Book III. VI tained by a straight line and the cir- VII. VIII. tained by two straight lines drawn IX. upon the circumference intercepted X. tained by two straight lines drawn from & 1o. I. bis. I. Let ABC be the given circle ; it is required to find its center. Draw within it any straight line AB, and bisect a it in D; from the point D drawb DC at right angles to AB, and produce it to E, and bisect CE in F: The point F is the center of the circle ABC. Forg FG For, if it be not, let, if poflible, G be the center, and join Book III. С c 8. D B E d 10. def. 17 right angle : But 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 shewn, that no other point but F is the center ; that is, F is the center 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 center of the circle is in the line which bisects the other. TF any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle. Let ABC be a circle, and A, B any two points in the circumference ; the straight line drawn С from A to B shall fall within the circle. For, if it do not, let it fall, if pos a 1.3 fible, without, as AEB; find - D 'the center of the circle ABC, and join AD, DB, and produce DF any straight line D meeting the circumference AB, to E: Then because DA is equal to DB, the F angle DAB is equal to the angle DBA; bs.si and because AE, a side of the triangle A E B DAE, E 3 + N.B. Whenever the expression “ straight lines from the center,” or “ drawn " from the center," occurs, it is to be understood that they are drawn to the circumference. C16. I. Book 11. DAE, is produced to B, the angle DEB is greater than the mangle DAE; but DAE is equal to the angle DBE; therefore the angle DEB is greater than the angle DBE : But to the great er angle the greater fide is opposite ; DB is therefore greater 19. I. than DE: But DB is equal to DF; wherefore DF is greater than DE, the less than the greater, which is impollible : Therefore the straight line drawn from A to B does not fall without the circle. In the same manner, it may be demonstrated that it does not fall upon the circumference; it falls therefore within it. Wherefore, if any two points, &c. Q. E. D. PRO P. III. THEO R. IF bisect a Itraight line in it which does not pass through the center, it shall cut it at right angles ; And if it cuts it at right angles, it shall biseat it. Let ABC be a circle ; and let CD, a straight line drawn through the center bisect any straight line AB, which does not pass through the center, in the point F: It cuts it also at right angles. & 1. 3• Take . E the center of the circle, and join EA, EB: Then, because AF is equal to FB, and FE common to the two triangles AFE, BFE, there are two sides in the one equal to two fides in the other, and the base EA is 0 angle AFE is equal to the angle BFE: E c 10. def. 1. angle : Therefore each of the angles AFE, BFE is a right angle; wherefore But let CD cut AB at right angles ; CD also bisects it, that is, AF is equal to FB. The fame construction being made, because EA, EB from d 5.1. the center are equal to one another, the angle EAF is equal a to the angle EBF; and the right angle AFE is equal to the right angle BFE: Therefore, in the two triangles EAF, EBF, there there are two angles in one equal to two angles in the other, Book III. and the fide 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 straight e 26.-8. line, &c. Q. E. D. IF in a circle two straight lines cut one another which do not both pass through the center, they do not bifeet each the other. Let ABCD be a circle, and AC, BD two straight lines in it which cut one another in the point E, and do not both pass through the center : AC, BD do not bisect one another. For, if it is possible, let AE be equal to EC, and BE to ED: If one of the lines pass thro' the center, it is plain that it cannot be bifected by the other which does not pass thro' the center : But, if neither of them pass thro' the center, take a F the center of the circle, and join EF : And because FE, a straight a 1. 3. F D line thro' the center, bifects another A AC which does not pass thro' the center, it shall cut it at right bangles ; B В E wherefore FEA is a right angle: Again, because the straight line FE bilects the straight line BD which does not pass thro' the center, it shall cut it at right angles; wherefore FEB is a right angle: And FEA was shewn to be a right angle; therefore FEA is equal to the angle FEB, the le's to the greater, which is impossible: Therefore AC, BD do not bifect one another. Wherefore, if in a circle, &c. Q. E. D. b 3. 3. I two circles cut one another, they shall not have the fame center. Let the two circles ABC, CDG cut one another in the points B, C; they have not the same center. E 4 For, Book II. I'or, if it be possible, let E be their center: Join EC, and draw any Itraight line EFG meet C A D F PRO P. VI. THE O R... F two circles touch one another internally, they shall not have the same center. Let the two circles ABC, CDE touch one another internally in the point C: They have not the same center. For, if they can, let it be F; join FC and draw any straight line FEB meeting them in E and B : And because F is the center of the circle ABC, CF is equal to FB: Alfo, because F is the center of the circle CDE, CF is equal to FE: And B F D E PROP. |