Question 82
Let $ABCD$ be a parallelogram with $AC \gt BD$, and let $O$ be the point of intersection of $AC$ and $BD$. The circle with center at $O$ and radius $OA$ intersects the extensions of $AD$ and $AB$ at points $G$ and $L$, respectively. Let $Z$ be the intersection point of lines $BD$ and $GL$. Prove that $\angle ZCA = 90^{\circ}$.