x. Similar segments of circles are those that contain equal angles. xi. A sector of a circle is formed of two radii and the arc included between them. To a pair of radii may belong either of the two conjugate arcs into which their ends divide the circle.—Newcomb. xii. Concentric circles are those that have the same centre. xiii. Points which lie on the circumference of a circle are said to be concyclic. xiv. A cyclic quadrilateral is one which is inscribed in a circle. xv. It will be proper to give here an explanation of the extended meaning of the word angle in Modern Geometry. This extension is necessary in Trigonometry, in Mechanics—in fact, in every application of Geometry, and has been partly given in I. Def. ix. Thus, if a line OA revolve about the point O, as in figures 1, 2, 3, 4, until it comes into the position OB, the amount of the rotation from OA to OB is called an angle. From the diagrams we see that in fig. 1 it is less than two right angles; in fig. 2 it is equal to two right angles; in fig. 3 greater than two right angles, but less than four; and in fig. 4 it is greater than four right angles. The arrow-heads denote the direction or sense, as it is technically termed, in which the line OA turns. It is usual to call the direction indicated in the above figures positive, and the opposite negative. A line such as OA, which turns about a fixed point, is called a ray, and then we have the following definition:— xvi. A ray which turns in the sense opposite to the hands of a watch describes a positive angle, and one which turns in the same direction as the hands, a negative angle. PROP. I.—Problem. To find the centre of a given circle (ADB). Sol.—Take any two points A, B in the circumference. Join AB. Bisect it in C. Erect CD at right angles to AB. Produce DC to meet the circle again in E. Bisect DE in F. Then F is the centre. Dem.—If possible, let any other point G be the centre. Join GA, GC, GB. Then in the triangles ACG, BCG we have AC equal to CB (const.), CG common, and the base GA equal to GB, because they are drawn from G, which is, by hypothesis, the centre, to the circumference. Hence [I. viii.] the angle ACG is equal to the adjacent angle BCG, and therefore [I. Def. xiii.] each is a right angle; but the angle ACD is right (const.); therefore ACD is equal to ACG—a part equal to the whole—which is absurd. Hence no point can be the centre which is not in the line DE. Therefore F, the middle point of DE, must be the centre. The foregoing proof may be abridged as follows:—Because ED bisects AB at right angles, every point equally distant from, the points A, B must lie in ED [I. x. Ex. 2]; but the centre is equally distant from A and B; hence the centre must be in ED; and since it must be equally distant from E and D, it must be the middle point of DE. Cor. 1.—The line which bisects any chord of a circle perpendicularly passes through the centre of the circle.