Conversely. If the two lines be AB, CD, which touch the circumference, and if, at the same time, the intercepted arcs EJF, EKF are equal, EF must be a diameter (Th. 1. 3.); and therefore AB, CD (Cor. 3. Th. 16. 3.), are parallel. But if only one of the lines, as AB, touch, while the other, GH, cuts the circumference, making the arcs EG, EH equal; then the diameter FE, which bisects the arc GEH, is perpendicular (Schol. Th. 13. 3.) to its chord GH: it is also perpendicular to the tangent AB; therefore AB, GH are parallel. If both lines cut the circle, as GH, JK, and intercept equal arcs GJ, HK; let the diameter FE bisect one of the chords, as GH: it will also bisect the arc GEH, so that EG is equal to EH; and since GJ is (by hyp.) equal to HK, the whole arc EJ is equal to the whole arc EK; therefore the chord JK is bisected by the diameter FE: hence, as both chords are bisected by the diameter FE, they are perpendicular to it; that is, they are parallel (Cor. Th. 20. 1.). SCHOLIUM. The restriction in the enunciation of the converse proposition, namely, that the lines do not cut each other within the circle, is necessary; for lines drawn through the points G, K, and J, H, will intercept equal arcs GJ, HK, and yet not be parallel, since they will intersect each other within the circle. PROP. XVIII. THEOR. If a straight line touch a circle, the straight line drawn from the centre to the point of contact, is perpendicular to the line touching the circle. A Let the straight line DE touch the circle ABC in the point C; take tne 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 must be (10. 1.) an acute angle; and to the greater angle the greater (12. 1.) side 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 shewn, that no other line but FC can be perpendicular to DE; FC is therefore perpendicular to DE. F B D E PROP. XIX. THEOR. 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 is in that line. 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 perpendicular (18. 3.) 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 F is not the centre of the circle ABC: In the same manner it may be shewn, that no other point which is not in CA, is the centre; that is, the centre is in CA. PROP. XX. THEOR. 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 BDC an angle at the centre, and BAC an angle at the circumference which have the same arc BC for the base; the angle BDC is double of the angle BAC. First, let D, the centre of the circle, be within the angle BAC, and join AD, and produce it to E: Because DA is equal to DB, the angle DAB is equal (3. 1.) to the angle DBA: therefore the angles DAB, DBA together are double of the angle DAB; but the angle BDE is equal (25. 1.) to the angles DAB, DBA; therefore also the angle BDE is double of the angle DAB; For the same reason, the angle EDC is double of the angle DAC: Therefore the whole angle BDC is double of the whole angle BAC. Again, let D, the centre of the circle, be without the angle BAC; and join AD and produce it to E. It may be demonstrated, as in the first case, that the angle EDC is double of the angle DAC, and that EDB, a part of the first, is double of DAB, a part of the other; therefore the remaining angle BDC is double of the remaining angle BAC. E B B E D D COR. If, in a circle, two chords drawn from a point in the circumference, be respectively equal to two chords drawn from another point, they shall include equal angles: for, the equal chords subtending equal arcs, (2. Cor. Th. 13. 3.), each angle must include the same portion of the circumference. PROP. XXI. THEOR. 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. 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 angle BFD is at the centre, and the angle BAD at the circumference, both having the same part of the circumference, viz. BCD, for their base; therefore the angle BFD is B double (20. 3.) of the angle BAD: for the same reason, the angle BFD is double of the angle BED: Therefore the angle BAD is equal to the angle BED. 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, B 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. PROP. XXII. THEOR. A The opposite angles of any quadrilateral figure inscribed in a circle, are together equal to two right angles. D 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. The angle CAB is equal (21. 3.) 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 CAB, ACB: To each of these equals add the angle ABC; and the angles ABC, A ADC, are equal to the angles ABC, CAB, BCA. But ABC, CAB, BCA are equal to two right angles (25. 1.); therefore also the B angles ABC, ADC are equal to two right angles; In the same manner, the angles BAD, DCB may be shewn to be equal to two right angles. COR. 1. If any side of a quadrilateral be produced, the exterior angle will be equal to the interior opposite angle. COR. 2. It follows, likewise, that a quadrilateral, of which the opposite angles are not equal to two right angles, cannot be inscribed in a circle. PROP. XXIII. THEOR 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. D 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 circles ACB, ADB, cut one another in the two points A, B, they cannot cut one another in any other point (10. 3.): 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, and similar segments of circles contain (12. def. 3.) equal angles, the angles ACB is equal to the angle ADB, the exterior to the interior, which is impossible (16. 1.) PROP. XXIV. THEOR. A B 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 as the point A be on C, and the straight li upon CD, the point B shall coincide with the point D, be cause AB is equal to CD: AB coinciding with CD, the segment AEB must (23. 3.) coincide with the segment CFD, and therefore is equal to it. PROP. XXV. THEOR. In equal circles, equal angles stand upon equal arcs, whether they be at the centres or circumferences. Let ABC, DEF be equal circles, and the equal angles BGC, EHF at their centres, and BAC, EDF at their circumferences: the arc BKC is equal to the arc ELF. Join BC, EF; and because the circles ABC, DEF are equal, the straight lines drawn from their centres are equal; therefore the two sides BG, GC, are equal to the two EH, HF; and the angle at G is equal to the angle at H; therefore the base BC is equal (1. 1.) to the base EF: and because the angle at A is equal to the angle at D, the segment BAC is similar (12. def. 3.) to the segment EDF; and they are upon equal straight lines BC, EF; but similar segments of circles upon equal straight lines are equal (24. 3.) to one another, therefore the segment BAC is equal to the segment EDF: but the whole circle ABC is equal to the whole DEF; therefore the remaining segment BKC is equal to the remaining segment ELF, and the arc BKC to the arc ELF.· PROP. XXVI. THEOR. In equal circles, the angles which stand upon equal arcs are equal to one another, whether they be at the centres or circumferences. Let the angles BGC, EHF at the centres, and BAC, EDF at the cir cumferences of the equal circles ABC, DEF, stand upon the equal arcs BC, EF: the angle BGC is equal to the angle EHF, and the angle BAC to the angle EDF. If the angle BGC be equal to the angle EHF, it is manifest (20. 3.) that the angle BAC is also equal to EDF. But, if not, one of them is the greater let BGC be the greater, and at the point G, in the straight line BG, make the angle (Prob. 9. 1.) BGK equal to the angle EHF. And because equal angles stand upon equal arcs (25. 3.), when they are at the centre, the arc BK is equal to the arc EF: but FF is equal to BC; therefore also BK is equal to BC, the less to the greater, which is impossible. Therefore the angle BGC is not unequal to the angle EHF; that is, it is |