For if the segment AEB be applied to the segment CFD, so as the point A be on C, and the straight line AB upon CD, the point в shall coincide with the point D, because AB is equal to CD: Therefore the straight line AB coinciding with CD, the segment AEB must (III. 23.) coincide with the segment CFD, and therefore is equal to it. Wherefore similar segments, &c. Q. E. D. 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 (1. 10.) AC in D, and from the point D draw (1. 11.) DB, at 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 (1. 6.) to DA, and therefore to DC; and because the three straight lines CA, DB, DC, are all equal; D is the centre of the circle (III. 9.): 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 de C scribed: And because the centre D is in AC, the segment ABC is a semicircle; but if the angles ABD, BAD are not equal to one another, at the point a, in the straight line BA, make (1. 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 BAE, the straight line BE is equal (1. 6.) 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 (1. 4.) to the base EC: But A E was shown to be equal to BE, wherefore also BE is equal to EC: And the three straight lines AE, E B, EC are therefore equal to one another; wherefore (III. 9.) 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. 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 (1. 4.) to the base EF: And because the angle at A is equal to the angle at D, the segment BAC is similar (III. Def. 11.) 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 the 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 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 (II. 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 the angle BGK equal to the angle EHF; but equal angles stand upon equal circumferences (III. 26.) 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: 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. In equal circles, equal straight lines cut off equal circumferences, the greater equal to the greater, and the less to the less. Let A B C, D E F be equal circles, and B C, E F equal straight lines in them, which cut off the two greater circumferences BA C, E D F, and the two less B G C, EHF; the greater B A C is equal to the greater EDF, and the less B G C to the less E H F. Take (III. 1.) K, L the centres of the circles, and join B K, KC, EL, LF: And because the circles are equal, the straight lines from their centres are equal; herefore B K, K C are equal to E L, L F; and the base BC is equal to the base EF; therefore the angle BKC is equal (1. 8.) to the angle ELF: But equal angles stand upon equal (III. 26.) circumferences, when they are at the centres; therefore the circumference BGC is equal to the circumference E H F. But the whole circle A B C is equal to the whole EDF; the remaining part therefore of the circumference, viz. BA C, 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, D E F be equal circles, and let the circumferences B G C, E H F also be equal; and join BC, EF: The straight line BC is equal to the straight line E F. Take (III. 1.) K, L the centres of the circles, and join B K, KC, E L, LF: and because the circumfer ence B G C is equal to the circumference E HF, the angle BKC is equal (III. 27.) to the angle ELF: And because the circles A B C, D E F are equal, the |