therefore the whole angle BEC is double of the whole angle BAC. Again, let the centre of the circle be without the angle BAC. It may be demonstrated, as in the first case, that the angle FEC is double of the angle FAC, and that FEB, a part of the first, is dou F B ble of FAB, a part of the other; therefore the remaining angle BEC is double of the remaining angle BAC. Therefore the angle at the centre, &c. Q. E. D. PROP. XXI. THEOR. The angles in the same segment of a circle are equal to See N. one another. Let ABCD be a circle, and BAD, BED angles in the same segment BAED: the angles BAD, BED shall be equal to one another. B E 71. 3. First, let the segment BAED be greater than a semicircle. Take+ F, the centre of the circle ABCD, and join BF, FD: and because the angle BFD is at the centre, and the angle BAD at the circumference, and that they have the same part of the circumference, viz. BCD for their base; therefore the angle BFD is double* of the angle *20. 3. BAD: for the same reason the angle BFD is double of the angle BED: therefore the angle BAD is equal† †7 Ax. to the angle BED. a semicircle on B Next, let the segment BAED be not greater than a semicircle. Draw AF to the centre, and produce it to C, and join CE: therefore the segment BADC is greater than a semicircle; and the angles in it BAC, BEC are equal, by the first case: for the same reason, because CBED is greater. than a semicircle, the angles CAD, CED, are equal: therefore the whole angle BAD is equal† †2 Ax. PROP. XXII. THEOR. The opposite angles of any quadrilateral figure inscribed in a circle, are together equal to two right angles. #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 are in the same segment CDAB; and the angle ADC to each of these equals add the angle 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 ACB fall within ADB: draw the straight line BCD, 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 segment 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 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. * *11. 1. +See Fig.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 10.1. right angles to AC, and join AB. First, let the angles ABD, BAD be equal+ to one another: then the straight 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. 2. ABD, and produce † BD, if necessary, to E, and join * 6. 1. + Constr. * 4. 1. † 1 Ax. * * 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 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 +SeeFig. 2. greater+ than the angle BAD, the centre E falls with *9. 3. out the segment ABC, which therefore is less than a +See Fig.3. 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. 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, † 1 Def. 3. 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 + Hyp. * 4. 1. † Hyp. * * * 24.3. upon equal straight lines are equal to one another, therefore the segment BAC is equal to the segment EDF: but the whole circle ABC is equal+ to the whole + Hyp. 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. †,3 Ax. H B CH 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. * If the angle BGC be equal to the angle EHF, it is manifest that the angle BAC is also equal to EDF. *20.3.& 7. But, if not, one of them must be greater than the other: Ax 1. let BGC be the greater, and at the point G, in the straight line BG, make the angle BGK equal to the angle EHF. Then because the angle BGK is equal to the angle EHF, and *23, 1. 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. † 1 Ax. 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 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. |