Problem #PRU-56582

Problem

The diagonals of the trapezium $ABCD$ with the bases $AD$ and $BC$ intersect at the point $O$; the points $B^{\prime }$ and $C^{\prime }$ are symmetrical to the vertices $B$ and $C$ with respect to the bisector of the angle $BOC$. Prove that $\mathrm{\angle }{C}^{\prime }AC=\mathrm{\angle }{B}^{\prime }DB$.