circumference BK is equal to the circumference 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. Proposition XXVIII. Theorem. In equal circles, equal straight lines cut off equal circumferences, 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. 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; 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. 3.) circumferences, when they are at the centres ; therefore the circumference BGC is equal to the circumference EAF. But the whole circle ABC is equal to the whole EDF; the remaining part, therefore of the circumference, viz. BAC, is equal to the remainining part EDF. Therefore, in equal circles, &c. Q. E. D. D B Proposition XXIX. Theorem. 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. 3.) K, L the centres of the circles, and joiu BK, KC,EL, LF: And because the circumferenice BGC is equal to the circumference EHF, the angle BKC is equal (27. 5.) to the angle FLF: And be. cause 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 BC is equal (4. 1.) to the base EF. Therefore, in equal circles, &c. Q. E. D. A Proposition XXX. Problem. 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 to AB, and join AD, DB ; the circumference ADB is bie sected 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 angle: tberefore the base AD is equal (4.1.) to the base BD. But equal straight lines cut off equal (28. 3.) 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. 1.3.) 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. Theorem. In a circle, the angle in a semicircle is a right angle; but the anglo 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 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 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 angle. And 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. 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 iris equal to the same two; and when the adjacent angles are equal, they are right angles. Proposition XXXII. Theorem. If a straight line touches 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 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 D draw (11. 1.) BA at right angles to EF, and take any point Cin 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. 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, ABĎ: 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. 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. Proposition XXXIII. Problem, Upon a given straight line to describe a segment of a circlo COletaining an angle equal to a given rectilineal angle. Lel AB be the given straight line, and the angle at C the given ree B H tilineal angle ; it is required to ciescribe 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; :herefore the angle AHB in a semicircle, is (31. 3.) 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 BAD 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 PG common to the triangles AFG, BFG, the two sides AF, FG are equal to the two BF, TG ; 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. 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 AHB : (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. 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. Proposition XXXIV. Problem. 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. Drav (17.3.) 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: a 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 (23.3.) to the angle in the alternate segment BAC of the circle : But the angle FBC is equal to the angle D; therefore the angle in the segment BAC is equal to the angle D; Wherefore the segment BAC is cut off from the given circle ABC, containing an angle equal to the given angle D. Which was to be done. Proposilion XXXV. Theorem. It two straight lines within a circle cut one another, the 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, within the circle ABCD, cut one another in the point E : the rectangle contained by AE, EC is equal to the rectangle contained by BE, ED. 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, EC is likewise equal to the rectangle BE, ED. 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 XD 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, EC are equal (3. 3.) to one another : And because the straight line BD is cut into two equal parts in the point F, and into two unequal in the point E, the rectangle BE, ED, together with the square of EF, is equal (5. 2.) to the square of FB ; that is, to the square of FA ; but the squares of AE, EF are equal (47. 1.) to the square of FA ; therefore the rectangle BE, FD, together with the square of EF, is equal to the squares of AE, EF: Take away the common square of EF, and the remaining rectangle BE, ED is equal to the remaining square of AE ; that is, to the rectangle AE, EC. Next, let BD, which passes through the centre, cut the other AC, which does not pass through the centre in E, but not at right angles : Then, as before, if BD be bisected in F, F is the centre of the circle. Join AF, and from F draw (12. 1.) FG perpendicular to AC ; therefore AG is equal (3.3.) to GC; wherefore the rectangle AE, EC, together with the square of EG, is equal (5. 2.) to the square of AG: To each of these equals add the square of GF; therefore the rectangle AE, EC, together with the squares of EG, GF, is equal to the squares of AG, GF: But the squares of EG, GF are equal (47. 1.) to the square of EF ; and the squares of AG, GF are equal to the square of AF: Therefore the rectangle AE, EC, together with the square of EF, is equal to the square of AF; that is to the square of FB : But the square of FB is equal (5. 2.) to the rectangle BE, ED, together with the square of EF; therefore the rectangle AE, EC, together with the square of EF is equal to the rectangle BE, ED, together with the square of EF: Take away the common square of EF, and the remaining rectangle AE, EC, is there. fore equal to the remaining rectangle BE, ED. Lastly, let neither of the straight lines AC, BD pass through the con |