An inner circle touches the outer one at point P. BC is any chord of the inner circle, which when extended, cuts the outer circle at points A and D. That is, the line segment ABCD is a chord of the outer circle. Prove that $\angle APB = \angle DPC$.
I've attached a (fancy) drawing of the problem.
I extended PB and PC to cut the outer circle at E and F respectively. It feels like $\triangle PAE \sim \triangle PDF$ (it would suffice for the proof), but I've been unable to prove so. This is my progress:
$\angle PEA = \angle PDA = \angle PDC$ (same chord PA subtends equal angles on same side of circumference)
$\angle PAD = \angle PAB = \angle PFD$ (chord PD)
And this is where I'm stuck. :(