are equal; therefore BK, KC, are equal to EL, LF, † 1 Def. 3. each to each and they contain equal angles; therefore * the base BC is equal to the base EF. equal circles, &c. Q. E. D. PROP. XXX. PROB. Therefore in 4. 1. 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 it in C; from the point C draw 10.1. CD at right angles+ to AB: the circumference ADB +11.1. shall be bisected in the point D. Join AD, DB: and 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, each to each; and the angle ACD is equal to the angle BCD, because each of them is a right angle: therefore the base AD is equal to the base BD. But equal straight lines cut off equal circumferences, the greater equal to the greater, * * D A C 4. 1. 28.3. and the less to the less; and AD, ĎB are each of them less than a semicircle, because DC* passes through Cor. 1.3. the centre: therefore the circumference AD is equal to the circumference 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 shall be a right angle; and the angle in the segment ABC, which is greater than a semicircle, shall be less than a right angle; and the angle in the segment ADC, which is less than a semicircle, shall be greater than a right angle. Join AE, and produce BA to F: and because BE is equal to EA, the angle EAB is equal to EBA; also, 15 Def. 1. 5.1. † 2 Ax. * 32. 1. † 1 Ax. because AE is equal to EC, the angle * * 10 Def.1. right angle: wherefore the angle BAC in a semicircle is a right angle. * 17. 1. 22.3. † 32. 1. +10 Def. 1. And because the two angles ABC, BAC of the triangle ABC are together less than two right angles, and that BAC has been proved to be a right angle; therefore 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 to two right angles: therefore the angles ABC, ADC, are equal to two right angles: and ABC has been proved to be 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. 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 which this line makes 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 DBF shall be equal to the angle which is in the segment DAB, and the angle DBE shall be equal to the angle in the segment DCB. * * E 19.3. 31. 3. 32. 1. From the point B draw BA at right angles to EF, 11. 1. and take any point C in the circumference DB, 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* in BA: therefore the angle ADB in a semicircle is a right angle: and consequently the other two angles BAD, ABD, are equal* to a right angle: but ABF is likewise a† right angle; + Constr. therefore the angle ABF is equal to the angles BAD, † 1 Ax. ABD: take from these equals the common angle ABD : therefore the remaining angle DBF is equal to the angle † 3 Ax. 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 to two right⚫ 22.3. angles but the angles DBF, DBE are likewise equal * to two right angles; therefore the angles DBF, DBE are equal to the angles BAD, BCD: and DBF has †1Ax. been proved equal to BAD; therefore the remaining angle DBE is equal + to the angle BCD in the alternate +2 Ax. segment of the circle. Wherefore, if a straight line, &c. Q. E. D. PROP. XXXIII. PROB. 13. 1. Upon a given straight line to describe a segment of a cir- See N. cle, which shall contain 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, which shall contain an angle equal to the angle C. First, let the angle at C be a right angle: bisect AB in F, and from the centre F, at the distance FB, describe the semicircle AHB: therefore the angle AHB in a semicircle is equal to the right angle at C. * H 10. 1. B 31. 3. * 23. 1. 11. 1. * 10. 1. 11. 1. * But if the angle C be not a right angle; at the point A, in the straight line AB, make the angle BAD equal to the angle C, and from the point A draw* AE at right angles to AD; bisect* AB in F, and from F draw 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, each to each: and +10 Def. 1. the angle AFG is equal to the angle BFG; therefore the base AG is equal to the base GB; and there * 4. 1. 3. * H E F B D fore the circle described from the centre G, at the distance GA, shall pass through the point B: let this be the circle AHB. The segment AHB shall contain an angle equal to the given rectilineal angle C. D H A F B E * Because from the point A, the extremity of the diameter AE, AD is drawn at right angles to AE, thereCor. 16. fore AD* 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: but the angle DAB is equal + to the angle C; therefore 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. * 32. 3. + Constr. † 1 Ax. * 17. 3. 23. 1. From a given circle to cut off a segment, 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 from the circle ABC a segment that shall contain an angle equal to the given angle D. Draw the straight line EF touching the circle ABC in the point B, and at the point B, in the straight line BF, make the angle FBC equal to the angle D: the segment BAC shall contain an angle equal to the given angle D. Because the straight line EF touches the circle ABC, and BC is drawn from the point of contact B, the angle FBC is equal to the angle in the alternate segment BAC of the circle: but the angle FBC is equal to the angle D; therefore the an- † gle in the segment BAC is equal to the angle D. Wherefore from the given circle ABC, the segment BAC is cut off, containing an angle equal to the given angle D. Which was to be done. PROP. XXXV. THEOR. 32. 3. Constr. 1 Ax. If two straight lines cut one another within a circle, the See N. rectangle contained by the segments of one of them, is equal to the rectangle contained by the segments of the other. Let the two straight lines AC, BD, cut one another in the point E within the circle ABCD: the rectangle contained by AE, EC shall be equal to the rectangle contained by BE, ED. B If AC, BD pass each of them through the centre, so that E is the centre; it is evident that AE, EC, BE, ED, being all+ equal, the rectangle AE, † 15 Def. EC is likewise equal to the rectangle BE, ED. * F 1. #3. 3. But let one of them BD pass through the centre, and cut the other AC, which does not pass through the centre, at right angles, in the point E: then, if BD be bisected in F, F is the centre of the circle ABCD: join AF: and because BD which passes through the centre, cuts the straight line AC, which does not pass through the centre, at right angles in E, AE is equal to EC: and because the straight line BD is cut into two equal parts in the point F, and into two unequal parts in the point E, the rectangle BE, ED, together with the square of EF, is equal* to the square of FB; that is, to the square of FA: but the of AE, EF, are equal squares to the square of FA; therefore the rectangle BE, ED, together with the square of EF, is equal to the † 1 Ax. squares of AE, EF: take away the common square of EF, and the remaining rectangle BE, ED is equal to † 3 Ax. the remaining square of AE; that is, to the rectangle AE, EC. E * 5.2. **47.1. Next, let BD, which passes through the centre, cut G |