Parallelogram ${ABCD}$ is given with ${AC>BD}$, and ${O}$ intersection point of ${AC}$ and ${BD}$. Circle with center at ${O}$and radius ${OA}$ intersects extensions of ${AD}$and ${AB}$at points ${G}$ and ${L}$, respectively. Let ${Z}$ be intersection point of lines ${BD}$and ${GL}$. Prove that $\angle ZCA={{90}^{{}^\circ }}$.