Two circles $c$ and $c'$ with centers $O$ and $O'$ lie completely outside each other. Points $A, B$, and $C$ lie on the circle $c$ and points $A', B'$, and $C$ lie on the circle $c'$ so that segment $AB\parallel A'B'$, $BC \parallel B'C'$, and $\angle ABC = \angle A'B'C'$. The lines $AA', BB$', and $CC'$ are all different and intersect in one point $P$, which does not coincide with any of the vertices of the triangles $ABC$ or $A'B'C'$. Prove that $\angle AOB = \angle A'O'B'$.