For, if the segment E F AEB be applied to the Tos segment CFD, so as the o N /~ point A be on C, and the - —— straight line AB upon A B C D CD, the point B shall coincide with the point D, because AB is equal to CD: therefore the straight line AB coinciding with CD, the segment AEB must (23. 3.) coincide with the segment CFD, and therefore is equal to it. Wherefore similar segments, &c. Q. E. D. 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. Bisect (10. 1.) AC in D, and from the point D draw (11. 1.) 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 (6. 1.) 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 (9. 3.): 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 if the angles ABD, BAD are not equal to one another, at the point A, in the straight line AB, make (23. 1.) 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 (6.1.) 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 (4.1.) 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 (9. 3.) 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. * See Note. 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 (4.1.) to the base EF; and because the angle at A is equal to the angle at D, the segment BAC is similar (11. def. 3.) to the segment EDF; and they are upon equal straight lines BC, EF; but similar segments of circles upon equal straight lines are equal (24. 3.) to one another; therefore the segment BAC is equal to the segment EDF; but the whole circle ABC is equal to the whole EDF; 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 circumference 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. 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 (23. 1.) the angle BGK equal to the angle EHF; but equal angles stand upon equal circumferences (26. 3.) 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. 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 (1.3.) 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 are equal: therefore BK, KC are equal to EL, LF; and the base BC is equal to the base EF; therefore the angle BKC is equal (8: 1.) to the angle ELF: but equal angles stand upon equal (26. 3.) 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; the remaining part therefore of the circumserence, 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. Take (1.3.) K, L, the centres of the circles, and join BK, KC, EL, LF; and because the circumference BGC is equal to the circumfeA D rence EHF, the angle BKC is equal (27. 3) 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 (4.1.) 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 (10. 1.) 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. Because AC is equal to CB, and CB common to the triangles ACD, BCD, the two sides AC, CD are equal D to the two BC, CD; and the angle ACD is equal to the angle BCD, because each of them is a right angle; therefore the base AD /* o Ş is equal (4. 1.) to the base BD: but equal A C B straight lines cut off equal (28. 3.) circumferences, the greater equal to the greater, and the less to the less, and AD, DB are each of them less than a semicircle; because DC passes through the centre (Cor. 1.3.): wherefore the circumference AD is equal to the cir cumference DB : therefore the given circumference is bisected in D. Which was to be done. PROP. XXXI. THEOR. In a circle, the angle in a semicircle is a right angle; but the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle. Let ABCD be a circle, of which the diameter is BC, and centre E; and draw CA, dividing the circle into the segments ABC, ADC, and join BA, AD, DC; the angle in the semicircle BAC is a right angle; and the angle in the segment ABC, which is greater than a semicircle, is less than a right angle; and the angle in the segment ADC, which is less than a semicircle, is greater than a right angle. Join AE, and produce BA to F; and because BE is equal to EA, the angle EAB is equal (5. 1.) to EBA.; also, because AE is equal to EC, the angle EAC is equal to ECA; F wherefore the whole angle BAC is equal to the two angles ABC, ACB; but FAC, the exterior angle of the triangle ABC, is equal (32. 1.) to the two angles ABC, ACB; therefore the angle BAC is equal to the D. angle FAC, and each of them is therefore a right (10. des. 1.) angle; wherefore the B angle BAC in a semicircle is a right angle. E C And because the two angles ABC, BAC, of the triangle ABC are together less (17. 1.) than two right angles, and that BAC is a right angle, ABC must be less than a right angle: and therefore the angle in a segment ABC greater than a semicircle, is less than a right angle. And because ABCD is a quadrilateral figure in a circle, any two of its opposite angles are equal (22.3.) to two right angles; therefore the angles ABC, ADC are equal to two right angles; and ABC is less than a right angle; wherefore the other ADC is greater than a right angle. Besides, it is manifest, that the circumference of the greater segment ABC falls without the right angle CAB, but the circumference of the less segment ADC falls within the right angle CAF. “And this is all that is meant, when in the Greek text, and the translations from it, the angle of the greater segment is said to be greater, and the angle of the less segment is said to be less, than a right angle.” CoR. From this it is manifest, that if one angle of a triangle be equal to the other two, it is a right angle, because the angle adjacent to it is equal to the same two; and when the adjacent angles are equal, they ate right angles. |