Iranian Mathematical Olympiad

Let $K$ and $T$ be the reflections of $P$ with respect to $M$ and $H$, respectively. According to Thales' Theorem, triangles $AKT$ and $MNH$ are similar. On the other hand, for triangles $ABD$ and $AEC$, $\angle EBD = \angle ECD$ and $\angle BAC$ appears in both triangles; therefore, these two triangles are similar. Now it suffices to prove that triangles $AKT$ and $ABD$ are similar, i.e. it must be shown that $\angle BAD = \angle KAT$ and $\frac{AB}{AD} = \frac{AK}{AT}$.

Note that these relations are equivalent to the similarity of triangles $ADT$ and $ABK$. But $T$ is the reflection of $P$ with respect to line $AC$, so triangles $ADT$ and $ADP$ are congruent and it suffices to show that triangles $APD$ and $ABK$ are similar.

Since diameters of quadrilateral $BPCK$ bisect each other, it is a parallelogram. Thus $BK \parallel CP$. This implies that
$$\angle ABK = \angle AEC = 180^\circ - \angle BEC = 180^\circ - \angle BDC = \angle ADB.$$
Furthermore, since $BPCK$ is a parallelogram, $BK = CP$.

In order to complete the proof it has to be shown that
$$\frac{AD}{DP} = \frac{AB}{BK} = \frac{AB}{CP}.$$

To prove this, the law of sines can be used in triangles $PDC$ and $ABD$. It is sufficient to prove that
$$\frac{\sin(\angle ADB)}{\sin(\angle ABD)} = \frac{\sin(\angle CDP)}{\sin(\angle DCP)}.$$
But $\angle ABD = \angle DCP$ and $\angle ADB = 180^\circ - \angle CDP$, which means the equation above holds, and this completes the proof.