to one another; wherefore E is the center of the circle. from Book III. the center E, at the distance of any of the three AE, EB, EC defcribe a circle, this fhall pafs thro' the other points; and the d. 9. 3 circle of which ABC is a fegment is defcribed., and it is evident that if the angle ABD be greater than the angle BAD, the center E falls without the fegment ABC, whic!. therefore is less than a femicircle. but if the angle ABD be lefs than BAD, the center E falls within the fegment ABC, which is therefore greater than a femicircle. wherefore a fegment of a circle being given, the circle is defcribed of which it is a fegment. Which was to be done. IN PROP. XXVI. THEOR. N equal circles, equal angles ftand upon equal circumferences, whether they be at the centers or circumferences. Let ABC, DEF be equal circles, and the equal angles BGC, EHF at their centers, and BAC, EDF at their circumferences. the circumference BKC is equal to the circumference ELF. Join BC, EF; and because the circles ABC, DEF are equal, the straight lines drawn from their centers are equal; therefore the two fides BG, GC, are equal to the two EH, HF; and the angle at G is equal to the angle at H; therefore the bafe BC is equal a. 4. 1. to the base EF. and because the angle at A is equal to the angle b at D, the fegment BAC is fimilar to the fegment EDF; and they b.11. Def. 3. are upon equal straight lines BC, EF; but fimilar fegments of circles upon equal ftraight lines are equal to one another; c. 24. 3. therefore the fegment BAC is equal to the fegment EDF. but the whole circle ABC is equal to the whole DEF, therefore the F Where Book III. remaining fegment BKC is equal to the remaining fegment ELF, and the circumference BKC to the circumference ELF. fore, in equal circles, &c. Q. E. D. a. 20. 3. IN PROP. XXVII. THEOR. N equal circles, the angles which ftand upon equal circumferences, are equal to one another, whether they be at the centers, or circumferences. Let the angles BGC, EHF at the centers, and BAC, EDF at the circumferences of the equal circles ABC, DEF stand upon the equal circumferences 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 manifeft that the angle BAC is alfo equal to EDF. but if not, one of them b. 23. I. a. 26. 3. b is the greater. let BGC be the greater, and at the point G, in the IN PROP. XXVIII. THEOR. N equal circles, equal ftraight lines cut off equal circumferences, the greater equal to the greater, and the lefs to the lefs. 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 lefs BGC, EHF. the greater BAC is equal to the greater EDF, and the lefs BGC to the lefs EHF. Book III. Take K, L the centers of the circles, and join BK, KC, EL, a. 1. 3; LF. and because the circles are equal, the ftraight lines from their centers are equal, therefore BK, KC, are equal to EL, LF; . IN N equal circles equal circumferences are fubtended Let ABC, DEF be equal circles, and let the circumferences BGC, EHF alfo be equal; and join BC, EF. the straight line BC is equal to the ftraight line EF. Book III. ว Take K, L the centers of the circles, and join BK, KC, EL, LF. and because the circumference BGC is equal to the circum b. 27.3. C. 4. I. 2. 10. I. ference EHF, the angle BKC is equal to the angle ELF. and because the circles ABC, DEF are equal, the ftraight lines from their centers are equal; therefore BK, KC are equal to EL, LF, and they contain equal angles. therefore the base BC is equal to the bafe EF. Therefore, in equal circles, &c. Q. E. D. ΤΟ PROP. XXX. PROB. C TO bifect a given circumference, that is, to divide it into two equal parts. it. Let ADB be the given circumference; it is required to bifect a Join AB, and bifect it in C; from the point C draw CD at right angles to AB, and join AD, DB. the circumference ADB is bifected in the point D. Because AC is equal to CB, and CD common to the triangles ACD, BCD, the two fides AC, CD b. 4. I. c. 28. 3. are equal to the two BC, CD; and Ꭰ lines cut off equal circumferences, the greater equal to the greater, and the lefs to the lefs, and AD, DB are each of them lefs than a d. Cor. 1. 3. femicircle; because DC paffes through the center 4. wherefore the circumference AD is equal to the circumference DB. therefore the given circumference is bifected in D. Which was to be done. I PROP. XXXI. THEOR. N a circle, the angle in a femicircle is a right angle; but the angle in a fegment greater than a femicircle is lefs than a right angle; and the angle in a fegment lefs than a femicircle is greater than a right angle. Let ABCD be a circle, of which the diameter is BC, and center E; and draw CA dividing the circle into the fegments ABC, ADC, and join BA, AD, DC. the angle in the femicircle BAC is a right angle; and the angle in the fegment ABC, which is greater than a femicircle, is less than a right angle; and the angle in the fegment ADC which is less than a femicircle is greater than a right angle. a F A Join AE, and produce BA to F; and because BE is equal to EA, the angle EAB is equal to EBA; alfo, becaufe 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 b to the two angles ABC, ACB; therefore the angle BAC is equal to the angle FAC, and each of them C is therefore a right angle. where B Book III. a. 5. 7. Cb. 37. I. E fore the angle BAC in a femicircle is a right angle. And because the two angles ABC, BAC of the triangle ABC c. 10. Def. 1, are together less than two right angles, and that BAC is a right d. 17. 1, angle, ABC must be less than a right angle; and therefore the angle in a fegment ABC greater than a femicircle, is less than a right angle. And becaufe ABCD is a quadrilateral figure in a circle, any two of its oppofite angles are equal to two right angles; there- e. 2 2.3 fore the angles ABC, ADC are equal to two right angles; and ABC is lefs than a right angle, wherefore the other ADC is greater than a right angle. Befides, it is manifeft, that the circumference of the greater fegment ABC falls without the right angle CAB, but the circum |