but the angle BEF is equal b to the angles EAB, EBA; 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. 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 of the angle GDC, and that GEB, a part of the first, is double of GDB, a part of the other; there- G fore the remaining angle BEC is double of the remaining angle BDC. Therefore the angle at the centre, &c. B F E B Q. E. D. 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, F and that they have the same part of the circumference, viz. Book III. BCD for their base; therefore the angle BFD is double a of the angle BAD: For the same reason, the angle BFD 20. 3. is double of the angle BED: Therefore the angie BAD is equal to the angle BED. B E D F But, if the segment BAED be not greater than a semicircle, let BAD, BED be angles in it; these also are equal to one another: 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 C 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. 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 32. 1. triangle are equala to two right angles, the three angles of the triangle CAB, viz. the angles CAB, ABC, BCA, are equal to two right angles: But the B 21. 3. 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 segA ment ADCB: Therefore the whole angle ADC is equal to the angles CAB, ACB: To each of these equals add the angle ABC; therefore the angles ABCCAB, BCA are equal to the angles ABC, ADC: 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, the angles BAD, DCB, may be shown Book III. to be equal to two right angles. Therefore, 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. 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 D 10. S. B. с b to the segment ADB, and that similar segments of circles contain b equal angles; the angle ACB is equal to the angle ADB, the exterior to the interior, which is impossible. « 16. 1. Therefore there cannot be two similar segments of a circle 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 See N 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 as the A point A be on E C C, and the straight line AB upon CD, the point B shall coincide with the point D, because AB is equal to CD; |