6. Therefore the whole angle BEC is double of the whole angle BAC. Case II.—Next, let the centre E of the circle be without the angle BAC. Construction.-Join AE, and produce it to meet the circumference in F. Demonstration.-1. It may be demonstrated, as in the first case, that the angle FEC is double of the angle FAC, and that FEB, a part of FEC, is double of FAB, a part of FAC; 2. Therefore the remaining angle BEC is double of the remaining angle BAC. Conclusion. Therefore, the angle at the centre, &c. Q. E. D. PROPOSITION 21.-THEOREM. another. (References, Prop. III. 1, 20.) Hypothesis.—Let ABCD be a circle, and BAD, BED, angles in the same segment BAED. Sequence. The angles BAD, BED, shall be equal to one another. Case I. First, let the segment BAED be greater than a semicircle. Construction.-Take F the centre of the circle ABCD (III. 1), and join BF, DF. Demonstration.-1. Because the angle BFD is at the centre, and the angle BAD at the circumference, and that they have B the same arc for their base, namely, BCD; 2. Therefore the angle BFD is double of the angle BAD. (III. 20.) 3. For the same reason, the angle BFD is double of the angle BED; 4. Therefore the angle BAD is equal to the angle BED. (ax. 7.) Case II.-Next, let the segment BAED be not greater than a semicircle. Construction.-Draw AF to the centre, and produce it to C, and join CE. Demonstration.-1. Then the segment BADC is greater than a semicircle, and BL therefore the angles BAC, BEC in it, are equal, by the first case. 2. For the same reason, because the segment CBED is greater than a semicircle, the angles CAD, CED, are equal. 3. Therefore the whole angle BAD is equal to the whole ungle BED. (ax. 2.) Conclusion.—Therefore, the angles in the same segment, &c. Q. E. D. PROPOSITION 22.—THEOREM. The opposite angles of any quadrilateral figure inscribed in a circle, are together equal to two right angles. (References—Prop. I. 32; III. 21.) Hypothesis.—Let ABCD be a quadrilateral figure inscribed in the circle ABCD. Sequence.—Any two of its opposite angles shall be together equal to two right angles. Construction.-Join AC, BD. Demonstration.-1. Because the three angles of every triangle are together equal to two right angles, (1. 32.) 2. The three angles of the triangle CAB, namely, CAB, ACB, ABC, are together equal to two right angles. 3. But the angle ČAB is equal to the angle CDB, because they are in the same segment CDAB; (III. 21.) 4. And the angle ACB is equal to the angle ADB, because they are in the same segment ADCB; 5. Therefore the two angles CAB, ACB, are together equal to the whole angle ADC. (ax. 2.). 6. To each of these equals add the angle ABC; 7. Therefore the three angles CAB, ACB, ABC, are equal to the two angles ABC, ADC. 8. But the angles CAB, ACB, ABC, are together equal to two right angles; (I. 32.) 9. Therefore also the angles ABC, ADC, are together equal to two right angles. 10. In like manner it may be shown that the angles BAD, BCD, are together equal to two right angles. Conclusion.—Therefore, the opposite angles, &c. Q. E. D. PROPOSITION 23.-THEOREM. Upon the same straight line, and on the same side of it, there cannot be two similar segments of circles, not coinciding with one another. (References—Prop. I. 16; III. 10; def. 11.) Hypothesis.—If it be possible, on the same straight line AB, and on the same side of it, let there be two similar segments of circles ACB, ADB, not coinciding with oue another. Construction. — 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 one of the segments must á fall within the other. Let ACB fall within ADB; draw the straight line BCD, and join AC, AD. Demonstration.-1. Because the segment ACB is assumed to be similar to the segment ADB (hyp.), and that similar segments of circles contain equal angles; (III. def 11.) 2. Therefore the angle ACB is equal to the angle ADB; that is, the exterior angle of the triangle ACD, equal to the interior and opposite angle; which is impossible. (İ. 16.) Conclusion. Therefore, there cannot be two similar segments of circles on the same straight line, and on the same side of it, which do not coincide. 