Saudi Arabia Mathematical Competitions 2012

Let $M$ be the midpoint of $AB$, and let $E$ be the point on line $BP$ such that $AE\parallel MP$. Then $P$ is the midpoint of $EB$. Since $\frac{AP}{PB}=\frac{CP}{PD}$, we have $\frac{AP}{PE}=\frac{CP}{PD}$. Also, note that

$\widehat{EPC}=\widehat{APD}=90^{\circ }$, so $\widehat{EPA}=\widehat{DPC}$. These last two facts imply that triangles $APE$ and $CPD$ are similar.



From this we conclude that $\widehat{PAE}=\widehat{PCD}$. Since $AE\parallel MP$, $\widehat{MPA}=\widehat{PAE}=\widehat{PCD}$. Take a point $O^{\prime }$ on ray $MP$ past $P$. Then

$$\widehat{O^{\prime }PD}=180^{\circ }-\widehat{APD}-\widehat{MPA}=90^{\circ }-\widehat{PCD}=\widehat{OPD}.$$

Therefore $M,P,O$ are collinear, which completes the proof.