therefore 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.-THEOREM. If any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle. (References-Prop. 1. 5, 16, 19; III. 1.) Let ABC be a circle, and A, B, any two points in the circumference. Then the straight line drawn from A to B shall fall within the circle. Find D the centre of the circle ABC, (III. 1) and join AD, DB; in the circumference AB take any point F; join DF, and let it cut the straight line AB in E. DEMONSTRATION Because DA is equal to DB, (1. def. 15) the angle DAB is equal to the angle DBA ; (1. 5) and because AE, a side of the triangle DAE, is produced to B, the exterior angle DEB is greater than the interior and opposite angle DAE; (1. 16) but the angle DAE was proved to be equal to the angle DBE; therefore the angle DEB is also greater than the angle DBE; but to the greater angle the greater side is opposite; (I. 19) therefore DB is greater than DE; but DB is equal to DF; (1. def. 15) therefore DF is greater than DE, wherefore the point E lies within the circle. In the same manner it may be proved that every point in AB lies within the circle; therefore the straight line AB lies within the circle. Wherefore, if any two points, &c. Q. E. D. PROP. III.—THEOREM. If a straight line drawn through the centre of a circle bisect a straight line in it which does not pass through the centre, it shall cut it at right angles; and conversely, if it cut it at right angles, it shall bisect it. (References-Prop. 1. 5, 8, 26; III. 1.) 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. Then CD shail cut AB at right angles. Take E the centre of the circle (III. 1), and join EA, EB. DEMONSTRATION Then, because AF is equal to FB (hyp.), and FE common to the two triangles AFE, BFE, there are two sides in the one equal to two sides in the other, each to each; and the base EA is equal to the base EB; (I. def. 15) therefore the angle AFE is equal to the angle BFE; (1.8) but when a straight line standing upon another straight line makes the adjacent angles equal to one another, each of them is a right angle;' (1. def. 10) therefore each of the angles AFE, BFE, is a right angle; wherefore the straight line CD, drawn through the centre, bisecting another AB that does not pass through the centre, cuts the same at right angles. But let CD cut AB at right angles. Then CD shall also bisect it, that is, AF shall be equal to FB. The same construction being made; because EA, EB, from the centre are equal to one another, the angle EAF is equal to the angle EBF; (1. 5) and the right angle AFE is equal to the right angle BFE; (I. def. 10) hence, in the two triangles EAF, EBF, there are two angles in the one equal to two angles in the other, each to each; and the side EF, which is opposite to one of the equal angles in each, is common to both; therefore the other sides are equal; (1. 26) therefore AF is equal to FB. Wherefore, if a straight line, &c. Q. E. D. PROP. IV. THEOREM If in a circle two straight lines cut one another which do not both pass through the centre, they do not bisect each other. (References Prop. III. 1, 3.) 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 centre. Then AC, BD, shall not bisect one another. CONSTRUCTION For, if it be 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 if neither of them pass through the centre, take F the centre of the circle, (III. 1) and join EF. Because FE, a straight line through the centre, is assumed to bisect another AC which does not pass through the centre, therefore it must cut it at right angles; (III. 3) wherefore FEA must be a right angle. Again, because the straight line FE is assumed to bisect the straight line BD which does not pass through the centre, therefore it must cut it at right angles; (III. 3) wherefore FEB must be a right angle; but FEA was shown to be a right angle; therefore FEA must be equal to the angle FEB, the less to the greater, which is impossible; therefore AC, BD, do not bisect one another. Wherefore, if in a circle, &c. Q. E. D. PROP. V.- THEOREM. If two circles cut one another, they shall not have the same centre. (References-I. def. 15.) Let the two circles ABC, CDG, cut one another in the points B, C. Then they shall not have the same centre. G E B CONSTRUCTION For, if it be possible, let E be their centre; join EC, and draw any straight line EFG meeting them in F and G. DEMONSTRATION And because E is assumed to be the centre of the circle ABC, EC must be equal to EF; (1. def. 15) again, because E is assumed to be the centre of the circle CDG, EC must be equal to EG; but EC was shown to be equal to EF; therefore EF must be equal to EG; the less to the greater, which is impossible; therefore E is not the centre of the circles ABC, CDG. Wherefore, if two circles, &c. Q. E. D. PROP. VI.-THEOREM. If one circle touch another internally, they shall not have the same centre. (References 1. def. 15.) Let the circle CDE, touch the circle ABC internally in the point C. Then they shall not have the same centre. C E D CONSTRUCTION For, if they can, let the centre be F; join FC, and draw any straight line FEB meeting them in E and B. DEMONSTRATION Because F is assumed to be the centre of the circle ABC, therefore CF must be equal to FB; (1. def. 15) also, because F is assumed to be the centre of the circle CDE, therefore CF must be equal to FE; but CF was shown to be equal to FB; |