4. DB is greater than DE; but DB is equal to DF (I. Def. 15); wherefore the less than the greater, which is impossible; therefore 6. 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; therefore 7. AB 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 conversely, 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, in the point F. Then CD shall cut AB at right angles. F D Construction. 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 1. The angle AFE is equal to the angle BFE (I. 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 (I. Def. 10); therefore 2. 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. Conversely, let CD cut AB at right angles. Then CD shall also bisect AB, that is, AF shall be equal to F The same construction being made, because EB, EA, from the centre are equal to one another (I. Def. 15); therefore 1. The angle EAF is equal to the angle EBF (I. 5) ; and the right angle AFE is equal to the right angle BFE (I. Def. 10), 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 the other sides are equal (I. 26); therefore 2. AF is equal to FB. Wherefore, if a straight line, &c. Q.E.D. PROPOSITION 4.-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. 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. E B D 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, find F the centre of the circle (III. 1), and join EF. Demonstration. Then because FE, a straight line drawn through the centre, bisects another AC which does not pass through the centre (hyp.); therefore FE cuts AC at right angles (III. 3); wherefore 1. FEA is a right angle. Again, because the straight line FE bisects the straight line BD, which does not pass through the centre (hyp.); therefore FE cuts BD at right angles (III. 3); wherefore 2. FEB is a right angle; but FEA was shown to be a right angle; therefore 3. The angle FEA is equal to the angle FEB (Ax. 1), the less equal to the greater, which is impossible; therefore 4. AC, BD do not bisect one another. 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. Construction. If possible, let E be the centre of the two circles; join EC, and draw any straight line EFG meeting the circumferences in F and G. Demonstration. And because E is the centre of the circle ABC; therefore 1. EF is equal to EC (I. Def. 15); A E again, because E is the centre of the circle CDG; therefore 2. EG is equal to EC (I. Def. 15) ; but EF was shown to be equal to EC; therefore 3. EF is equal to EG (Ax. 1), the less line equal to the greater, which is impossible. Therefore 4. E is not the centre of the circles ABC, CDG. Wherefore, if two circles, &c. Q.E.D. PROPOSITION 6.-Theorem. If one circle touch another internally, they shall not have the sam、 centre. Let the circle CDE touch the circle ABC internally in the point C. Then they shall not have the same centre. D B Construction. If possible, let F be the centre of the two circles; join FC, and draw any straight line FEB, meeting the circumferences in E and B. Demonstration. And because F is the centre of the circle ABC, 1. FB is equal to FC (I. Def. 15); C B also, because F is the centre of the circle CDE, 2. FE is equal to FC (I. Def. 15) ; but FB was shown to be equal to FC; therefore 3. FE is equal to FB (Ax. 1), the less line equal to the greater, which is impossible; therefore 4. F is not the centre of the circles ABC, CDE. Therefore, 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 fromit to the circumference, the greatest is that in which the centre is, and the other part of that diameter is the least; and, of the rest, that which is nearer to the line which passes through the centre, is always greater than one more remote; and, from the same point, there can be drawn only two equal tsraight lines to the circumference, one upon each side of the diameter. Let ABCD be a circle, and AD its diameter, in which let any point F be taken which is not the centre; let the centre be E. Then, of all the straight lines FB, FC, FG, &c., that can be drawn |