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. C E B For if AB do not fall within the circle, let it fall, if possible, without, as AEB; find (III. 1.) D the centre of the circle ABC, and join AD, 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, (I. 5.) 1. The angle DAB is equal to the angle DBA; and because AE, a side of the triangle DAE, is produced to B, (I. 16.) The angle DEB is greater than the angle DAE; but DAE is equal to the angle DBE; therefore 2. 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; 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. PROP. III.-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; it cuts it also at right angles. с Take (III. 1.) E the centre of the circle, and join EA, EB. Then, because AF is equal (Hyp.) 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 EA is equal (I. Def. 15.) to the base EB; therefore (I. 8.) 1. 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 (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 AB that does not pass through the centre, cuts the same at right angles. But let CD cut AB at right angles; CD also bisects it, that is, AF is equal to FB. The same construction being made; because EA, EB, from the centre are equal to one another, (I. 5.) 1. The angle EAF is equal to the angle EBF; and the right angle AFE is equal (I. Def. 10.) 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 (I. 26.); therefore 2. AF is equal to FB. Wherefore, if a straight line, &c. Q.E.D. PROP. 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; AC, BD, do not bisect one another. A B E с 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, take (III. 1.) F the centre of the circle, and join EF. And because FE, a straight line through the centre, bisects another AC which does not pass through the centre, it shall cut it (III. 3.) at right angles; 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, it shall cut it at right angles (III. 3.); wherefore 2. FEB is a right angle: and FEA was shown to be a right angle; therefore PROP. V.-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 ; they have not the same centre. с E B G 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, (I. Def. 15.) CE is equal to EF: 1. the less to the greater, which is impossible; therefore Q.E.D. PROP. VI.-THEOREM. 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 have not the same centre. For, if they can, let the centre he 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, (I. Def. 15.) 1. CF is equal to FB; also, because F is the centre of the circle CDE, CF is equal to FE: 2. but CF was shown to be equal to FB; therefore (Ax. 1.) the less to the greater, which is impossible; wherefore PROP. VII.-THEOREM. 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 is the greatest, and FD, the other part of the diameter AD, is the least: and of the others, FB is greater than FC, and FC than FG. A B Ꭰ Join BE, CE, GE; and because two sides of a triangle are greater than the third, (I. 20.) 1. BE, EF, are greater than BF; but AE is equal (I. Def. 15.) to EB; therefore 2. 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 the angle BEF is greater than the angle CEF; therefore (I. 24.) 3. The base BF is greater than the base FC: for the same reason, 4. CF is greater than GF. Again, because GF, FE, are greater (I. 20.) than EG, and EG is equal to ED; therefore AE, EF, that is, AF, is greater than BF. 5. GF, FE, are greater than ED: take away the common part FE, and (Ax. 5.) 6. The remainder GF is greater than the remainder FD. Therefore FA is the greatest, and FD the least of all the straight lines from F to the circumference; and BF is greater than CF, and CF than GF. Also there can be drawn only two equal straight lines from the point F to the circumference, one upon each side of the shortest line FD. At the point E, in the straight line EF, make (I. 23.) the angle FEH equal to the angle GEF, and join FH. Then because GE is equal (I. Def. 15.) to EH, and EF common to the two triangles GEF, HEF, the two sides GE, EF, are equal to the two HE, EF, each to each; and the angle GEF is equal (Constr.) to the angle HEF; therefore (I. 4.) 1. The base FG is equal to the base FH: but 2. Besides FH, no other straight line can be drawn from F to the circumference equal to FG ; for, if there can, let it be FK; and because FK is equal to FG, and FG to GH, 3. FK is equal to FH; that is, a line nearer to that which passes through the centre, is equal to one which is more remote; which is impossible. Therefore, if any point be taken, &c. Q.E.D. PROP. VIII.-THEOREM. If any point be taken without a circle, and straight lines be drawn from it to the circumference, whereof one passes through the centre; of those which fall upon the concave circumference, the greatest is that which passes through the centre; and of the rest, that which is nearer to the one passing through the centre is always greater than one more remote: but of those which fall upon the convex circumference, the least is that between the point without the circle and the diameter; and of the rest, that which is nearer to the least is always less than one more remote: and only two equal straight lines can be drawn from the same point to the circumference, one upon each side of the least line. |