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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 $|\angle APQ| = |\angle QPB|$.

$\hspace{60pt}$ enter image description here

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

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    dtldarek, thanks. I think I saw it on AoPS actually. @Phira, I did not know I knew so much. :) (I did not know that interesting theorem, thx for the pointer.)2012-05-23

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$Q$ 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.