EUCLID'S ELEMENTS OF GEOMETRY. Book III. DEFINITIONS. I. Equal circles are those of which the diameters are equal, or from the centres of which the straight lines to the circumferences are equal. "This is not a definition but a theorem, the truth of which is evident; for, if the circles be applied to one another, so that their centres coincide, the circles must likewise coincide, since the straight lines from the centres are equal." II. A straight line is said to touch a circle, when it meets the circle, and being produced does not cut it. O III. Circles are said to touch one another, which meet, but do not cut one another. IV. Straight lines are said to be equally distant from the centre of a circle, when the perpendiculars drawn to them from the centre are equal. V. And the straight line on which the greater perpendicular falls is said to be farther from the centre. 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 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. And 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. PROP. 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 (I. 10.) AB in D; from the point D draw (I. 11.) DC at right angles to AB, and produce it to E, and bisect CE in F: the point F is 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 (Constr.) 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; and the base GA is equal (I. Def. 15.) to the base GB, because they are drawn from the centre G; therefore (I. 8.) 1. The angle ADG is equal to the angle GDB: 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 (I. Def. 10.); therefore 2. The angle GDB is a right angle: but FDB is likewise (Constr.) a right angle; wherefore 3. The angle FDB is equal to the angle GDB, the greater to the less, which is impossible; therefore 4. G is not the centre of the circle ABC. In the same manner it can be shown, that no other point but F is the centre; that is 5. 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. 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 (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; 2. 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 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. A 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. 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, |