Book III. 1 B. An arch of a circle is any part of the circumference. PROP. Ι. PROB. To find the centre of a given circle. 1 Let ABC be the given circle; it is required to find its Draw within it any straight line AB, and bisect a it in D; Book III. from the point D drawb DC at right angles to AB, and pro- a 10. 1. duce it to E, and bifect CE in F: The point F is the centre b 11.1. of the circle ABC. centre.. Draw For, if it be not, let, if possible, G be the centre, and join GA, GD, GB: Then, because DA is equal to DB, and DG common to the two triangles ADG, BDG, the two fides AD, DG are equal to the two BD, DG, each to each; and the base C GA is equal to the base GB, because c 8. 1. G F A D B d 7. def. 3. Cor. From this it is manifest that if in a circle a straight line bisect another at right angles, the centre of the circle is in the line which bisects the other. PROP. II. THEOR. F any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle. ン Book III. Let ABC be a circle, and A, B any two points in the cir cumference; the straight line drawn from A to B shall fall within the circle. C Take any point in AB as E; find D the centre of the circle ABC; join AD, DB and DE, and let DE meet the circumference in F. Then D a 5. I. because DA is equal to DB, the A angle DAB is equal to the angle DBA; and because AE, a fide of the triangle DAE, is produced to B, the angle DEB is b 16.1. greaterb than the angle DAE; but DAE is equal to the angle DBE; therefore the angle DEB is greater than the angle DBE: Now to the greater angle the greater fide is opC 19. 1. pofited; DB is therefore greater than DE: but BD is equal to DF; wherefore DF is greater than DE, and the point E is therefore within the circle. The same may be demonstrated of any other point between A and B, therefore AB is within the circle. Wherefore, if any two points, &c. Q. E. D. ' E B F PROP. III. THEOR. a 1. 3. F a ftraight line drawn through the centre of a circle bisect a straight line in the circle, which does not pass through the centre, it will cut that line at right angles; and, if it cut it at right angles, it will bifect it. Let ABC be a circle, and let CD, a straight line drawn through the centre, bisect any straight line AB, which does not pass through the centre, in the point F: It cuts it alfo at right angles. Takea E the centre of the circle, and join EA, EB. Then, because AF is equal to FB, and FE common to the two triangles |