THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD

Let $\{D\}=(BP\cap \mathcal{C})$ and $\{M\}=DE\cap AC$.

Then $\angle BDE=\angle BCE=180^{\circ }-\angle ACB-\angle PCE=180^{\circ }-\angle ABC-\angle CBE=\angle APE$, hence $\Delta MPD\sim \Delta MEP$ (AA), which leads to $\angle MPD\equiv \angle MEP$ and $M{P}^2=MD\cdot ME$.

On the other hand, from the power of the point $M$ with respect to the circle $\mathcal{C}$ it follows that $M{C}^2=MD\cdot ME$, i.e. $M{P}^2=M{C}^2$, which means that $M$ is the midpoint of $[PC]$.

It follows that $\angle MEP\equiv \angle MPD\equiv \angle MQP$, i.e., the quadrilateral $MQEP$ is cyclic. We obtain that $\angle MEQ\equiv \angle MPQ\equiv \angle MEP$, and the conclusion.