BOOK III. BCD for their base; therefore the angle BFD is double a of the angle BAD: For the same reason, the angle BFD is double of the angle BED: Therefore the angle BAD is equal to the angle BED. • 20. 3. B E But, if the segment BAED be not greater than a semicircle, let BAD, BED be angles in it; these also are equal to one another: 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 F equal: Therefore the whole angle BAD is equal to the whole angle BED. Wherefore the angles in the same segment, &c. Q. E.D. PROP. XXII. THEOR. THE opposite angles of any quadrilateral figure described in a circle, are together equal to two right angles. Let ABCD be a quadrilateral figure in the circle ABCD; any two of its opposite angles are together equal to two right angles. Join AC, BD; and because the three angles of every 32. 1. triangle are equal a 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 A B b21. 3. angle CAB is equal to the angle CDB, because they are in the same segment BADC, and the angle ACB is equal to the angle ADb, because they are in the same segment ADCB: Therefore the whole angle ADC is equal to the angles CAB, ACB: To each of these equals add the angle ABC; therefore the angles ABC, CAB, BCA are equal to the angles ABC, ADC: But ABC, CAB, BCA, are equal to two right angles; therefore also the angles ABC, ADC, are equal to two right angles: In the same manner, the angles BAD, DCB, may be shown Book III. to be equal to two right angles. Therefore, the opposite angles, &c. Q. E. D. PROP. XXIII. THEOR. UPON the same straight line, and upon the same See N. side of it, there cannot be two similar segments of circles, not coinciding with one another. If it be possible, let the two similar segments of circles viz. ACB, ADB be upon the same side of the same straight line AB, 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 pointa: One of the segments must therefore fall within the other: Let ACB fall within ADB, and draw the straight line BCD, and join CA, DA: And A because the segment ACB is similar B C a 10.3. to the segment ADB, and that similar segments of circles contain bequal angles; the angle ACB is equal to the an- 11. Def. 3. gle ADB, the exterior to the interior, which is impossible. 16. 1. Therefore there cannot be two similar segments of a circle 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 See N. lines, are equal to one another. Let AEB, CFD be similar segments of circles upon the equal straight lines AB, CD; the segment AEB is equal to the segment CFD. For if the segment AEB be applied to the segment CFD, so as the A point A be on E C, and the straight line AB upon CD, the point B shall coincide with the point D, because AB is equal to CD; BOOK III. Therefore the straight line AB coinciding with CD, the segment AEB musta coincide with the segment CFD, and therefore is equal to it. Wherefore similar segments, &c. Q.E.D. 23.3. PROP. XXV. PROB. See N. A SEGMENT of a circle being given, to describe the circle of which it is the segment. a 10.1. Let ABC be the given segment of a circle; it is required to describe the circle of which it is the segment. Bisecta AC in D, and from the point D drawb DB at 11.1. right angles to AC, and join AB: First, let the angles ABD, BAD be equal to one another; then the straight 6. 1, line BD is equal to DA, and therefore to DC; and be cause the three straight lines DA, DB, DC, are all equal; 9. 3. D is the centre of the circled. 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 is a semicircle: But e if the angles ABD, BAD are not equal to one another, at 23. 1. the point A in the straight line AB make 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 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; 4. 1. therefore the base AE is equalf to the base EC: but AE was shown to be equal to EB, wherefore also BE is equal to EC: And the three straight lines AE, EB, EC are therefore equal to one another; wherefored E is the centre Book III. 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. 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 the equal angles BGC, EHF at their centres, and BAC, EDF at their circumferences: The circumference BKC is 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; and the angle at G is equal to the angle at H; therefore the base, BC is equal to the base EF. And because 4. 1. the angle at A is equal to the angle at D, the segment BAC is similarb to the segment EDF; and they are upon 11 Def. 3. equal straight lines BC, EF: But similar segments of circles upon equal straight lines are equal to one another, 24. 3. therefore the segment BAC is equal to the segment EDF: G C C BOOK III. 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 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 ma20. 3. nifesta that the angle BAC is also equal to EDF. But, if A B H C E F not, one of them is the greater: Let BGC be the greater, 23. 1. and at the point G, in the straight line BG, make the angle BGK equal to the angle EHF; but equal angles € 26. 3. stand upon equal circumferences, when they are at the centre; therefore the circumference BK is equal to the cir cumference 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: And the angle at A is half of the angle BGC, and the angle at D half of the angle EHF: Therefore the angle at A is equal to the angle at D. Wherefore, in equal circles, &c. Q. E.D. |