Angles in Same Segment of Circle are Equal

Proof

 * Euclid-III-21.png

Let $ABCD$ be a circle, and let $\angle BAD, \angle BED$ be angles in the same segment $BAED$.

Let $F$ be the center of $ABCD$, and join $BF$ and $FD$.

From the Inscribed Angle Theorem:
 * $\angle BFD = 2 \angle BAD$
 * $\angle BFD = 2 \angle BED$

So $\angle BAD = \angle BED$.

Hence the result.