coincide with the segment CFD, and therefore is equal to it. Wherefore, similar segments, &c. Q. E. D. PROP. XXV. PROB. 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. 8 Ax. * 11. 1. Bisect AC in D, and from the point D draw* DB * 10. 1. at right angles to AC, and join AB. First, let the angles ABD, BAD be equal to one another; then the straight line DB is equal to DA, and also to DC, * 6. 1. since AC is bisected in D; and because the three straight lines DA, DB, DC, are all equal, therefore D is the centre of the circle.* From the centre D, at * 9. 3. 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, the BAD are not equal to one another, at the point A, in the straight line AB make* the angle BAE equal to * 23. 1. the angle ABD, and produce BD, if necessary, to meet AE in E, and join EC: then, because the angle ABE is equal to the angle BAE, the straight line BE is equal to EA: and because AD is equal to DC, and * 6. 1. DE common to the triangles ADE, CDE, the two * 4. 1. * 9. 3. * * sides AD, DE are equal to the two CD, DE, each to each; and the angle ADE is equal to the angle CDE, Const. for each of them is a right angle; therefore the base AE is equal to the base EC: but EA was proved to * 1 Ax. be equal to EB, wherefore also EB is equal to EC: and the three straight lines EA, EB, EC are therefore equal to one another; therefore* E is the centre of the circle. From the centre E, at the distance of any of the three EA, 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: therefore, a segment of a circle being given, the circle is described of which it is a segment. Which was to be done. კ. • Hyp. PROP. XXVI. THEOR. In equal circles, equal angles stand upon equal circumferences, whether they be at the centres or circumferences. Let ABC, DEF be equal circles, and let the angles BGC, EHF at their centres be equal, or BAC, EDF at their circumferences, be equal to each other: the circumference BKC shall be equal to the circumference ELF. Join BC, EF; and because the circles ABC, DEF are equal, the straight lines drawn from their centres * 1 Def. are equal:* therefore the two sides BG, GC are equal to the two EH, HF; and the angle at G is equal to the angle at H; therefore the base BC is equal to Hyp. the base EF: and because the angle at A is equal to * 4. 1. 3. the angle at D, the segment BAC is similar to the *11 Def. segment EDF; and they are upon equal straight lines BC, EF; but similar segments of circles upon equal straight lines, are equal to one another; therefore the * 24. 3. segment BAC is equal to the segment EDF: but the whole circle ABC is equal to the whole DEF; therefore 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. THEOR. In equal circles, the angles which stand upon equal circumferences are equal to one another, whether they be at the centres or circumferences. Let ABC, DEF be equal circles, and let the angles BGC, EHF at their centres, or BAC, EDF at their circumferences, stand upon the equal parts BC, EF of the circumferences: the angle BGC shall be 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 that the angle BAC is also equal to EDF: * 20. 3. but, if not, one of them must be greater than the other. Let BGC be the greater, and at the point G, in the straight line BG, make the angle B BGK equal to the H G CE F 23. 1. Hyp. * 1 Ax. angle EHF; but equal angles stand upon equal cir26. 3. cumferences, when they are at the centre; therefore the circumference BK is equal to the circumference EF: but EF is equal* to BC; therefore also BK is equal✶ to BC, the less to the greater, which is impossible: therefore the angle BGC is not unequal to the angle EHF; that is, it is equal to it: but the angle at A is half of the angle BGC, and the angle at D half * 7 Ax. of the angle EHF: therefore the angle at A is equal* to the angle at D. Wherefore, in equal circles, &c. Q. E. D. * 20. 3. * 1. 3. 3. PROP. XXVIII. THEOR. In equal circles, equal straight lines cut off equal circumferences, the greater equal to the greater, and the less to the less. 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 shall be equal to the greater EDF, and the less BGC to the less EHF. Take* K, L, the centres of the circles, and join BK, KC, EL, LF: and because the circles are equal, the 1 Def. straight lines from their centres are equal; therefore BK, KC are equal to EL, LF; and the base BC is * Hyp. equal to the base EF; therefore the angle BKC is A D 8. 1. equal to the angle ELF: 26. 3. equal circumferences, when they are at the centres; therefore the circumference BGC is equal to the cirHyp. cumference EHF: but the whole circle ABC is equal* to the whole EDF; therefore the remaining part of the circumference, viz. BAC, is equal to the remaining * 3 Ax. part EDF. Therefore, in equal circles, &c. Q. E. D. PROP. XXIX. THEOR. 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 shall be equal to the straight line EF. A D * 27. 3. Take* K, L, the centres of the circles, and join BK, * 1. 3. KC, EL, LF; then, because the circumference BGC is equal to the circumference EHF, the angle BKC is equal to the angle ELF: and because the circles ABC, DEF are equal, the straight lines from their centres are equal: therefore BK, KC B are equal to EL, LF, and CE F * 1 Def. 3. they contain equal angles: therefore the base BC is equal to the base EF. Therefore, in equal circles, &c. * 4. 1. Q. E. D. PROP. XXX. PROB. 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 in C; from the point C * 10. 1. draw CD at right angles to AB: the circumference * 11. 1. ADB shall be bisected in the point D. Join AD, DB; and because AC is equal to CB, and CD common to the triangles ACD, BCD, the two sides |