4. Therefore the base BC is equal to the base EF. (I. 4.) 5. And because the angle at A is equal to the angle at D, (hyp.) 6. The segment BAC is similar to the segment EDF; (III. def. 11.) 7. And they are on equal straight lines BC, EF. 8. But similar segments of circles on equal straight lines are equal to one another; (III. 24.) 9. Therefore the segment BAC is equal to the segment EDF. 10. But the whole circle ABC is equal to the whole circle DEF; (hyp.) 11. Therefore the remaining segment BKC is equal to the remaining segment ELF; (ax. 3.) 12. Therefore the arc BKC is equal to the arc ELF. Conclusion.-Therefore, in equal circles, &c. Q. E. D. PROPOSITION 27.-THEOREM. In equal circles, the angles which stand upon equal arcs are equal to one another, whether they be at the centres or at the circumferences. (References-Prop. I. 23; III. 20, 26.) Hypothesis. Let ABC, DEF, be equal circles, and let the angles BGC, EHF, at their centres, and the angles BAC, EDF, at their circumferences, stand on equal arcs BC, EF. Sequence. The angle BGC shall be equal to the angle EHF, and the angle BAC equal to the angle EDF. Construction.-If the angle BGC be equal to the angle EHF, it is manifest that the angle BAC is also equal to the angle EDF. (III. 20, ax. 7.) But, if not, one of them must be the greater. Let BGC be the greater, and at the point G, in the straight line BG, make the angle BGK equal to the angle EHF. (I. 23.) Demonstration.-1. Because the angle BGK is equal to the angle EHF, and that in equal circles equal angles stand on equal arcs, when they are at the centres; (III. 26.) 2. Therefore the arc BK is equal to the arc EF. 3. But the arc EF is equal to the arc BC; (hyp.) 4. Therefore the arc BK is equal to the arc BC (ax. 1); the less to the greater, which is impossible. 5. Therefore the angle BGC is not unequal to the angle EHF, that is, it is equal to it. 6. And the angle at A is half of the angle BGC, and the angle at D is half of the angle EHF; (III. 20.) 7. Therefore the angle at A is equal to the angle at D. (ax. 7.) Conclusion. Therefore, in equal circles, &c. Q. E. D. PROPOSITION 28.-THEOREM. In equal circles, equal chords cut off equal arcs, the greater equal to the greater, and the less equal to the less. (References-Prop. I. 8; III. 1, 26, def. 1.) Hypothesis.-Let ABC, DEF, be equal circles, and BC, EF, equal chords in them, which cut off the two greater arcs BAC, EDF, and the two less arcs BGC, EHF. Sequence. The greater arc BAC shall be equal to the greater arc EDF, and the less arc BGC equal to the less are EHF. Construction.-Take K, L, the centres of the circles (III. 1), and join BK, KC, EL, LF. A K D Demonstration.-1. Because the circles ABC, DEF, are equal, their radii are equal; (III. def. 1.) 2. Therefore the two sides BK, KC, are equal to the two sides EL, LF, each to each; 3. And the base BC is equal to the base EF; (hyp.) 4. Therefore the angle BKC is equal to the angle ELF. (I. 8.) 5. But in equal circles equal angles stand on equal arcs, when they are at the centres; (III. 26.) 6. Therefore the arc BGC is equal to the arc EHF. 7. But the whole circle ABC is equal to the whole circle DEF; (hyp.) 8. Therefore the remaining arc BAC is equal to the remaining arc EDF. (ax. 3.) Conclusion. Therefore, in equal circles, &c. Q. E. D. PROPOSITION 29.-THEOREM. In equal circles equal arcs are subtended by equal chords. (References-Prop. I. 4; III. 1, 27.) Hypothesis. Let ABC, DEF, be equal circles, and let BGC, EHF, be equal arcs in them, and join BC, EF. Sequence. The chord BC shall be equal to the chord EF. Construction. Take K, L, the centres of the circles (III. 1), and join BK, KC, EL, LF. E D Demonstration.-1. Because the arc BGC is equal to the arc EHF (hyp.), the angle BKC is equal to the angle ELF. (III. 27.) 