VII. The angle of a segment is that which is contained by the straight line and the circumference. VIII. An angle in a segment is the angle contained by two straight lines drawn from any point in the circumference of the segment, to the extremities of the straight line which is the base of the segment. IX. An angle is said to insist or stand upon the circumference intercepted between the straight lines that contain the angle. X. A sector of a circle is the figure contained by two straight lines drawn from the centre, and the circumference between them. XI. Similar segments of circles are those in which the angles are equal, or which contain equal angles. Draw within it any straight line AB, and bisect AB in D; (1. 10.) from the point D draw DC at right angles to AB, (1.11.) produce CD to E, and bisect CE in F. Then the point F shall be the centre of the circle ABC. For, if it be not, let, if possible, G be the centre, and join GA, GD, GB. Then, because DA 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, (1. def. 15.) because they are drawn from the centre G: therefore the angle ADG is equal to the angle GDB: (1.8.) 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; (1. def. 10.) therefore the angle GDB is a right angle: but FDB is likewise a right angle; (constr.) wherefore the angle FDB is equal to the angle GDB, (ax. 1.) the greater equal to the less, which is impossible; therefore G is not the centre of the circle ABC. In the same manner it can be shewn that no other point out of the line CE is the centre; and since CE is bisected in F, any other point in CE divides CE into unequal parts, and cannot be the centre. Therefore no point but 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 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. Let ABC be a circle, and A, B any two points in the circumference. The straight line drawn from A to B shall fall within the circle. 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, (111. 1.) and join DA, DB; in the circumference AB take any point F, join DF, and produce it to meet AB in E. Then, because DA is equal to DB, (1. def. 15.) therefore the angle DAB is equal to the angle DBA; (1. 5.) but DAE was proved equal to the angle DBE; therefore the angle DEB is greater than the angle DBE; but to the greater angle the greater side is opposite, (1. 19.) therefore DB is greater than DE: but DB is equal to DF; (1. def. 15.) wherefore 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 demonstrated that it does not fall upon the circumference; therefore it falls within it. Wherefore, if any two points, &c. Q. E.D. PROPOSITION III. THEOREM. If a straight line drawn through the centre of a circle bisect a straight line in which it does not pass through the centre, it shall cut it at right angles: and conversely, if it cuts 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. Take E the centre of the circle, (III. 1.) and join EA, EB. 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; (1. 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 AB, that is, AF shall be equal to FB. The same construction being made. Because EA, EB, from the centre are equal to one another, (1. def. 15.) the angle EAF is equal to the angle EBF; (1. 5.) and the right angle AFE is equal to the right angle BFE: (1. 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; (1. 26.) therefore AF is equal to FB. Wherefore, if a straight line, &c. Q.E.D. PROPOSITION 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. 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. A E For, if it is possible, let AE be equal to EC, and BE to ED. 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, (111. 1.) and join EF. 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: (111. 3.) 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: (111. 3.) but FEA was shewn to be a right angle; therefore the angle FEA is equal to the angle FEB, (ax. 1.) PROPOSITION V. THEOREM. If two circles cut one another, they shall not have the same centre. A E B For, if it be possible, let E be their centre: join EC, and draw any straight line EFG meeting them in F and G. And because E is the centre of the circle ABC, therefore EC is equal to EF: (1. def. 15.) again, because E is the centre of the circle CDG, therefore EC is equal to EG: (1. def. 15.) but EC was shewn to be equal to EF; therefore EF is equal to EG, (ax. 1.) the equal 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. PROPOSITION VI. THEOREM. If one circle touch another internally, they shall not have the same centre. Let the circle CDE touch the circle ABC internally in the point C. They shall not have the same centre. D B For, if they have, let it be F: join FC, and draw any straight line FEB, meeting them in E and B. also, because F is the centre of the circle CDE, |