PRO P. XXIII. PROBLEM. To bisect a given arc, that is, to divide it into two equal parts. Let ADB be the given arc; it is required to divide it into two equal parts. Draw the right line AB, which bisect in c (I. 10.); and, from the point c, erect the perpendicular CD (1. 11.); then will the arc ADB be bisected in the point D, as was required. For, join the points AD, DB: then, since the two fides AC, CD, of the triangle ADC, are equal to the two fides BC, CD of the triangle BDC, and the angle ACD to the angle BCD (1. 8.), the base AD will be equal to the base dB (I. 4.) And, because pc, or pc produced, passes through the centre of the circle (Ill. I Cor.), the segments ADE, DBF will be each of them less than a semicircle. But equal chords are subtended by equal arcs, the greater equal to the greater, and the less to the less (III. 22.); whence the chord AD being equal to the chord DB, the arc AED will be equal to the arc DFB. Q. E. I. SCHOLIUM. An arc of a circle cannot, in general, be trisected, or divided into three equal parts, by any known method, which is purely geometrical. PROP. XXIV. THEOREM. The angle formed by a tangent to a circle and a chord drawn from the point of contact, is equal to the angle in the alternate segment. B - Let ec be a tangent to the circle AFDE, and as a chord. drawn from the point of contact; then will the angle cae be equal to the angle Afe in the alternate segment. For draw the diameter AD (III. 1.) and join the points F, D: Then, because bc is a tangent to the circle, and AD is a line drawn through the centre, from the point of contact, the angle dac will be a right angle (III. 12.). And, because App is a semi-circle, the angle DFA will also be a right angle (III. 16.), and equal to the angle DAC..' But since all angles in the fame segment of a circle are equal to each other (III. 15.), the angle die will be equal to the angle DAE. If, therefore, from the equal angles DAC, DFA, there be taken the equal angles DAE, DFE, the remaining angle cae will be equal to the remaining angle Afe. Q. E. D. H4 PROP PROP. XXV. PROBLEM. Upon a given right line to describe a segment of a circle, that shall contain an angle equal to a given rectilineal angle. Let AB be the given right line, and c the given rectie lineal angle; it is required to describe a segment of a circle upon the line AB that shall contain an angle equal to c. Make the angle BAD equal to c (1. 20.), and, from the point A, draw AE at right angles to AD (I. 11.), and make the angle ABF equal to the angle FAB (1. 20.) Then, since the angles ABF, FAB, are equal to each other, and less than two right angles, the sides AF, FB will meet, and be equal to each other (I. 25 Cor. and I.6.) From the point F, therefore, at the distance FA, or FB, describe the circle AEB, and ABEA will be the segment required. For let af be produced to cut the circle in E; and join the points E, B. Then, because AD is perpendicular to the diameter AE, it will be a tangent to the circle at the point A (III. 10.) And, because AD touches the circle, and AB is a chord drawn from the point of contact, the angle BAD will be equal to the angle afl in the alternate segment WII. 24.) But the angle BAD is equal to the angle c, by construction; consequently the angle AEB is also equal to the angle c. Q. E. I, SCHOLIUM. When the given angle is a right angle, a semi-circle described upon the given line will be the seg. ment required (III. 16.) PRO P. XXVI. PROBLEM. To cut off a segment from a given circle, that shall contain an angle equal to a given rectilineal angle. Let ABC be a given circle, and D a given rectilineal angle; it is required to cut off a segment from the circle ABC, that shall contain an angle equal to D. Draw the light line EF, to touch the circle ABC in the point A (III. 10.), and make the angle FAB equal to the angle D (I. 20.); then will APCA be the fegment required. For, fince EF is a tangent to the circle, and AB is a chord drawn from the point of contact, the angle FAB will be equal to the angle ace in the alternate segment (III. 24.) But But the angle FAB is equal to the angle D, by construction ; consequently the angle ACB, in the segment ABCA, is also equal to the angle D. Q. E. I. PRO P. XXVII. THEOREM. If two right lines in a circle intersect each other, the rectangle contained under the segments of the one, will be equal to the rectangle contained under the segments of the other. Let AB, CD be any two right lines, in the circle ACBD, intersecting each other in the point F; then will the rectangle contained under the parts AF, FB of the one, be equal to the rectangle contained under the parts CF, FD of the other. For, through the point F, draw the diameter hi (III. 1.); and, from the centre E, draw eg at right angles to AB (I. 12.), and join Ae: Then, fince Aer is a triangle, and the perpendicular eG divides the chord ab into two equal parts (III. 3.), the line FB will be equal to the difference of the segments AG, GF. And, because e is the centre of the circle, and se is equal to Ei or EH, the line Fi will be equal to the sum of |