A cute geometry problem about angle trisectors.

Question

Here is a cute geometry problem I saw some time ago. I know the solution, I just wanted to share ;-) (Please, don't be mad at me.)

Consider an acute triangle $\triangle ABC$. Let $AP$, $AQ$ and $BP$,$BQ$ be the angle trisectors as shown on the picture below. Prove that $|\angle APQ| = |\angle QPB|$.

$\hspace{60pt}$

Edit: There does exist one very simple and elegant solution, so don't be stumbled if you happen to guess/derive it.

Answer 1

$Q$ is the intersection of the two angle bisectors of the triangle $ABP$ at $A$ and $B$. Therefore, $Q$ is the incenter of $ABP$ and $PQ$ is the bisector of $\angle BPA$ as claimed.