centre E falls within the segment ABC, which is therefore greater than a semicircle. 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. Let ABC, DEF be equal circles, having the equal angles BGC, EHF at their centres, and BAC, EDF at their circumferences. The arc BKC shall be equal to the arc ELF. PROOF.-Because the circles ABC, DEF are equal (Hyp.), the straight lines from their centres are equal (III. def. 1); Therefore the two sides BG, GC are equal to the two sides Triangles EH, HF, each to each; And the angle at G is equal to the angle at H (Hyp.); And because the angle at A is equal to the angle at D (Hyp.), BGC and equal in every re spect. B The segment BAC is similar to the segment EDF (III. def. 11), .seg ments BAC and EDF are similar and on And they are on equal straight lines BC, EF. But similar segments of circles on equal straight lines are equal equal to one another (III. 24); straight lines. Therefore the segment BAC is equal to the segment EDF... are But the whole circle ABC is equal to the whole circle equal. DEF (Hyp.); Therefore the remaining segment BKC is equal to the remaining segment ELF (Ax. 3). ... arc BKC = arc ELF. Therefore the arc BKC is equal to the arc ELF. 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. 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. The angle BGC shall be equal to the angle EHF, and the angle BAC equal to the angle EDF. G 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). PROOF.-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); Therefore the arc BK is equal to the arc EF. Therefore the angle BGC is not unequal to the angle And the angle at A is half of the angle BGC, and the angle at D is half of the angle EHF (III. 20); Therefore the angle at A is equal to the angle at D (Ax. 7). 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. 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. The greater arc BAC shall be equal to the greater arc EDF, and the less arc BGC equal to the less arc EHF. CONSTRUCTION.-Take K, L, the centres of the circles Take K (III. 1), and join BK, KC, EL, LF. and L the centres. K E PROOF.-Because the circles ABC, DEF are equal, their radii are equal (III. def. 1). Therefore the two sides BK, KC are equal to the two sides EL, LF, each to each; And the base BC is equal to the base EF (Hyp.); ELF (I. 8). But in equal circles equal angles stand on equal when they are at the centres (III. 26); Therefore the arc BGC is equal to the arc EHF. DEF (Hyp.); Therefore the remaining arc BAC is equal to the remaining arc EDF (Ax. 3). Therefore, in equal circles, &c. Q.E.D. Take K Proposition 29.—Theorem. In equal circles equal arcs are subtended by equal chords. Let ABC, DEF be equal circles, and let BGC, EHF be equal arcs in them, and join BC, EF. The chord BC shall be equal to the chord EF. CONSTRUCTION.-Take K, L, the centres of the circles and L the (III. 1), and join BK, KC, EL, LF. centres. PROOF.-Because the arc BGC is equal to the arc EHF (Hyp.), the angle BKC is equal to the angle ELF (III. 27). And because the circles ABC, DEF are equal (Hyp.), their radii are equal (III. def. 1). Therefore the two sides BK, KC are equal to the two sides EL, LF, each to each; and they contain equal angles; Therefore the base BC is equal to the base EF (I. 4). 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. Let ADB be the given aro. It is required to bisect it. CONSTRUCTION.-Join AB, and bisect it in C (I. 10). From the point C draw CD at right angles to AB (I. 11), and join AD and DB. Then the arc ABD shall be bisected in the point D. PROOF.-Because AC is equal to CB (Const.), and CD is common to the two triangles ACD, BCD; The two sides AC, CD are equal to the two sides BC, CD, each to each; And the angle ACD is equal to the angle BCD, because each of them is a right angle (Const.); Therefore the base AD is equal to the base BD (I. 4). And each of the arcs AD, DB is less than a semicircle, 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. 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. The angle in the semicircle BAC shall be a right angle; The angle in the segment ABC, which is greater than a semicircle, shall be less than a right angle; The angle in the segment ADC, which is less than a semicircle, shall be greater than a right angle. CONSTRUCTION. Join AE, and produce BA to F. EBA (I. 5); And, because EA is equal to EC, Therefore the whole angle BAC is equal to the two angles ABC, ACB (Ax. 2). But FAC, the exterior angle of the triangle ABC, is equal to the two angles ABC, ACB (I. 32). E base AD= base BD. < BAE + < EAC, or < BAC= < ABC + ACB= < FAC and .. a Therefore the angle BAC is equal to the angle FAC right angle. (Ax. 1), |