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; B and the base EA is equal to the base EB; [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.D2 [I, 26. If in a circle two straight lines cut one another, which do not 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. B E C [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; therefore the angle FEA is a right angle. [Hypothesis. [III. 3. straight line BD, which does not pass through the centre, FE cuts BD at right angles; [Hyp. [III. 3. Again, because the straight line FE bisects the therefore the angle FEB is a right angle. But the angle FEA was shewn to be a right angle; 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. Q.E.D. PROPOSITION 5. 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. 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. A 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. Q.E.D. 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. D [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 F is not the centre of the circles ABC, CDE. Wherefore, if two circles &c. Q.E.D. PROPOSITION 7. THEOREM. If any point be taken in the diameter of a circle which is not the centre, of all the straight lines which can be drawn from this point to the circumference, the greatest is that in which the centre is, and the other part of the diameter is the least; and, of any others, that which is nearer to the straight line which passes through the centre, is always greater than one more remote; and from the same point there can be drawn to the circumference two straight lines, and only two, which are equal to one another, one on each side of the shortest line. Let ABCD be a circle and AD its diameter, in which let any point F be taken which is not the centre; let E be the centre of all the straight lines FB, FC, FG, &c. that can be drawn from F to the circumference, FA, which passes through E, shall be the greatest, and FD, the other part of the diameter AD, shall be the least; and of the others FB shall be greater than FC, and FC than FG. Join BE, CE, GE. Then, because any two sides of a triangle are greater than the third side, [I. 20. that is, AF is greater than BF. Again, because BE is equal to CE, [I. Definition 15. and EF is common to the two triangles BEF, CEF; the two sides BE, EF are equal to the two sides CE, EF, each to each; but the angle BEF is greater than the angle CEF; therefore the base FB is greater than the base FC. [I. 24. In the same manner it may be shewn that FC is greater than FG. Again, because GF, FE are greater than EG, [I. 20. and that EG is equal to ED; [I. Definition 15. Take away the common part FE, and the remainder GF is greater than the remainder FD. Therefore FA is the greatest, and FD the least of all the straight lines from F to the circumference; and FB is greater than FC, and FC than FG. Also, there can be drawn two equal straight lines from the point A to the circumference, one on each side of the shortest line FD. For, at the point E, in the straight line EF, make the angle FEH equal to the angle FEG, and join FH. Then, because EG is equal to EH, [I. 23. [I. Definition 15. and EF is common to the two triangles GEF, HEF; the two sides EG, EF are equal to the two sides EH, EF, each to each; and the angle GEF is equal to the angle HEF ; [Constr. therefore the base FG is equal to the base FH. [I. 4. But, besides FH, no other straight line can be drawn For, if it be possible, let FK be equal to FG. from F to the circumference, equal to FG. Then, because FK is equal to FG, and FH is also equal to FG, therefore FH is equal to FK; [Hypothesis. [Axiom 1. that is, a line nearer to that which passes through the centre is equal to a line which is more remote ; which is impossible by what has been already shewn. Wherefore, if any point be taken &c. Q.E.D. PROPOSITION 8. THEOREM. If any point be taken without a circle, and straight lines be drawn from it to the circumference, one of which passes through the centre; of those which fall on the concave circumference, the greatest is that which passes through the centre, and of the rest, that which is nearer to the one passing through the centre is always greater than one more remote; but of those which fall on the |