• $AB=EF$ is equivalent to $\angle AFB = \angle EFA$
• Prove the angle condition using $\angle BOA = 2\angle BEA$

Assume that the center of $\alpha$ is $O$. We have

$\angle FBE = \angle PBE = 180^\circ - \angle EPB - \angle PEB$

$=180^\circ-\angle APB-\angle PEB = \angle AOB-\angle PEB $

$=2\angle AEB - \angle AEB = \angle AEB$

So, on circle $ABEF$, $\angle AEB = \angle FBE \implies AB=EF$.