Angles in Same Segment of Circle are Equal
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In the words of Euclid:
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.