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, B for their base; therefore the angle BFD is 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 to C, and B 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. F D F 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. 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, ADC, are equal to the angles ABC, CAB, BCA. But ABC, CAB, BCA are equal to two right angles (32. 1.); therefore also the 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. be If it be possible, let the two similar segments of circles, viz. ACB, ADB, 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 (9. def. 3.) equal angles, the angle 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 coinciding with CD, the segment AEB must (23. 3.) coincide with the segment CFD, and therefore is equal to it. A segment of a circle being given, to describe the circle of which it is the segment. Let ABC be the given segment of a circle; it is required to describe the circle of which it is the segment. Bisect (10. 1.) AC in D, and from the point D draw (11. 1.) DB at right angles to AC, and join AB: First, let the angles ABD, BAD be equal to one another; then the straight line BD is equal (6. 1.) to DA, and therefore to DC; and because the three straight lines DA, DB, DC, are all equal; D is the centre of the circle (9. 3.); from the centre D, at the distance of any of the three DA, DB, DC, describe a circle; this shall pass through the other points; and the circle of which ABC is a segment is described and because the centre D is in AC, the segment ABC is a semicircle. Next, let the angles ABD, BAD be unequal; at the point A, in the straight line AB, make (23. 1.) the angle BAE equal to the angle ABD, and produce BD, if necessary, to E, and join EC: and because the angle ABE is equal to the angle BAE, the straight line BE is equal (6. 1.) to EA and because AD is equal to DC, and DE common to the triangles ADE, CDE, the two sides AD, DE are equal to the two CD, DE, each to each; and the angle ADE is equal to the angle CDE, for each of them is a right angle; therefore the base AE is equal (4. 1.) to the base EC: but AE was shewn to be equal to EB, wherefore also BE is equal to EC: and the three straight lines AE, EB, EC are therefore equal to one another; wherefore (9. 3.) E is the centre of the circle. From the centre E, at the distance of any of the three AE, EB, EC, describe a circle, this shall pass through the other points; and the circle of which ABC is a segment is described: also, it is evident, that if the angle ABD be greater than the angle BAD, the centre E falls without the segment ABC, which therefore is less than a semicircle; but if the angle ABD be less than BAD, the centre E falls within the segment ABC, which is therefore greater than a semicircle Wherefore, a segment of a circle being given, the circle is described of which it is a segment. PROP. XXVI. 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 (4. 1.) to the base EF: and because the angle at A is equal to the angle at D, the segment BAC is similar (9. 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. XXVII. 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 circumferences 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 (23. 1.) BGK equal to the angle EHF. And because equal angles stand upon equal arcs (26. 3.), when they are at the centre, the arc BK is equal to the arc EF: but EF 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 equal to it and the angle at A is half the angle BGC, and the angle at D half of the angle EHF; therefore the angle at A is equal to the angle at D. PROP. XXVIII. THEOR. In equal circles, equal straight lines cut off equal arcs, the greater equal to the greater, and the less to the less. Let ABC, DEF be equal circles, and BC, EF equal straight lines in them, which cut off the two greater arcs BAC, EDF, and the two less BGC, EHF: the greater BAC is equal to the greater EDF, and the less BGC to the less EHF. Take (1. 3.) K, L, the centres of the circles, and join BK, KC, EL, LF; and because the circles are equal, the straight lines from their centres are equal; therefore BK, KC are equal to EL, LF; but the base BC is also equal to the base EF; therefore the angle BKC is equal (8. 1.) to the angle ELF and equal angles stand upon equal (26. 3.) arcs, when they are at the centres; therefore the arc BGC is equal to the arc EHF. But the whole circle ABC is equal to the whole EDF; the remaining part, therefore, of the circumference viz. BAC, is equal to the remaining part EDF. In equal circles equal arcs are subtended by equal straight lines. Let ABC, DEF be equal circles, and let the arcs BGC, EHF also be equal; and join BC, EF: the straight line BC is equal to the straight line EF. Take (1. 3.) K, L the centres of the circles, and join BK, KC, EL, LF : and because the arc BGC is equal to the arc EHF, the angle BKC is equal (27. 3.) to the angle ELF: also because the circles ABC, DEF are equal, their radii are equal: therefore BK, KC are equal to EL, LF: and they contain equal angles; therefore the base BC is equal (4. 1.) to the base EF. |