2. E. D. PROPOSITION 24.—THEOREM. Similar segments of circles upon equal straight lines are equal to one another. (Reference—Prop. III. 23.) Hypothesis. — Let AEB, CFD, be similar segments of circles upon the equal straight lines AB, CD, Sequence. — The segment AEB shall be equal to the segment CFD. Demonstration. — 1. For if the segment AEB A be applied to the segment CFD, so that the point A may be on the point C, and the straight line AB on the straight line CD, 2. Then the point B shall coincide with the point D, because AB is equal to CD. 3. And the straight line AB coinciding with CD, the segment AEB must coincide with the segment CFD (III. 23); and is therefore equal to it. Conclusion.—Therefore, similar segments, &c. Q. E. D. PROPOSITION 25.-PROBLEM. A segment of a circle being given, to describe the circle of vhich it is the segment. (References—Prop. I. 4, 6, 10, 11, 23; III. 9.) Given. --Let ABC be the given segment of a circle. Sought. It is required to describe the circle of which ABC is a segment. Construction.–1. Bisect AC in D. (1. 10.) 2. From the point D draw DB at right angles to AC (I. 11); and join AB. MDC Case I.–First, let the angles ABD, BAD, be equal to one another. Then D shall be the centre of the circle required. Proof.-1. Because the angle ABD is equal to the angle BAD; (hyp.) 2. Therefore DB is equal to DA. (I. 6.) 5. Therefore the three straight lines DA, DB, DC, are all equal; 6. And therefore D is the centre of the circle. (III. 9.) 7. Hence, if from the centre D, at the distance of any of the three lines, DA, DB, DC, a circle be described, it will pass through the other two points, and be the circle required. Case II.—Next, let the angles ABD, BAD, be not equal to one another. Construction.—1. At the point A, in the straight line AB, make the angle BAE equal to the angle ABD; (I. 23.) 2. Produce BD, if necessary, to E, and join EC. Then E shall be the centre of the circle required. Proof.-1. Because the angle BAE is equal to the angle ABE (const.), EA is equal to EB. (I. 6.) 2. And because AD is equal to CD (const.), and DE is common to the two triangles ADE, CDE; 3. The two sides AD, DE, are equal to the two sides CD, DE, each to each; 4. And the angle ADE is equal to the angle CDE, for each of them is a right angle; (const.) 5. Therefore the base EA is equal to the base EC. (I. 4.) 6. But EA was shown to be equal to EB; 7. Therefore EB is equal to EC. (ax. 1.) 8. Therefore the three straight lines EA, EB, EC, are all equal; *9. And therefore E is the centre of the circle. (III. 9.) 10. Hence, if from the centre E, at the distance of any of the three lines, EA, EB, EC, a circle be described, it will pass through the other two points, and be the circle required. 11. In the first case, it is evident that, because the centre D is in AC, the segment ABC is a semicircle. 12. In the second case, if the angle ABD be greater than BAD, the centre E falls without the segment ABC, which is therefore less than a semicircle; 13. But if the angle ABD be less than the angle BAD, the centre E falls within the segment ABC, which is therefore greater than a semicircle. Conclusion.—Therefore, a segment of a circle being given, the circle has been described of which it is a segment. Q.E.F. PROPOSITION 26.—THEOREM. In equal circles, equal angles stand upon equal arcs, whether they be at the centres or at the circumferences. (References—Prop. I. 4; III. 24, def. 1, 11.) Hypothesis.—Let ABC, DEF, be equal circles, having the equal angles BGC, EHF, at their centres, and BÁC, EDF, at their circumferences. Sequence. The arc BKC shall be equal to the arc ELF. Construction.-Join BC, EF. Demonstration.-1. Because the circles ABC, DEF, are equal (hyp.), the straight lines from their centres are equal; (III. def. 1.) 2. Therefore the two sides BG, GC, are equal to the two sides EH, HF, each to each; 3. And the angle at G is equal to the angle at H; (hyp.) |