EUCLID'S ELEMENTS, BOOK III, Definitions, 1. Equal circles are those of which the diameters are equal, or from the centres of which the straight lines to the circum ferences are equal. 2. A straight line is said to touch a circle, or to be a tangent to it, when it meets the circle, and being produced does not cut it. 3. Circles are said to touch one another, which meet, but do not cut one another. 4. 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. 5. And the straight line on which the greater perpendicular falls, is said to be farther from the centre. 6. A segment of a circle is the figure contained by a straight line and the circum ference which it cuts off. 7. The angle of a segment is that which is contained by the straight line and the circumference. 8. 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. 9. An angle is said to insist or stand upon the circumference intercepted between the straight lines that contain the angle. 10. A sector of a circle is the figure contained by two straight lines drawn from the centre and the circumference between them. 11. Similar segments of circles are those which contain equal angles. [Any portion of the circumference is called an arc, and the chord of an arc is the straight line joining its extremities.] Proposition 1.—Problem. - To find the centre of a given circle. Let ABC be the given circle. It is required to find its centre. CONSTRUCTION.—Draw within the circle any chord AB, and bisect it in D (I. 10). From the point D draw DC at right angles to AB (I. 11). Produce CD to meet the circumference in E, and bisect CE in F (I. 10). Then the point F shall be the centre of the circle ABC. PROOF.—For if F be not the centre, if A possible let G be the centre; and join GA, GD, GB. Then, because DA is equal to DB (Const.), and Du common to the two triangles ADG, BDG; The two sides AD, DG are equal to the two sides BD, DG, each to each; And the base GA is equal to the base GB, being radii of the same circle; Therefore the angle ADG is equal to the angle BDG (I. 8). ---LADG But when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of the angles is called a right angle (1. Def, 10). Suppose G the centre. :.LGDB Therefore the angle GDB is a right angle. Therefore the angle GDB is equal to the angle FDB (Ax. 11), the less to the greater; which is impossible. Therefore G is not the centre of the circle ABC. In the same manner it may be shown that no point which is not in CE can be the centre. And since the centre is in CE, it must be in F, its point of bisection. Therefore F is the centre of the circle ABC: which was to be found. COROLLARY.–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. Proposition 2.—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 and B any two points in the circumference. The straight line drawn from A to B shall fall within the circle. CONSTRUCTION.–Find D the centre of the circle ABC (III. 1), and join DA, DB. In AB take any point E; join DE, and produce it to the circumference in F. PROOF.-Because DA is equal to DB, the angle DAB is equal to the angle DBA (1. 5). And because AE, a side of the triangle DAE, is produced to B, the exterior angle DEB is greater than the interior and opposite angle DAE (I. 16). But the angle DAE was proved to be equal to the angle DBE; Therefore the angle DEB is also greater than DBE. But the greater angle is subtended by the greater side (I. 19); .:DB>DE. Therefore DB is greater than DE, ZDBE = Z DAB, and... Z DEB DBE. But DB is equal to DF; therefore DF is greater than DE, and the point E is therefore within the circle. In the same manner it may be proved that every point in Therefore the straight line AB lies within the circle. Proposition 3.—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. CD shall cut AB at right angles, CONSTRUCTION.—Take E, the centre of the circle (III. 1), and join EA, EB. Proof.—Because AF is equal to FB (Hyp.), and FE common to the two triangles AFE, BFE, and the base EA equal to the base EB (I. Def. 15), Therefore the angle AFE is equal to the angle BFE (I. 8); Therefore each of the angles AFE, BFE is a right angle (I. def. 10); Therefore the straight line CD, drawn through the centre, respect. bisecting another, AB, that does not pass through the centre, also cuts it at right angles. Conversely, let CD cut AB at right angles. CD shall bisect AB; that is, AF shall be equal to FB —the same construction being made. PROOF.-Because the radii EA, EB are equal, the angle EAF is equal to the angle EBF (I. 5), And the angle AFE is equal to the angle BFE (Hyp.), Therefore, in the two triangles EAF, EBF, there are two angles in the one equal to two angles in the other, each to Triangles AFE and BEF are equal in every each, and the side EF, which is opposite to one of the equal Therefore their other sides are equal (I. 26); Proposition 4.-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 at the point E, and do not both pass through the centre. AC, BD shall not bisect one another. 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. If they But if neither of them pass through the centre, if possible, bisect each let AE be equal to EC, and BE to ED. other, CONSTRUCTION.-Take F, the centre of the circle (III. 1), and join EF. EF bisects Because FE, a straight line drawn AC at through the centre, bisects another line right angles. AC, which does not pass through the centre (Hyp.), therefore it cuts it at right angles (III. 3); Therefore the angle FEA is a right > angle. And EF Again, because the straight line FE bisects the straight 3D line BD, which does not pass through the centre (Hyp.), angle.. therefore it cuts it at right angles (III. 3); Therefore the angle FEB is a right angle. Therefore the angle FEA is equal to the angle FEB, the ...(FEA less to the greater, which is impossible; Therefore AC, BD do not bisect each other." bisects BD at right ELFEB. |