The points $A, B, C, D$ lie in that order on a circle. The segments $AC$ and $BD$ intersect at the point $P$. The point $B'$ lies on the line $AB$ such that $A$ is between $B$ and $B'$ and $|AB'| = |DP |$. The point $C'$ lies on the line $CD$ such that $D$ is between $C$ and $C'$ lies and $|DC' | = |AP|$. Prove that $\angle B'PC' = \angle ABD'$. =