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 (9. 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 DEF; therefore the remaining segment BKC is equal to the remaining segment ELF, and the arch BKC to the arch ELF. Wherefore in equal circles, &c. Q. E. D. In equal circles, the angles which stand upon equal arches 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 arches 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. 3.) 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 (23. 1.) BGK equal to the angle EHF. And because equal angles stand upon equal arches (26.3.), when they are at the centre, the arch BK is equal to the arch 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 arches, 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 arches 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; but the base BC is also equal to the base EF; therefore the angle BKC is equal (8. 1.) to the angle ELF: and equal angles stand upon equal (26.3.) arches, when they are at the centres; therefore the arch BGC is equal to the arch EHF. But the whole circle ABC is equal to the whole IDF; 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 arches are subtended by equal straight lines. Let ABC, DEF be equal circles, and let the arches 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 arch BGC is equal to the arch EHF, the angle BKC is equal (27.3.) to the angle ELF: also because the circles ABC, DEF are equal, their radii 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 arch, that is, to divide it into two equal parts, Let ADB be the given arch; 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 arch ADB is bisected in the point D. Because AC is equal to CB, and CD common to the triangle ACD, BCD, the two sides AC, CD are equal to D the two BC, CD; and the angle ACD is C B arches, 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 arch AD is equal to the arch DB: and therefore the given arch ADB 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; draw CA dividing the circle into the segments ABC, ADC, and join BB, 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 EΒΑ; F also, because AE is equal to EC, the And because the two angles ABC, BAC of the triangle ABC are together less (17. 1.) than two right angles, and 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. Also 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. Therefore, in a circle, &c. Q. E. D. 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. PROP. XXXII. THEOR. If a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle, the angles made by this line with the line which touches the circle, shall be equal to the angles 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 arch 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. 3.) in BA; therefore the angle ADB, in a semicircle, is a right (31.3.) angle, and consequently the other two angles BAD, ABD are equal (32. 1.) 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; and there will remain the angle DBF equal to the angle BAD, which is in the alternate seg ment of the circle, And because ABCD is a quadrilateral figure in a circle, the opposite angles BAD, BCD are equal (22. 3.) 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, &c. Q. E. D. PROP. 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; bisect (10. 1.) AB in F, and from the centre F, at the dis H 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 BAD equal to the angle C, and from the point A draw (11. 1.) AE H at right angles to AD; bi sect (10. 1.) AB in F, and from F draw (11. 1.) FG at right angles to AB, and join G GB: Then because AF is qual (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. 3.) 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 AНВ (32. 3.); but the angle DAB is equal to the angle C, therefore also the angle C is equal to the angle in the segment AHB : |