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> In the cyclic quadrilateral $ABCD$, the sides $AB, DC$ meet at $Q$, the side $AD,BC$ meet at $P$, $M$ is the midpoint of $BD$. If $\angle APQ = 90$°, prove that $PM$ is perpendicular to $AB$. 
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I feel like this problem should be relatively simple in the end but I'm not getting it yet. So far I've noticed that $\angle DCB$ is equal to $\angle BAP$; $\angle ADC, \angle CBQ$ and $\angle PBA$ are equal; $\angle BAD$ and $\angle QCB$ too are equal. The triangles $QAD$ and $QCB$ are similar, and the triangles $APB$ and $CPD$ as well. triangles $DPM$ and $MPB$ should have the same area (although I believe M could play a more important role). I've looked for similar right triangles which didn't work for me, and have tried marking the intersection of $MP$ and $AB$ as $H$ and let $x=\angle PHA$ and $y=\angle PHB$ and tried to show $x=y$, but that hasn't worked either. Any help would really be appreciated! 

[![Here's a sketch I drew...](https://i.stack.imgur.com/xfN15.png)](https://i.stack.imgur.com/xfN15.png)