Book III. to CD: Therefore the straight line AB coinciding with CD, the segment AEB must coincide with the segment CFD, and therefore is equal to it. Wherefore, "similar segments,' &c. Q. E. D. a 23. 3. See N. a 10. 1. c 6. 1. 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. Bisecta AC in D, and from the point D draw DB at right b 11. 1. 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 must pass through the other points; consequently the circle of which ABC is a segment is described: And because the centre D is in AC, the d 9. 3. f 6. 1. f segment ABC is a semicircle: But if the angles ABD, BAD are not equal to one another, at the point A in the straight e 23. 1. line AB, make the angle BAE equal to the angle ABD, 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; therefore the base AE is equal 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: wherefore E is the centre g 4. 1. d 9. 3. d of the circle. From the centre E, at the distance of any of Book III. the three AE, EB, EC, describe a circle, this must pass through the other points; and thus 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 66 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 the a 4. 1. angle at A is equal to the angle at D, the segment BAC is similar to the segment EDF; and they are upon equal b 11. def. 3. straight lines BC, EF; but similar segments of circles upon equal straight lines are equal to one another, therefore the c 24. 3. segment BAC is equal to the segment EDF: But the whole Book III. 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. a 20. 3. 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, that the angle BAC is also equal to EDF. But, if not, a b 23. 1. c 26. 3. 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, 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 the angle BGC, and the angle at D half the angle EHF: Therefore, the angle at A is equal to the angle at D. Wherefore, " in equal circles," &c. Q. E. D. Book III. 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 is equal to the greater EDF, and the less BGC to the less EHF. Take a K, L, the centres of the circles, and join BK, KC, a 1. 3. EL, LF: And because the circles are equal, the straight lines from their centres are equal; therefore BK, KC are equal to EL, LF; and the base BC is also equal to the base EF; therefore the angle BKC is equal to the angle ELF: But equal b 8. 1. angles stand upon equal circumferences when they are at c 26. 3. the centres therefore the circumference BGC is equal to the circumference EHF. But the whole circle ABC is equal to the whole EDF: the remaining part therefore 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 is equal to the straight line EF. F Book III. a 1. 3. a Take K, L, the centres of the circles, and join BK, KC, EL, LF: and because the circumference BGC is equal to the b 27. 3. c 4. 1. a 10. 1. 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 are equal to EL, LF, and they contain equal angles: Therefore the base BC is equal to the base EF. Therefore, "in equal circles," &c. 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, draw CD at right angles to AB, and join AD, DB; the circumference ADB is bisected in the point D. A C B Because AC is equal to CB, and CD common to the triangles ACD, BCD, the two sides AC, CD are equal to the two BC, CD; and the angle ACD is also equal to the angle BCD, because each of them is a right angle: therefore the base AD is equal to the base BD. But equal straight lines cut off equal circumferences, the greater equal to the greater, and the less to the less, and AD, DB are each of them less than a semid Cor. 1. 3. circle; because DC passes through the centre: Wherefore the circumference AD is equal to the circumference DB; therefore the given circumference ADB is bisected in D. Which was to be done. b 4. 1. c 28. 3. |