Diagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB \parallel DC$ intersect each other at the point $O$. Using a similarity criterion for two triangles,show that $\frac{OA}{OC} = \frac{OB}{OD}$.

(N/A) Given: $ABCD$ is a trapezium with $AB \parallel CD$ and diagonals $AC$ and $BD$ intersecting at $O$.
To prove: $\frac{OA}{OC} = \frac{OB}{OD}$.
Proof:
Consider $\triangle OAB$ and $\triangle OCD$.
$1$. $\angle AOB = \angle COD$ (Vertically opposite angles).
$2$. $\angle OAB = \angle OCD$ (Alternate interior angles,since $AB \parallel CD$ and $AC$ is a transversal).
$3$. $\angle OBA = \angle ODC$ (Alternate interior angles,since $AB \parallel CD$ and $BD$ is a transversal).
Therefore,by $AAA$ similarity criterion,$\triangle OAB \sim \triangle OCD$.
Since the triangles are similar,their corresponding sides are proportional:
$\frac{OA}{OC} = \frac{OB}{OD}$.
Hence proved.