11. Similar segments of circles are those in which the angles are equal, or which contain equal angles. PROPOSITION 1.- Problem. To find the centre of a given circle. Construction. Draw within it any straight line AB, and bisect AB in D (I. 10); from the point D draw DC at right angles to AB (I. 11); produce it to E, and bisect CE in F. Then the point F shall be the centre of the circle ABC. Proof. For, if it be not, let, if possible, G be the centre, and join GA, GD, GB. Then, because DÅ is equal to DB (constr.), and DG common to the two triangles ADG, BDG, the two sides AD, DG, are equal to the two BD, DG, each to each ; and the base GA is equal to the base GB, (I. Def. 15), because they are drawn from the centre G; therefore 1. The angle ADG is equal to the angle GDB (I. 8), but when a straight line standing upon another straight line makes the adjacent angles equal to one another, each of the angles is a right angle (I. Def. 10); therefore 2. The angle GDB is a right angle ; but FDB is likewise a right angle (constr.); wherefore 3. The angle FDB is equal to the angle GDB (Ax. I), the greater to the less, which is impossible; therefore 4. G is not the centre of the circle ABC. In the same manner it can be shown, that no other point but F is the centre; that is, 5. 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 bisects another at right angles, the centre of the circle is in the 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, B, any two points in the circumference. Then the straight line drawn from A to B shall fall within the circle. с E Construction. For if AB do not fall within the circle, let it fall, if possible, without, as AEB; find D, the centre of the circle ABC (III. 1), and join DA, DB; in the circumference AB, take any point F, join DF, and produce it to meet AB in E. Demonstration. Then because DA is equal to DB (I. Def. 15), therefore 1. The angle DAB is equal to the angle DBA (I. 5); and because AE, a side of the triangle DA E, is produced to B, 2. The exterior angle DEB is greater than the interior and opposite angle DAE (I. 16); but DAE was proved equal to the angle DBE; therefore 3. The angle DEB is greater than the angle DBE ; but to the greater angle, the greater side is opposite (I. 19) ; therefore 4. DB is greater than DE; but DB is equal to DF (I. Def. 15); wherefore 5. DF is greater than DE; 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. 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 A B that does not pass through the centre, cuts the same at right angles. Conversely, let CD cut AB at right angles. 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. Then AC, BD, shall not bisect one another. Construction. For, if it he 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. E B 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. Let the two circles ABC, CDG, cut one another in the points B, C. Then 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 |