Prove that these two angles are equal 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. :(  

 A: This is a standard question, which uses the idea of homothety.
Hint: Show that $EF$ is parallel to $AD$. This is immediate by homothety.
Hence this gives us a trapezoid that is inscribed in a circle, hence is isosceles, so $AE=FD$, and we are done.

Your original intuition on similar triangles is incorrect, which is why your proof couldn't proceed.
However, what is true is that $PBA \sim PDF$. Show this. If you can do it directly, this is another proof.
Hint: You have already shown that $\angle PAD = \angle PFD$. Now do the other angle by angle chasing.
A: Hint: Note that $PE:PB=PF:PC=r_1:r_2$, hence $EF\|BC$, hence $AE=DF$, hence the claim
A: Have you learned about inversion? If you invert about point P, the result is trivial. The two circles become parallel lines, your blue line becomes a circle that passes through P, and the other lines through P remain unchanged.
A: I came across the property - The angle between a tangent and a chord is equal to the inscribed angle on the opposite side of the chord.
This gives for chord PA: $\angle XPA = \angle PDA$ and for chord PB: $\angle XPB = \angle PCB$.
Using these gives the required result:
$\angle APB = \angle XPB - \angle XPA = \angle PCB - \angle PDA = (\angle DPC + \angle PDC) - \angle PDA = \angle DPC$
