Draw within it any straight line AB, The point F shall be the centre For if F be not the centre, D B E if possible, let G be the centre; and join GA, GD, GB. Then, because DA is equal to DB, [Construction. and DG is common to the two triangles ADG, BDG ; the two sides AD, DG are equal to the two sides BD, DG, each to each ; and the base GA is equal to the base GB, because they are drawn from the centre G; [I. Definition 15. therefore the angle ADG is equal to the angle BDG. [I. 8. But when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of the angles is called a right angle; [I. Definition 10. therefore the angle BDG is a right angle. But the angle BDF is also a right angle. [Construction. Therefore the angle BDG is equal to the angle BDF, [Ax. 11. the less to the greater; which is impossible. Therefore G is not the centre of the circle ABC. In the same manner it may be shewn that no other point out of the line CE is the centre; and since CE is bisected at F, any other point in CE divides it into unequal parts, and cannot be the centre. Therefore no point but F is the centre ; that is, F is the centre of the circle ABC: which was to be found. COROLLARY. 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 straight line which bisects the other. PROPOSITION 2. THEOREM. If 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 and B any two points in the circumference: the straight line drawn from A to B shall fall within the circle. For if it do not, let it fall, if possible, without, as AEB. Find D the centre of the circle ABC; [III. 1. and join DA, DB; in the arc AB take any point F, join DF, and produce it to meet the straight line AB at E. Then, because DA is equal to ᎠᏴ, [I. Definition 15. the angle DAB is equal to the angle DBĄ. E B [I. 5. And because AE, a side of the triangle DAE, is produced to B, the exterior angle DEB is greater than the interior opposite angle DAE. [I. 16. But the angle DAE was shewn to be equal to the angle DBE; therefore the angle DEB is greater than the angle DBE. But the greater angle is subtended by the greater side; [I. 19. therefore DB is greater than DE. But DB is equal to DF; [I. Definition 15. therefore DF is greater than DE, the less than the greater; which is impossible. Therefore the straight line drawn from A to B does not fall without the circle. In the same manner it may be shewn that it does not fall on the circumference. Therefore it falls within the circle. Wherefore, if any two points &c. Q.E.D. PROPOSITION 3. 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 if it cut it at right angles it shall 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, at the point F: CD shall cut AB at right angles. Take E the centre of the circle; and join EA, EB. [III.1. Then, because AF is equal to FB, [Hypothesis. and FE is common to the two triangles AFE, BFE ; the two sides AF, FE are equal to the two sides BF,FE, each to each; and the base EA is equal to the base EB; F [I. Def. 15. therefore the angle AFE is equal to the angle BFE. [I. 8. But when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of the angles is called a right angle; [I. Definition 10. therefore each of the angles AFE, BFE is a right angle. Therefore the straight line CD, drawn through the centre, bisecting another AB which does not pass through the centre, also cuts it at right angles. But let CD cut AB at right angles: CD shall also bisect AB; that is, AF shall be equal to FB. The same construction being made, because EA, EB, drawn from the centre, are equal to one another, [I. Def. 15. the angle EAF is equal to the angle EBF. [I. 5. And the right angle AFE is equal to the right angle BFE. Therefore 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 their other sides are equal; therefore AF is equal to FB. Wherefore, if a straight line &c. Q.E.D. [I. 26. If in a circle two straight lines cut one another, which do not both pass through the centre, they do not bisect one another. Let ABCD be a circle, and AC, BD two straight lines in it, which cut one another at the point E, and do not both pass through the centre: AC, BD shall not bisect one another. If one of the straight 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, if possible, let AE be equal to EC, and BE equal to ED. Take F the centre of the circle and join EF. E [III. 1. Then, because FE, a straight line drawn through the centre, bisects another straight line AC which does not pass through the centre; FE cuts AC at right angles; [Hypothesis. [III. 3. therefore the angle FEA is a right angle. Again, because the straight line FE bisects the straight line BD, which does not pass through the centre, FE cuts BD at right angles; therefore the angle FEB is a right angle. But the angle FEA was shewn to be a right angle; [Hyp. [III. 3. therefore the angle FEA is equal to the angle FEB,[Ax. 11. the less to the greater; which is impossible. Therefore AC, BD do not bisect each other. Wherefore, if in a circle &c. PROPOSITION 5. Q.E.D. THEOREM. If two circles cut one another, they shall not have the same centre. Let the two circles ABC, CDG cut one another at the points B, C: they shall not have the same centre. For, if it be possible, let E be their centre; join EC, and draw any straight line EFG meeting the circumferences at F and G. Then, because E is the centre of the circle ABC, EC is equal to EF. [I. Definition 15. Again, because E is the centre of the circle CDG, EC is equal to EG. [I. Definition 15. But EC was shewn to be equal to EF; therefore EF is equal to EG, the less to the greater; which is impossible. [Axiom 1. Therefore E is not the centre of the circles ABC, CDG. Wherefore, if two circles &c. PROPOSITION 6. Q.E.D. THEOREM. If two circles touch one another internally, they shall not have the same centre. Let the two circles ABC, CDE touch one another internally at the point C: they shall not have the same centre. For, if it be possible, let F be their centre; join FC, and draw any straight line FEB, meeting the circumferences at E and B. Then, because F is the centre of the circle ABC, FC is equal to FB. [I. Def. 15. Again, because F is the centre of the circle CDE, FC is equal to FE. E [I. Definition 15. But FC was shewn to be equal to FB; therefore FE is equal to FB, [Axiom 1. the less to the greater; which is impossible. Therefore Fis not the centre of the circles ABC, CDE. Wherefore, if two circles &c. Q.ED. |