B 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 center 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. PROPOSITION XXVI. Theor. In equal circles, equal angles stand upon equal circum ferences, 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 (Hyp.), the straight lines drawn from their centres are equal (1 Def. u.): 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. 111.) 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. 111.) to one another; therefore the segment BAC is equal to the segment EDF: But the whole circle ABC is equal (Hyp.) to the whole DEF; therefore the remaining segment BKC is equal (3 Ax.) to the remaining segment ELF, and the circumference BKC to the circumference ELF. Wherefore, in equal circles, etc. Q. E. D. PROPOSITION XXVII. THEOR. In equal circles, the angles which stand upon equal circum ferences 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 circunferences 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 (20. 111.) 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. Now equal angles stand upon equal circumferences (26.111.), when they are at the centre; therefore the circumference BK is equal to the circumference EF: A H 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, etc. Q. E, D. PROPOSITION XXVIII. F H Theor. In equal circles, equal straight lines cut off equal circum ferences, 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. III.) K, L, the centres of the circles, and join BK, KC, EL, LF; And because the circles are equal (Hyp.), 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. 11.) circumferences, when they are at the centres; therefore the circumference BGC is equal to the cir. cumference EHF: But the whole circle ABC is equal (Hyp.) to the whole EDF; the remaining part therefore of the circumference, viz, BAC, is equal (3 Ax.) to the remaining part EDF. Therefore, in equal circles, etc. Q. E, D. PROPOSITION 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. II.) K, L, the centres of the circles, and join BK, KC; EL, LF: And because the circumference BGC is equal to the circumference EHF, the angle BKC is equal (27. 11.) 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 BCS is equal (4. 1.) to the base EF. Therefore, in equal circles, etc. Q. E, D D PROPOSITION 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 (11. 1.) to AB, and join AD, DB: the circumference ADB is bisected in the point D. 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 equal to the angle BCD, because each of them is a right (Constr.) angle; therefore the base AD is equal (4. 1.) to the base BD. But equal straight lines cut off equal (28. 11.) 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. 11.); wherefore the circumference AD is equal to the circumference DB. Therefore the given circumference is bisected in D. Which was to be done. PROPOSITION 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 (15 Def. 1.) 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; 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 angle FAC, and each of them is therefore a right (10 Def. 1.) angle: wherefore the angle BAC in a semicircle is a right angle. And because the two angles ABC, BAC of the B 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 augle. And because ABCD is a quadrilateral figure in a circle, any two of its opposite angles are equal (22. 111.) to two right angles; the fore 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 are right angles. PROPOSITION XXXII. Theor. If a straight line touch a circle, and from the point of con tact a straight line be drawn cutting the circle, the angles made by this line with the line touching the circle shall be equal to the angles which are in the alternate segments of the circle. Let the straight line EF touch the circle ABCD in B, and from the point B let the straight line BD be drawn, cutting the circle : the angles which BD makes with the touching line EF shall be equal to the angles in the alternate segments of the circle; that is, the angle FBD is equal to the angle which is in the segment DAB, and the angle DBE to the angle in the segment BCD. From the point B draw (11. 1.) BA at right angles to EF, and take any point C in the circumference BD, and join AD, DC, CB: And because the straight line EF touches the circle ABCD in the point B, and BA is drawn at right angles to the touching line from the point of contact B, the centre of the circle is (19. 111.) in BA; therefore the angle ADB in a semicircle is a right (31. III.) angle; and consequently the other two angles BAD, ABD are equal to a right angle: But ABF is likewise a right angle; therefore the angle ABF is equal to the angles BAD, ABD: Take from these equals the common angle ABD; therefore the remaining angle DBF is equal to the angle BAD, which is in the alternate segment of the circle. And because ABCD is a quadrilateral figure in a circle, the opposite angles BAD, BCD are equal (22. 11.) to two right angles ; therefore the angles DBF, DBE, being likewise equal (13. 1.) to two right angles, are equal to the angles BAD, BCD: and DBF has been proved equal to BAD; therefore the remaining angle DBE is equal to the angle BCD in the alternate segment of the circle. Wherefore, if a straight line, etc. Q. E. D. B VOL. II. A F PROPOSITION XXXIII. PROB. Upon a given straight line to describe a segment of a circle, containing an angle equal to a given rectilineal angle. Let AB be the given straight line, and the angle at C the given rectilineal angle; it is required to describe upon the given straight line AB a segment of a circle, containing an angle equal to the angle C. First, let the angle at C be a right angle; and bisect (10. 1.) AB in F, and from the centre F, at the distance FB, describe the semicircle AHB; therefore the angle AHB in a semicircle is (31. 111.) equal to the right angle at C. But if the angle C be not a right angle, at the point A, in the straight line AB, make (23. 1.) the angle BAĎ equal to the angle C, and from the point A draw (11. 1.) AE at right angles to AD: bisect (10. 1.) AB in F, and from F draw (11. 1.) FG at right angles to AB, and join GB. And because AF is equal to FB, and FG common to the triangles AFG, BFG, the two sides AF, FG are equal to the two BF, FG; and the angle AFG is equal to the angle BFG; therefore the base AG is equal (4. 1.) to the base GB; and the circle described from the centre G, at the distance GA, shall pass through the point B: Let this be the circle AHB : and because from the point A, the extremity of the diameter AE, AD is drawn at right angles to AE, therefore AD (Cor. 16. 11.) touches the circle: And because AB, drawn from the point of contact A, cuts the circle, the angle DAB is equal to the angle in the alternate segment AHB (32. in.). But the angle DAB is equal to the angle C; therefore also the angle C is equal to the angle in the segment AHB. Wherefore, upon the given straight line AB, the segment AHB of a circle is described, which contains an angle equal to the given angle at C. Which was to be done. A B В н D E PROPOSITION XXXIV. Prob. To cut off a segment from a given circle, which shall contain an angle equal to a given rectilineal angle. Let ABC be the given circle, and D the given rectilineal angle; it is required to cut off a segment from the circle ABC that shall contain an angle equal to the given angle D. Draw (17. III.) the straight line EF touching the circle ABC in the point B; and at the point B, in the straight line BF, make (23. 1.) the angle FBC equal to the angle D: Therefore, because the straight line EF touches the circle ABC, and BC is drawn from the point of contact B, the angle FBC is equal |