Dem.—1. Suppose the angles at the centres to be given equal. Now because the circles are equal their radii are equal (Def. i.). Therefore the two triangles AOB, DHE have the sides AO, OB in one respectively equal to the sides DH, HE in the other, and the angle AOB equal to DHE (hyp.). Therefore [I. iv.] the base AB is equal to DE.
Again, since the angles ACB, DFE are [xx.] the halves of the equal angles AOB, DHE, they are equal [I. Axiom vii.]. Therefore (Def. x.) the segments ACB, DFE are similar, and their chords AB, DE have been proved equal; therefore [xxiv.] the segments are equal. And taking these equals from the whole circles, which are equal (hyp.), the remaining segments AGB, DKE are equal. Hence the arcs AGB, DKE are equal.
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.