About the point $P$ inside a square $ABCD$ satisfying $BP = BC$ and $\angle APD = 90^\circ$ I'm trying to solve the following problem from the Polish XI Olimpiada Matematyczna Gimnazjalistów, question 2 from this document. (The deadline of 12 October 2015 has now passed.)

I managed to solve it using analytical approach, but I have to solve it the "normal way" - in this case using angles.
I almost found the proof, but I need a help in showing 1 thing. Take a look at my approach, where I found almost all angles:

I don't know how to show, that angles $DCQ$ is equal to angle $PBC$. When we prove this, we'll prove, that $AP=2PD$.

Consider the circle $\Gamma_1$ centred at $B$ passing through $A, P, C$, and the circle $\Gamma_2$ through $A, D, P$.  Since $\angle DPA$ is a right angle, $AD$ is a diameter of $\Gamma_2$.  Therefore the radius of $\Gamma_1$ is twice that of $\Gamma_2$.
Using your angle chase, $AP$ subtends an angle of $180-2\alpha$ at the centre of $\Gamma_1$, and hence an angle of $90-\alpha$ at the circumference.  We also see that $DP$ subtends an angle of $90-\alpha$ at the circumference of $\Gamma_2$.  Hence the ratio of $AP$ to $DP$ is the ratio of the two radii, which is $2$.
