A cute geometry problem about angle trisectors.

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.

• Am I missing something? Doesn't it immediately follow from the fact that the angle bisectors of the angles of the triangle $APB$ are concurrent and concur at $Q$. Since $QA$ is the angle bisector of $\angle PAB$, and $QB$ is the angle bisector of $\angle PBA$, we have that $PQ$ is the angle bisector of $\angle APB$. – user17762 May 23 '12 at 22:56
• @Marvis, please, do not spoil the fun for others ;-) (Yes, it is that simple.) – dtldarek May 23 '12 at 23:01
• What's wrong with spoiling other's fun? Besides, surely the idea of math.se is to post aswers. Maybe @Marvis should post the comment as an answer. – George Lowther May 23 '12 at 23:13
• @GeorgeLowther That is what I meant, the comments below the question are often for clarification and other similar things (and so this is spoiling the fun for those who read comments, but not answers). Posting this as an answer would be definitely a better idea ;-) – dtldarek May 23 '12 at 23:20
• I remember a similar problem but not where I saw it. Can you remember where you saw this one? – plm May 23 '12 at 23:27

$Q$ 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.