In equal circles the arcs, which are cut off by equal chords, shall be equal , the major arc equal to the major arc, and the minor to the minor. A к B Let ABC, DEF be equal circles ; and let the chord BC= the chord EF. and the minor arc BGC = the minor arc EHF. Construction. and join BK, KC, EL, LF. Proof. Because the Os ABC, DEF are equal, .:. their radii are equal. BK = EL, KC=LF, Hyp. .:. the BKC = the < ELF. I. 8. the arc BGC = the arc EHF; for these arcs subtend equal angles at the centre; III. 26. and they are the minor arcs. But the whole Oce ABGC = the whole Oce DEHF; Hyp. .. the remaining arc BAC = the remaining arc EDF: and these are the major arcs. Q.E.D. [For Exercises see p. 212.] PROPOSITION 29. THEOREM. In equal circles the chords, which cut off equal arcs, shall be equal. A D Construction. Find K, L the centres of the circles. Join BK, KC, EL, LF. because the arc BGC = the arc EHF, Hence in the Aø BKC, ELF, BK= EL, being radii of equal circles ; Because KC=LF, for the same reason, and the BKC = the ELF; Proved. i. BC= EF. I. 4. Q.E.D. EXERCISES. ON PROPOSITIONS 26, 27. 1. If two chords of a circle are parallel, they intercept equal arcs. 2. The straight lines, which join the extremities of two equal arcs of a circle towards the same parts, are parallel. 3. In a circle, or in equal circles, sectors are equal if their angles at the centres are equal. 4. If two chords of a circle intersect at right angles, the opposite arcs are together equal to a semi-circumference. 5. If two chords intersect within a circle, they form an angle equal to that subtended at the circumference by the sum of the arcs they cut off. 6. If two chords intersect without a circle, they form an angle equal to that subtended at the circumference by the difference of the arcs they cut off. 7. If AB is a fixed chord of a circle, and P any point on one of the arcs cut off by it, then the bisector of the angle APB cuts the conjugate arc in the same point, whatever be the position of P. 8. Two circles intersect at A and B ; and through these points straight lines are drawn from any point P on the circumference of one of the circles : shew that when produced they intercept on the other circumference an arc which is constant for all positions of P. 9. A triangle ABC is inscribed in a circle, and the bisectors of the angles meet the circumference at X, Y, Z. Find each angle of the triangle XYZ in terms of those of the original triangle. ON PROPOSITIONS 28, 29. 10. The straight lines which join the extremities of parallel chords in a circle (i) towards the same parts, (ii) towards opposite parts, are equal. 11. Through A, a point of intersection of two equal circles, two straight lines PAR, XAY are drawn : shew that the chord PX is equal to the chord QY. 12. Through the points of intersection of two circles two parallel straight lines are drawn terminated by the circumferences: shew that the straight lines which join their extremities towards the same parts are equal. 13. Two equal circles intersect at A and B; and through A any straight line PÂQ is drawn terminated by the circumferences: shew that BP=BQ. 14. ABC is an isosceles triangle inscribed in a circle, and the bisectors of the base angles meet the circumference at X and Y. Shew that the figure BXAYC must have four of its sides equal. What relation must subsist among the angles of the triangle ABC, in order that the figure BXAYC may be equilateral ? Note. We have given Euclid's demonstrations of Propositions 26, 27 ; but it should be noticed that these propositions also admit of proof by the method of superposition. To illustrate this method we will apply it to Proposition 26. PROPOSITION 26. [ALTERNATIVE PROOF.] In cqual circles, the arcs which subtend equal angles, whether at the centres or circumferences, shall be equal. A F к S Let ABC, DEF be equal circles ; and let the 2$ BGC, EHF at the centres be equal, and consequently the LS BAC, EDF at the Ocos equal. 11. 20. Then shall the arc BKC=the arc ELF. .. Proof. For if the O ABC be applied to the O DEF, so that the centre G may fall on the centre H, then because the circles are equal, Hyp. their O ces must coincide ; hence by revolving the upper circle about its centre, the lower circle remaining fixed, B may be made to coincide with E, and consequently GB with HE. And because the LBGC=the L EHF, :: GC must coincide with HF : and since GC=HF, : C must fall on F. Now B coincides with E, and C with F, and the Oce of the O ABC with the Oce of the O DEF; : the arc BKC must coincide with the arc ELF. .. the arc BKC=the arc ELF. Q. E.D. Hyp. Нур. Let ADB be the given arc. It is required to bisect the arc ADB. At C draw CD at rt. angles to AB, meeting the given arc at D. 1. 11. Then shall the arc ADB be bisected at D. Join AD, BD. In the As ACD, BCD, Constr. Because and CD is common; and the 2 ACD = the - BCD, being rt. angles : AD= BD. 1. 4. And since in the O ADB, the chords AD, BD are equal, ... the arcs cut off by them are equal, the minor arc equal to the minor, and the major arc to the major : and the arcs AD, BD are both minor arcs, for each is less than a semi-circumference, since DC, bisecting the chord AB at rt. angles, must pass through the centre of the circle. III. 1. Cor. ... the arc AD= the arc BD: that is, the arc ADB is bisected at D. Q.E.F. .. III. 28. a EXERCISES. 1. If a tangent to a circle is parallel to a chord, the point of contact will bisect the arc cut off by the chord. 2. Trisect a quadrant, or the fourth part of the circumference, cf a circle. a |