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.