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.
Cor. 2.—The locus of the centres of the circles which pass through two fixed points is the line bisecting at right angles that connecting the two points.
Cor. 3.—If A, B, C be three points in the circumference of a circle, the lines bisecting perpendicularly the chords AB, BC intersect in the centre.
PROP. II.—Theorem.