Belarusian Mathematical Olympiad

Let $D$ be symmetric to $A$ with respect to the vertex $C$ (see the Fig.). The segment $BC$ is the altitude and the median in the triangle $ABD$ so the triangle $ABD$ is isosceles. Likewise, the triangle $ADP$ is isosceles. Let $P^{\prime }$ be the intersection point of the medians $DM$ and $BC$ of the triangle $ABD$. Since the point of intersection divides the medians in the ratio $2:1$, points $P$ and $P^{\prime }$ coincide. Since $C$ belongs to the bisector of the angle $\angle DPA$, we see that $C$ is equidistant to the lines $DP$ and $PA$. If $C$ belongs to the bisection of the angle $\angle PZY$, then $C$ is equidistant to the lines $PZ$ and $ZY$, therefore, $C$ is equidistant to the lines $YP$ and $YX$, i.e. $C$ belongs to the bisection of the angle $\angle PYX$. In the same way, one can prove the converse proposition.