1. 1.3 *5.1. the same manner it can be shown 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 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 in a circle a straight line bisect another at right angles, the centre of the circle is in the line which bisects the other. 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, 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 * : +15Def. 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 +16. 1. 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 *19.1. greater side is * opposite, therefore DB is greater +15Def. than DE: but DB is equal to DF; 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 demon 1 strated that it does not fall upon the circumference; therefore it falls within it. Wherefore, if any two points, &c. Q. E. D. 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 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: it shall cut AB at right angles. +15Def. 1. Take E the centre of the circle, and join EA, *1. 3., EB. Then, because AF is equal† to FB, and †Hyp 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; 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 of them is a right* angle; therefore each of the an * *8 1. B *10 Det. 1. gles 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. +15 Def. 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 the centre are equal † to one an- 1. other, the angle EAF is equal to the angle *5. 1. EBF; and the right angle AFE is equal to 10 Def the right angle BFE: therefore, in the two tri 1. angles, 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; there26.1. fore the other sides are equal; therefore AF is equal to FB. Wherefore, if a straight line, &c. Q. E. D. * 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 pass through 1.3. the centre, take* F the centre A 3.3. of the circle, and join EF; and BE * F Hyp. other+ AC which does not pass through the centre, it cuts it at right angles: wherefore FEA is a right angle: again, because the straight Hyp. line FE bisects+ the straight line BD, which does not pass through the centre, it cuts it 3.3. at right angles: wherefore FEB is a right angle: but FEA was shown to be a right angle; 1AX. 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. Q. E. D. 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 F and G: and because E is the centre of the cir cle ABC, EC is equal to EF: again, because E is the centre of the circle CDG, EC is equal t to EG: but EC was shown to be equal to EF; therefore EF is equal to EG, the less to the greater, which is impossible. D +15 Def 1. +15 Def E 1. B +1 Ax. Therefore E is not the centre of the circles ABC, CDG. Wherefore, if two circles, &c. Q. E. D. 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. 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 centre of the circle ABC, FC is equal † to FB; also, because Fis the centre of the circle CDE, FC is equal to FE: but FC was F D B +15 Def. 1. +15 Def. shown to be equal to FB; therefore FE is 1. equal to FB, the less to the greater, which is † 1 Ax. H *20. 1. impossible: therefore F is not the centre of the circles ABC, CDE. Therefore, if two circles, &c. Q. E. D. PROP. 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. Join BE, CE, GE: and because two sides of a triangle are greater than the third, therefore BE, EF are greater than BF: but AE is +15 Def. equal to BE; therefore AE, EF, that is, AF is greater than BF: again, because BE is equal to CE, and FE common to the triangles BEF, CEF, the two sides BE, EF are equal to the two CE, EF, each to each; but +9Ax.1. the angle BEF is greater † than the angle CEF; therefore the * *24. 1. base BF is greater than the base CF: for the same reason CF is greater than GF: again, be* 20. 1. cause GF, FE are greater than EG, and EG |