32. 1. * 21.3. † 2 Ax. +2 Ax. † 1 Ax. See N. * 10 3. + Hyp. 16.1. Let ABCD be a quadrilateral figure in the circle ABCD; any two of its opposite angles shall together be equal to two right angles. * Join AC, BD: and because the three angles of every triangle are equal to two right angles, the three angles of the triangle CAB, viz. the angles CAB, ABC, BCA, are equal to two right angles: but the angle CAB is equal to the angle CDB, because they B are in the same segment CDAB; and the PROP. XXIII. THEOR. Upon the same straight line, and upon the same side of it, there cannot be two similar segments of circles, not coinciding with one another. If it be possible, upon the same straight line AB, and upon the same side of it, let there be two similar segments of circles, ACB, ADB, not coinciding with one another. Then, because the circle ACB cuts the circle ADB in the two points A, B, they cannot cut one another in any other point; therefore one of the segments must fall within the other: let ACB fall within ADB: draw the straight line BCD, and join CA, DA. And because the segment ACB is similar to the segment ADB, and that similar segments of circles contain equal angles; therefore the angle ACB is equal to the angle ADB, the exterior to the interior, which is impossible *. Therefore there cannot be two similar segments of circles upon the same side of the same line, which do not coincide. Q. E. D. PROP. XXIV. THEOR. Similar segments of circles upon equal straight lines are See N. equal to one another. Let AEB, CFD be similar segments of circles upon the equal straight lines AB, CD: the segment AEB shall be equal to the segnient CFD. E For if the segment AEB be applied to the segment CFD, so that the point A may be on C, and the straight line AB upon CD, the point B shall coincide with the point D, be 1 * cause AB is equal to CD: therefore the straight line AB coinciding with CD, the segment AEB must CO- * 23. 3. incide with the segment CFD, and therefore is equal † † 8 Ax. to it. Wherefore similar segments, &c. Q. E. D. PROP. XXV. PROB. A segment of a circle being given, to describe the circle of See N. which it is the segment. * 10. 1. * 1. 11.1. * 6. 1. Let ABC be the given segment of a circle; it is required to describe the circle of which it is the segment. Bisect AC in D, and from the point D draw* DB at right angles to AC, and join AB. First, let the angles ABD, BAD be equal† to one another: then the straight † See Fig. line BD is equal * to DA, and therefore to DC: and because the three straight lines DA, DB, DC, are all equal, 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, the segment ABC, 9.3. is a semicircle. But if the angles ABD, BAD are not equal to one another, at the point A, in the straight See Fig. 2. and 3. 23. 1. line AB, make the angle BAE equal to the angle + See Fig. ABD, and produce † BD, if necessary, to E, and join 2. * 6. 1. 4. 1. † 1 Ax. + 2. 9. 3. * * EC. And 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 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 a Constr. right angle; therefore the base AE is equal to the base EC: but AE was shewn to be equal to EB; wherefore also BE is equal + to EC; and therefore the three straight lines AE, EB, EC are equal to one another : wherefore* 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, t 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. See Fig. + See Fig. 3. † 1 Def. 3. + 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, and 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 † are equal: therefore the two sides BG, GC are equal to the two EH, HF, each to each: and the angle at G is equal to the angle at H; therefore the base BC is equal to the base EF. And because the angle at A is equal to the angle at D, the segment BAC is 11 Def. 3. similar to the segment EDF; and they are upon equal straight lines BC, EF: but similar segments of circles * 4. 1. † Hyp. * * upon equal straight lines are equal to one another, * 34. 3. therefore the segment BAC is equal to the segment EDF: but the whole circle ABC is equal + to the whole † Hyp. DEF; therefore the remain ing 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. H B CE 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, and BAC, EDF at their circumferences, stand upon the equal circumferences, BC, EF: the angle BGC shall be equal to the angle EHF, and the angle BAC to the angle EDF. † 3 Ax. Ax. 1. If the angle BGC be equal to the angle EHF, it is manifest that the angle BAC is also equal to EDF. 20.5.&7 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 BGK 23. 1. equal to the angle EHF. Then because the angle BGK is equal to the angle EHF, and that equal angles stand upon equal circumferences*, * 26.3. when they are at the centres; therefore the circumference BK is equal to the circumference EF: but EF is equal to BC; therefore also BK is equal + to BC, Hyp. the less to the greater, which is impossible: therefore +1 Ax. the angle BGC is not unequal to the angle EHF; that is, it is equal to it: and the angle at A is half of the † 20.3. angle BGC, and the angle at D half of the angle EHF; 'therefore the angle at A is equal to the angle at D. † 7 Ax. Wherefore, in equal circles, &c. Q. E. D. 1. 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 straight lines from their centres +1 Def. 3. are equal: therefore BK, KC + Hyp. *8. 1. * 26.3. + Hyp. † 3 Ax. 1. 3. 27.3. are equal to EL, LF, each to * * D upon equal circumferences, when they are at the centres; therefore the circumference BGC is equal to the circumference 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 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. Take* K, L, the centres of the circles, and join BK, KC, EL, LF: and 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 * B D K |