Let $\vartriangle ABC$be a triangle and let $\Gamma$ its circumscribed circle. Let $M$ be the midpoint of the side $BC$ and let $D$ be the point of intersection of the line $AM$ with $\Gamma$. By $D$ a straight line is drawn parallel to $BC$, which intersects $\Gamma$ at a point $E$. Let $N$ be the midpoint of the segment $AE$ and let $P$ be the point of intersection of $CN$ with $AM$. Show that $AP = PC$.