Problem 1385

1. Identify the given elements and their properties:
- Parallelogram $ABCD$ with $AC > BD$.
- $O$ is the intersection point of diagonals $AC$ and $BD$.
- Circle centered at $O$ with radius $OA$ intersects extensions of $AD$ and $AB$ at points $G$ and $L$, respectively.
- $Z$ is the intersection point of lines $BD$ and $GL$.

2. **Consider the line $l$ perpendicular to $AC$ at $C$:**
- Let $l$ intersect $AB$ at $X$ and $AD$ at $Y$.

3. **Prove that $GL$ and $BD$ intersect $XY$ at the same point:**
- Since $O$ is the center of the circle and $OA = OG = OL$, $G$ and $L$ lie on the circle centered at $O$ with radius $OA$.
- $O$ is the midpoint of both diagonals $AC$ and $BD$ because diagonals of a parallelogram bisect each other.

4. **Use the properties of the circle and the perpendicular line $l$:**
- Since $l$ is perpendicular to $AC$ at $C$, $l$ is the perpendicular bisector of $AC$.
- Therefore, $X$ and $Y$ are symmetric with respect to $AC$.

5. **Show that $GL$ and $BD$ intersect $XY$ at the same point:**
- Since $G$ and $L$ are on the circle centered at $O$ with radius $OA$, and $O$ is the midpoint of $AC$ and $BD$, the line $GL$ will be symmetric with respect to $AC$.
- The intersection point $Z$ of $BD$ and $GL$ will lie on the line $XY$ because $XY$ is perpendicular to $AC$ and passes through $C$.

6. **Conclude that $\angle ZCA = 90^\circ$:**
- Since $Z$ lies on $XY$ and $XY$ is perpendicular to $AC$ at $C$, $\angle ZCA = 90^\circ$.

$\blacksquare$