It may be demonstrated, as in the first case, that 1. The angle FEC is double of the angle FAC, and that 2. FEB, a part of FEC, is double of FAB, a part of FAC; 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. B C First, let the segment BAED be greater than a semicircle. Take (III. 1.) F the centre of the circle ABCD, and join BF, FD. And because the angle RFD 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 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. 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, upon the same side of the same straight line AB, not coinciding with one another. be A 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: CB 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 (III. 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. BROP. XXIV.-THEOREM. 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 A 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. PROP. XXV.-PROBLEM. 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 (I. 10.) AC in D, and from the point D draw (I. 11.) DB, at right angles to AC; and join AB. First, let the angles ABD, BAD, be equal to one another; then (I. 6.) The straight line BD is equal to DA, and therefore 1. to DC; and because the three straight lines DA, DB, DC, are all equal; (III. 9.) 2. D is the centre of the circle. 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, 3. The segment ABC is a semicircle. But if the angles ABD, BAD, are not equal to one another: At the point 4, in the straight line BA, make (I. 23.) 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, (I. 6.) 1. The straight line BE is equal 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 (Constr.) a right angle; therefore (I. 4.)` 2. The base AE is equal to the base EC: but AE was shown to be equal to BE, wherefore also 3. BE is equal to EC: and the three straight lines AE, EB, EC, are therefore equal to one another; wherefore (III. 9.) 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. And 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. Which was to be done. PROP. XXVI.-THEOREM. In equal circles, equal angles stand upon equal circumferences, whether they be at the centres or circumferences. Let ABC, DEF, be equal circles, having the equal angles BGC, EHF, at their centres, and BAC, EDF, at their circumferences: the circumference BKC is equal to the circumference ELF. H B Join BC, EF. And because the circles ABC, DEF, are equal, the straight lines drawn from their centres (III. Def. 1.) are equal: therefore 1. The two sides BG, GC, are equal to the two EH, HF, each to each; and the angle at G is equal (Hyp.) to the angle at H; therefore (I. 4.) 2. The base BC is equal to the base EF. And because the angle at 4 is equal (Hyp.) to the angle at D, (III. Def. 11.) 3. The segment BAC is similar to the segment EDF; and they are upon equal straight lines BC, EF; but similar segments of circles upon equal straight lines are equal (III. 24.) to one another; therefore 4. The segment BAC is equal to the segment EDF: but the whole circle ABC is equal (Hyp.) to the whole DEF; therefore (Ax. 3.) 5. The remaining segment BKC is equal to the remaining segment ELF, and the circumference BKC to the circumference ELF. Wherefore, in equal circles, &c. Q.E.D. |