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