PROP. I. PROB. To find the centre of a given circle. Let ABC be the given circle; it is required to find its centre. Draw any straight line AB, terminating in the circumference, and bisect* it in D; from the point D *10. 1. draw* DC at right angles to AB, prolong it to E, and bisect CE in F: the point F shall be the centre of the circle ABC. #11. 1. For, if it be not, let, if possible, G, a point out of the line CE, be the centre, and join GA, GD, GB: then, because DA is equal† to DB, and +Const. 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 112 Def. 1. GA is equal to the base GB, because they are drawn from the centre G therefore the angle ADG is equal to the angle GDB: #8. 1. FG E B but when a straight line standing upon another straight line makes the adjacent angles equal to *8. Def. 1. one another, each of the angles is a* right angle; therefore the angle GDB is a right angle: but +Const. FDB is likewise at right angle; wherefore the angle GDB is equal to the angle FDB, the less to the greater, which is impossible; therefore G, a point out of the line CE, cannot be the centre of the +1 Ax. 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. circle ABC. And since CE is bisected in F, any other point in CE divides it 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 a straight line, terminating in the circumference of a circle, be bisected by a perpendicular, that perpendicular will pass through the centre. PROP. II. THEOR. 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 it do not, let it fall, if possible, #1. 3. without, as AEB; find D the +12 Def. 1. because DA is equalt to DB, the A #5. 1. +16. 1. C EB 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 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 +12 Def. 1. equal† to DF; wherefore DF is greater than DE, the less than the greater, which is impossible; therefore #19. 1. the straight line drawn from A to B does not fall without the circle. In like 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. 1 PROP. 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. #1.3. +Hyp. Take E the centre of the circle, and join EA, EB. Then because AF is equalf 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 to each; and the base +12 Def. 1. EA is equal to the base EB; therefore the angle AFE is equal* to the angle BFE: but when a straight line standing upon another straight line makes the adjacent angles equal to one another, each #8. 1. *8 Def. 1. of them is a right angle; there fore 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. 112 Def. 1. #5. 1. The same construction being made, because the lines EA, EB, from the centre are equal,† the angle EAF is equal to the angle EBF; and the right 11 Ax. 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 the other sides are equal; therefore AF is equal to FB. Wherefore, if a straight line, &c. Q. E. D. #26. 1. PROP. 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 is 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 *1. 3. +Hyp. *3.3. B F pass through the centre, take* F the centre of the circle, and join EF; and because FE, a straight A line through the centre, bisects another+ AC which does not pass through the centre, it cuts it at right angles: wherefore FEA is a right angle: again, because the straight line FE bisects† the straight line BD, which does not pass through the centre, it cuts it at right angles: wherefore FEB is a right angle: but FEA was shown to be a right angle; therefore FEA is equal to the angle FEB, the less to the greater, which is impossible: therefore AC, BD do not bisect one another. Wherefore, if in a circle, &c. +Hyp. *3. 3. +1 Ax. Q. E. D. PROP. 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 point C; they shall not have the same centre. For, if it be possible, let E be their centre: join EC, and draw a straight line EFG meeting them in F and G: and because E is the centre of the circle +12 Def. 1. +12 Def. 1. ABC, EC is equal to EF: again, because E is the centre to EG: but EC was shown to be equal to EF: therefore EF ti Ax. is equal to EG, the less to the greater, which is impossible. B G Therefore E is not the centre of the circles ABC, CDG. Wherefore, if two circles, &c. Q. E. D. PROP. VI. THEOR. If two circles touch one another, they shall not have the same centre. Let the two circles ABC, CDE touch one another in the point C: they shall not have the same centre. For if they have, let it be F: join FC and draw a straight line FEB, meeting them in E and B: and +12 Def. 1. because F is the centre of the circle ABC, +12 Def. 1. +1 Ax. FB; therefore FE is equal to D 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. |