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. B A Ꭰ C First, let the angles ABD, BAD, be equal to one another; then (I. 6.) 1. The straight line BD is equal to DA, and therefore 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: B B E E A D At the point 4, in the straight line B4, 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 BÃE, (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.) 4. E is the centre of the circle. 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. B L 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; C 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 A 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. PROP. XXVII.-THEOREM. In equal circles, the angles which stand upon equal circumferences are equal to one another, whether they be at the centres or circumferences. Let the angles BGC, EHF, at the centres, and BAC, EDF, at the circumferences of the equal circles, ABC, DEF, stand upon the equal circumferences BC, EF; the angle BGC is equal to the angle EHF, and the angle BAC to the angle EDF. If the angle BGC be equal to the angle EHF, it is manifest (III. 20.) that the angle BAC is also equal to EDF. But if not, one of them is the greater; let BGC be the greater, and at the point G, in the straight line BG, make (I. 23.) the angle BGK equal to the angle EHF; but equalangles stand upon equal circumferences (III. 26.) when they are at the centre; therefore 1. The circumference BK is equal to the circumference E EF: but EF is equal to BC; therefore also (Ax. 1.) 2. BK is equal to BC, the less to the greater, which is impossible: therefore 3. The angle BGC is not unequal to the angle EHF; that is, it is equal to it: and the angle at 4 is half of the angle BGC, and the angle at D half of the angle EHF; therefore (Ax. 7.) 4. The angle at A is equal to the angle at D. Wherefore, in equal circles, &c. Q.E.D. B PROP. XXVIII.-THEOREM. In equal circles, equal straight lines cut off equal circumferences, the greater equal to the greater, and the less to the less. K Let ABC, DEF, be equal circles, and BC, EF, equal straight lines in them, which cut off the two greater circumferences BAC, EDF, and the two less BGC, EHF: the greater BAC is equal to the greater EDF, and the less BGC to the less EHF. A G E D L H Take (III. 1.) K, L, the centres of the circles, and join BK, KC, EL, LF. And because the circles are equal, the straight lines from their centres (III. Def. 1.) are equal; therefore 1. BK, KC, are equal to EL, LF, each to each; and the base BC is equal (Hyp.) to the base EF; therefore (I. 8.) The angle BKC is equal to the angle ELF: 2. but equal angles stand (III, 26.) upon equal circumferences, when they are at the centres; therefore 3. The circumference BGC is equal to the circumference EHF: but the whole circle ABC is equal (Hyp.) to the whole EDF; the remaining part therefore of the circumference, viz. 4. BAC is equal to the remaining part EDF. Therefore, in equal circles, &c. Q.E.D. PROP. XXIX. -THEOREM. In equal circles equal circumferences are subtended by equal straight lines. Let ABC, DEF, be equal circles, and let the circumferences BGC, EHF, also be equal; and join BC, EF: the straight line BC is equal to the straight line EF. G Η Take (III. 1.) K, L, the centres of the circles, and join BK, KC, EL, LF, and because the circumference BGC is equal to the circumference EHF, (III. 27.) 1. The angle BKC is equal to the angle ELF: and because the circles ABC, DEF, are equal, the straight lines from their centres (III. Def. 1.) are equal; therefore 2. BK, KC, are equal to EL, LF, each to each; and they contain equal angles; therefore (I. 4.) PROP. XXX. -PROBLEM. To bisect a given circumference, that is, to divide it into two equal parts. Let ADB be the given circumference; it is required to bisect it. Join AB, and bisect it (I. 10.) in C; from the point C draw (I. 11.) CD at right angles to AB; the circumference ADB is bisected in the point D. and D B Join AD, DB. And because AC is equal to CB, and CD common to the triangles ACD, BCD, 1. The two sides AC, CD, are equal to the two BC, CD, each to each; 2. The angle ACD is equal to the angle BCD, because each of them is a right angle; therefore (I. 4.) 3. The base AD is equal to the basé BD. But equal straight lines cut off (III. 28.) equal circumferences, the greater equal to the greater, and the less to the less; and 4. AD, DB, are each of them less than a semicircle, because DC passes through the centre (III. 1. Cor.) Wherefore 5. The circumference AD is equal to the circumference DB. Therefore the given circumference is bisected in D. Which was to be done. PROP. XXXI.-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. B Let ABCD be a circle, of which the diameter is BC, and centre E; and draw CA dividing the circle into the segments ABC, ADC, and join BA, AD, DC: the angle in the semicircle BAC is a right angle; and the angle in the segment ABC, which is greater than a semicircle, is less than a right angle; and the angle in the segment ADC, which is less than a semicircle, is greater than a right angle. |