See N. * 10. 1. * 11. 1. + Constr. VI. A segment of a circle is the figure contained by a straight line and the circumference it cuts off. VII. "The angle of a segment is that which is 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. And an angle is said to insist or stand upon X. A sector of a circle is the figure contained by XI. Similar segments of circles are those PROPOSITION I. PROBLEM.-To find the centre of a given circle. Let ABC be the given circle; it is required to find its centre. Draw within it any straight line AB, and bisect * it in D; from the point D, draw DC at right angles to AB, and produce it to E, and bisect CE in F: the point F shall be the centre ABC. of the circle FG A D B For if it be not, let, if possible, G be the centre, and join GA, GD, GB: then, because DA is equal † to DB, and DG common to the two triangles ADG, BDG, the two sides AD, DG are equal to the two BD, DG, each to each; † 15 Def. 1. and the base GA is equal † to the base GB, because they are • 10 Def. 1. drawn from the centre G: therefore the angle ADG is COR. 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 line which bisects the other. Constr. † Ax. 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. A 1. 3. D E B † 15 Def. 1. 5. 1. For if it do not, let it fall, if possible, without, as AEB: find D the centre of the circle ABC, and join DA, DB; in the circumference AB take any point F, join DF, and produce it to E. Then, because DA is equal to DB, the angle DAB is equal to the angle DBA and because AE, a side of the triangle DAE, is produced to B, the exterior angle DEB is greater than the angle DAE: but 16. 1. 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, therefore DB is greater than DE: but DB is equal to DF; wherefore DF is greater than 15 Def. 1. DE, the less than the greater; which is impossible: therefore the straight line drawn from A to B does not fall without the N. B. Whenever the expression "straight lines from the centre" or "drawn from the centre" occurs, it is to be understood that they are drawn to the circumference. 19. 1. 1. 3. † Hyp. * 8. 1. 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. THEOR.-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, in the point F: it shall cut AB at right angles. Take E the centre of the circle, and join EA, EB. Then, because AF is equal † to FB, and FE common to the two triangles AFE, BFE, there are two sides in the one, equal to two sides in the other, each + 15 Def. 1. to each; and the base EA is equal to the * E 10 Def. 1. them is a right angle; 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: CD shall also bisect it, that is, AF shall be equal to FB. The same construction being made, because EA, EB, from † 15 Def. 1. the centre, are equal to one another, the angle EAF is equal to the angle EBF; and the right angle AFE is equal + † 10 Def. 1. to the right angle BFE; therefore, in the two triangles EAF, 5. 1. * 26. 1. 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; therefore AF is equal to FB. Wherefore if a straight line, &c. Q. E. D. PROPOSITION IV. THEOR.-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: AC, BD shall not bisect one another. 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, B 1.3. ↑ Hyp. take F the centre of the circle, and join A EF: and because FE, a straight line through the centre, bisects another + AC which does not pass through the centre, it cuts it at right angles; • 3.3. wherefore FEA is a right angle: again, because the straight line FE bisects the straight line BD, which does not pass ↑ Hyp. through the centre, it cuts it at right angles; wherefore FEB is a right angle: but FEA was shewn to be a right angle: therefore FEA is equal to the angle FEB, the less to the +1 Ax. greater; which is impossible: therefore AC, BD do not bisect one another. Wherefore, if in a circle, &c. Q. E. D. PROPOSITION V. THEOR.—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 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 them in 3. 3. the 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. THEOR.-If two circles touch one another internally, they shall not have the same centre. Let the two circles ABC, CDE touch one another internally in the point C: they shall not have the same centre. F B D For if they have, let it be F: join FC, and draw any straight line FEB, meeting them in E and B: and because F is the † 15 Def. 1. centre of the circle ABC, FC is equal to FB: also, because F is the centre of the circle CDE, FC is equal to FE: but FC was shewn to be equal to FB; therefore FE is equal † to FB, the less to the greater; which is impossible: therefore F is not the centre of the circles ABC, CDE. Therefore, if two circles, &c. Q. E. D. † 15 Def. 1. † 1 Ax. * 20. 1. PROPOSITION VII. THEOR.-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 it to the circumference, the greatest is that in which the centre is, and the other part of that diameter is the least; and, of any others, 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 straight lines that are equal to one another, one upon 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 the centre be E of all the straight lines FB, FC, FG, &c. that can be drawn from F to the circumference, FA 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 greater than FG. BA E K G DH Join BE, CE, GE: and because two sides of a triangle are greater *than the third, therefore BE, EF are † 15 Def. 1. greater than BF: but AE is equal † to BE; therefore AE, |