2. And because the circles ABC, DEF, are equal (hyp.), their radii are equal; (III. def. 1.) 3. Therefore the two sides BK, KC, are equal to the two sides EL, LF, each to each; and they contain equal angles; 4. Therefore the base BC is equal to the base EF. (I. 4.) Conclusion. Therefore, in equal circles, &c. Q. E. D. PROPOSITION 30.-PROBLEM. To bisect a given arc, that is, to divide it into two equal parts. (References-Prop. I. 4, 10, 11; III. 1 cor., 28.) Given.-Let ADB be the given arc. Sought. It is required to bisect it. Construction.-1. Join AB, and bisect it in C; (I. 10.) 2. From the point C draw CD at right angles to AB (I. 11), and join AD and DB. Then the arc ADB shall be bisected in the point D. Proof.-1. Because AC is equal to CB (const.), and CD is common to the two tri- A angles ACD, BCD; В 2. The two sides AC, CD, are equal to the two sides BC, CD, each to each; 3. And the angle ACD is equal to the angle BCD, because each of them is a right angle; (const.) 4. Therefore the base AD is equal to the base BD. (I. 4.) 5. But equal chords cut off equal arcs, the greater equal to the greater, and the less equal to the less; (III. 28.) 6. And each of the arcs AD, DB, is less than a semicircle, because DC, if produced, is a diameter; (III. 1, cor.) 7. Therefore the arc AD is equal to the arc DB. Conclusion.-Therefore, the given arc is bisected in D. Q. E. F. PROPOSITION 31.-THEOREM. In a circle, the angle in a semicircle is a right angle; but the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle. (References-Prop. I. 5, 17, 32; III. 22.) Hypothesis.-Let ABC be a circle, of which BC is a diameter, and E the centre; and draw CA, dividing the circle into the segments ABC, ADC, and join BA, AD, DC. Sequence.-1. The angle in the semicircle BÁC shall be a right angle; 2. The angle in the segment ABC, which is greater than a semicircle, shall be less than a right angle; 3. The angle in the segment ADC, which is less than a semicircle, shall be greater than a right angle. Ꮐ Construction.-Join AE, and produce BA to F. Demonstration.-1. Because EA is equal to EB, (I. def. 15.) 2. The angle EAB is equal to the angle EBA; (I. 5.) 3. And, because EA is equal to EC, 4. The angle EAC is equal to the angle ECA; 5. Therefore the whole angle BAC is equal to the two angles ABC, ACB. (ax. 2.) 6. But FAC, the exterior angle of the triangle ABC, is equal to the two angles ABC, ACB; (I. 32.) 7. Therefore the angle BAC is equal to the angle FAC, (ax. 1.) B 8. And therefore each of them is a right angle. E (I. def. 10.) 9. Therefore the angle in a semicircle BAC is a right angle. 10. And because the two angles ABC, BAC, of the triangle ABC, are together less than two right angles (I. 17), and that BAC has been shown to be a right angle; 11. Therefore the angle ABC is less than a right angle. 12. Therefore the angle in a segment ABC, greater than a semicircle, is less than a right angle. 13. And because ABCD is a quadrilateral figure in a circle, any two of its opposite angles are together equal to two right angles; (III. 22.) 14. Therefore the angles ABC, ADC, are together equal to two right angles. 15. But the angle ABC has been shown to be less than a right angle; 16. Therefore the angle ADC is greater than a right angle. 17. Therefore the angle in a segment ADC, less than a semicircle, is greater than a right angle. Conclusion. Therefore, the angle, &c. Q. E. D. Corollary. From this demonstration it is manifest that if one angle of a triangle be equal to the other two, it is a right angle. For the angle adjacent to it is equal to the same two angles; (I. 32.) And when the adjacent angles are equal, they are right angles. (I. def. 10.) PROPOSITION 32.-THEOREM. The angles contained by a tangent to a circle and a chord drawn from the point of contact, are equal to the angles in the alternate segments of the circle. |