2. The demonstration of this case is included in the foregoing. Cor. 1.—If the opposite angles of a cyclic quadrilateral be equal, one of its diagonals must be a diameter of the circumscribed circle. Cor. 2.—Parallel chords in a circle intercept equal arcs. Cor. 3.—If two chords intersect at any point within a circle, the sum of the opposite arcs which they intercept is equal to the arc which parallel chords intersecting on the circumference intercept. 2. If they intersect without the circle, the difference of the arcs they intercept is equal to the arc which parallel chords intersecting on the circumference intercept. Cor. 4.—If two chords intersect at right angles, the sum of the opposite arcs which they intercept on the circle is a semicircle. PROP. XXVII.—Theorem. In equal circles (ACB, DFE), angles at the centres (AOB, DHE), or at the circumferences (ACB, DFE), which stand on equal arcs (AB, DE), are equal. Dem.—If possible let one of them, such as AOB, be greater than the other, DHE; and suppose a part such as AOL to be equal to DHE. Then since the circles are equal, and the angles AOL, DHE at the centres are equal (hyp.), the arc AL is equal to DE [xxvi.]; but AB is equal to DE (hyp.). Hence AL is equal to AB—that is, a part equal to the whole, which is absurd. Therefore the angle AOB is equal to DHE. 2. The angles at the circumference, being the halves of the central angles, are therefore equal. PROP. XXVIII.—Theorem. In equal circles (ACB, DFE), equal chords (AB, DE) divide the circumferences into arcs, which are equal each to each—that is, the lesser to the lesser, and the greater to the greater. Dem.—If the equal chords be diameters, the Proposition is evident. If not, let O, H be the centres. Join AO, OB, DH, HE; then because the circles are equal their radii are equal (Def. i.). Hence the two triangles AOB, DHE have the sides AO, OB in one respectively equal to the sides DH, HE in the other, and the base AB is equal to DE (hyp.). Therefore [I. viii.] the angle AOB is equal to DHE. Hence the arc AGB is equal to DKE [xxvi.]; and since the whole circumference AGBC is equal to the whole circumference DKEF, the remaining arc ACB is equal to the remaining arc DFE. Exercises. 1. The line joining the feet of perpendiculars from any point in the circumference of a circle, on two diameters given in position, is given in magnitude. 2. If a line of given length slide between two lines given in position, the locus of the intersection of perpendiculars to the given lines at its extremities is a circle. (This is the converse of 1.) PROP. XXIX.—Theorem. In equal circles (ACB, DFE), equal arcs (AGB, DCK) are subtended by equal chords.