ELEMENTS OF GEOMETRY. BOOK III. DEFINITIONS. A. THE radius of a circle is the straight line drawn from Book III. the centre to the circumference. Book III. B. V. tained by a straight line and the VI. contained by two straight lines VII. upon the arch intercepted between VIII. tained by two straight lines drawn IX. are those in which the an- PROP. I. PROB. Book III. To find the centre lof a given circle. Let ABC be the given circle; it is required to find its centre. Draw within it any straight line AB, and bisecta it in D; a 10. 1. from the point D drawb DC at right angles to AB, and pro- b 11. 1. duce it to E, and bisect CE in F. The point F is the centre of the circle ABC. 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 sides AD, DG are equal to the two BD, DG, each to each; and the base GA is equal to the base GB, because they are radii of the same circle ; therefore the angle ADG is equals c 8.1. to the angle GDB; therefore the angle GDB is a right angled. But d7. Def. 1. FDB is likewise a right angle; wherefore the angle FDB is equal to the angle GDB, the greater to the less, which is impossible. Therefore G is not the centre of the circle ABC. In the same manner it can be shown that no other point but F is the centre; that is, F is the centre of the circle ABC. Which was to be found. 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. IF any two points be taken in the circumference of a circle, the straight line which joins them will fall within the circle. Book III. Let ABC be a circle, and A, B any two points in the cir cumference; the straight line drawn circle. D 7В b 5. 1. angle DAB is equalb to the angle F the triangle DAE, is produced to B, the angle DEB is c 16. 1. greater than the angle DAE. But DAE is equal to the angle DBE; therefore the angle DEB is greater than the d 19. 1. angle DBE; therefore DB is greater than DEd. 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. PROP. III. THEOR. IF a straight 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 bisect 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 also at right angles. Takea E the centre of the circle, and join EA, EB. Then, because AF is qual to FB, and FE common to the two a 1. 3. triangles AFE, BFE, there are two sides in the one equal to Book III.) two sides in the other, but the base EA is equal to the base EB; therefore the angle AFE is equal to the b 8.1. angle BFE; therefore each of the angles AFE, BFE is a right angle; c7. Def. 1. wherefore the straight line CD, drawn through the centre, bisecting another, AB, which does not pass A B D The same construction being made, because the radii EA, EB are equal to each other, the angle EAF is equals to d 5. 1. the angle EBF; and the right angle AFE is equal to the right angle BFE; therefore, in the two triangles EAF, EBF, there are two angles in one equal to two angles in the other; and the side EF, which is opposite to one of the equal angles in each, is common to both; therefore the other sides are equale; AF therefore is equal to FB. Wherefore, if a e 26. 1, straight line, &c. Q. E. D. PROP. IV. THEOR. IF in a circle two straight lines cut each other, which do not both pass through the centre, they do not bisect each other. Let ABCD be a circle, and AC, BD two straight lines in it, which cut each other in the point E, and do not both pass through the centre; AC, BD do not bisect each other. For, if it is possible, let AE be equal to EC, and BE to ED. If one of the lines pass through the centre, it is plain that it cannot be bisected by the other, which does not pass through the centre. But AA if neither of them pass through the F D centre, takea F the centre of the circle, and join EF. Because B E FE, a straight line through the centre, bisects another AC, which a 1. 3. K |