Question 84
A parallelogram $ABCD$ with obtuse angle $\angle ABC$ is given. After rotating the triangle $ACD$ around the vertex $C$, we get a triangle $CD' A'$, such that the points $B$, $C$ and $D'$ are collinear. The extension of the median of the triangle $CD' A'$ that passes through $D'$ intersects the straight line $BD$ at the point $P$. Prove that $PC$ is the bisector of the angle $\angle BP D'$.