equal circles, and the remaining segments are equal, and therefore the arches AGC and DKF are equal. In the same manner it can be demonstrated that the arches AGC and DKF are equal if the given angles at the circumferences ABC and DEF are acute, by drawing OA and OC and also HD, HF. But if the given angles at the circumferences are either right or obtuse, bisect them and the halves of them are equal, and it can be proved, as above, tbat the arches upon which these halves stand are equal, whence it follows that the arches on which the given angles stand are equal. PROP. XXVII. THEOR. Fig. 38. In equal circles (ABC, DEF) the angles (ABC and DEF) which stand upon equal arches, are equal, whether they be at the centres or at the circumferences. For, if it be possible, let one of them, DEF be greater than the other, and make the angle DEG equal to ABC. Because in the equal circles ABC and DEF, the (1) Cons. angle ABC is equal to DEG (1), the arches ÁHC (2) Prop.26 and DKG are equal (2), but AHC and DKF are also equal (3), and therefore DKG is equal to DKF, a (5) Hypoth. part equal to the whole, which is absurd : neither angle therefore is greater than the other, and therefore they are equal. Cor. From the preceding proposition it is evident, that in a circle right lines which intercept equal arches are parallel, because the alternate angles are equal, and vice versa. PROP. XXVIII. THEOR. Fig. 39. In equal circles (ABC, DEF) equal right lines (AC and DF) cut off equal arches, the greater equal to the greater (ABC to DEF), the less to the less (AGC to DHF.) If the equal right lines be diameters, the proposition is evident. If not, let K and L be the centres of the circles, and draw the lines KA, KC, LD, and LF. B. 3. Because the circles are equal (1) AK and KC are (1) hyp.tk. equal to LD and LF, and also AC and DF are equal (1), therefore the angle AKC is equal to the angle DLF (2), and therefore the arch AGC is equal to the (2) Prop. 8. arch DHF (3); and since the circles are equal, take (3) Prop.26 away these equal arches from them, and the remain-B. 3. ing arches ABC and DEF are equal. ) 8 PROP. XXIX. THEOR. In equal circles (ABC, DEF) the right lines ( AC and Fig. 29. DF), which subtend equal arches, are equal. If the equal arches be semicircles, the proposition Butif not, let K and L be the centres of the circles, (1) kypoth. the angles AKC and DLF are equal (2), but in the (2) Prop.27 triangles AKC and DLF the sides AK and KC are B. 3. equal to DL and LF (3), and therefore the bases AC (3) Hyp. & and DF are equal (4). Def.1.B. So (4) Prop. 4. Schol. Whatever has been demonstrated in the B. 1. preceding propositions of equal circles, is also true of the same circle. To bisect a given arch ( ABC). Fig. 40. Draw the right lines AB and CB. (1) Cons. B. 3. 45. B. 1. CL a prec. PROP. XXXI. THEOR. the angle in a segment greater than a semicircle is acute, See N. and the angle in a segment less than a semicircle is obtuse, Fig. 41. Part 1. The angle ABC in a semicircle is a right angle. Let O be the centre of the circle and draw OB and OA are equal, the angles OAB and OBA are also (1)Prop. 5. equal (1); in the same manner it can be proved that the angles OCB and OBC are equal; therefore the and BAC, and therefore the angle ABC is a right Part 2. The angle ABC in a segment greater than a semicircle is acute. (3) Part. Draw AD a diameter of the circle, and also the lines CD, CA. Because in the triangle ACD the angle (4) Cor. P. 17.B.1. ACD in a semicircle is a right angle (3), the angle ADC is acute (4), but the angles ADC and ABC are (6) Prop.24 in the same segment ABDC, and therefore equal (6), therefore the angle ABC is acute. Fig. 45. Part 3. The angle ABC in a segment less than a semicircle is obtuse. Take in the opposite circumference any point D, and draw DA and DC. Because in the quadrilateral figure ABCD the op(6) Prop.22 posite angles B and D are equal to two right angles (6), but the angle D is less than a right angle (7), (7) Part the angle ABC must be obtuse. Fig. 41. Cor. 1. Hence can be derived a method of drawing through the extremity B of any right line BC a perpendicular to it; take any point 0 outside the given line, describe a circle passing through B and cutting the line BC in any point C, draw XC and join the points A and B by the right line AB, this line is perpendicular to CB, for the angle ABC in a semicircle is right. Cor. 2. Hence also can be drawn a tangent to a circle from a given point without it: draw a right line B. 3. .B. 3. prec. . from the given point to the centre of the circle bisect it, and from thepoint of bisection as a centre describe a circle through the given point; the right line drawn from either intersection of this circle with the given circle is a tangent; for it is perpendicular to the radius draw to the point where it meets the circle, because the angle in a semicircle is a right angle. PROP. XXXII. THEOR. If a right line (EF) be a tangent to a circle and Fig. 44, from the point of contact a right line (AC) be drawn cutting the circle, the angle (FAC) made by this line with the tangent is equal to the angle (ABC) in the alternate segment of the circle. If the secant should pass through the centre, it is (1) Prop.18 evident that the angles are equal, for each of them $ 31. B. 3. is a right angle (1). But if not, draw through the point of contact the line AB perpendicular to the tangent EF and join BC. Because the right line EF is a tangent to the circle and AB is drawn through the point of contact perpendicular to it, AB passes through the centre (2), (2) Prop.19 B. 3. and therefore the angle ACB is a right angle (3); (3) Prop.31 therefore in the triangle ABC the sum of the angles B. 3. ABC and BAC is equal to a right angle (4) and there- (4) Cor. 1. fore equal to the angle BAF, take away the common P. 32.B. 1. . angle BAC, and the remaining angle CAF is equal to the angle ABC, in the alternate segment. The angles EAC and ADC are also equal. Draw the right lines AD and DC, because in the quadrilateral ABCD the opposite angles ABC and ADC taken together are equal to two right angles (5), (5) Prop.22 the sum of the angles EAC and FAC is equal to the B. 3. (6) Prop.13 sum of ABC and ADC (6), take away the equals FAC 8.1. B. . and ABC (7), and the remaining angle EAC is equal(7) Part to the angle ADC in the alternate segment. prec. L. PROP. XXXIII. PROB. 46. a Fig. 45.4 On a given right line (AB) to describe a segment of a circle, that shall contain an angle equal to a given See N angle (V). Fig. 45. If the given angle V be a right angle, bisect the given line in 0, from the centre 0, describe a circle with the radius OA, the circle is divided by the given (1) Prop.31 line into two semicircles, therefore each of them conB. 3. tains an angle equal to the given right angle (1). given line AB at either extremity of it A, an angle EAB O with the radius 0 A passes through B, because OA (2)Const. and OB are equal (2), and its segment ACB contains Prop. 6. an angle equal to the given acute angle V, and its obtuse angle V. (3) Const. of Because EF is a tangent to the circle at A (3), and Prop. 16.9 from it is drawn AB cutting the circle, the angle in B. 3. the segment ACB is equal to the angle EAB (4), and (4) Prop.32 therefore to the given acute angle V (5); and also the (5) Constr, angle in the segment AGB is equal to the angle FAB (4), and therefore to the given obtuse angle V (5). Schol. In the same manner a circle can be described, which shall touch a given line EF, and pass through a given point B without that line. B. 1. с B. 3.. PROP. XXXIV. PROB. Fig. 47. To cut off from a given circle (ABC) a segment which shall contain an angle equal to a given angle (V.). Draw FA a tangent to the circle at any point A, at the point of contact A make with the line AF an anġle FAC equal to the given one V; the segment ABC contains an angle equal to the given V. a |