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 $AP=2PD$.

I would go about it a bit differently. If $Q$ is the midpoint of $\overline{AP}$, $\angle BQA$ is a right angle, $|BA|=|AD|$ (why?), and $\angle ABQ=\angle ADP$ (why?), so the triangles $\triangle ABQ$ and $\triangle ADP$ are congruent. Thus, $|AQ|=|DP|$, and therefore ... ?
