Diagonals of convex quadrilateral $ABCD$ intersect at $O$. If $OA \times OD = OB \times OC$,prove that $AB \parallel CD$.

(N/A) Given: In quadrilateral $ABCD$,diagonals $AC$ and $BD$ intersect at $O$ such that $OA \times OD = OB \times OC$.
Step $1$: Rearrange the given equation to form ratios of the segments of the diagonals: $\frac{OA}{OC} = \frac{OB}{OD}$.
Step $2$: Consider $\triangle AOB$ and $\triangle COD$. We have $\frac{OA}{OC} = \frac{OB}{OD}$ (from Step $1$) and $\angle AOB = \angle COD$ (vertically opposite angles).
Step $3$: By the $SAS$ similarity criterion,$\triangle AOB \sim \triangle COD$.
Step $4$: Since the triangles are similar,their corresponding angles are equal,so $\angle OAB = \angle OCD$ and $\angle OBA = \angle ODC$.
Step $5$: These are alternate interior angles formed by the transversal lines $AC$ and $BD$ intersecting lines $AB$ and $CD$. Since the alternate interior angles are equal,$AB \parallel CD$.