The diagonals of the trapezium \(ABCD\) with the bases \(AD\) and \(BC\) intersect at the point \(O\); the points \(B'\) and \(C'\) are symmetrical to the vertices \(B\) and \(C\) with respect to the bisector of the angle \(BOC\). Prove that \(\angle C'AC = \angle B'DB\).