In the given figure,if $AB \parallel DC$ and $AC$ and $PQ$ intersect each other at the point $O$,prove that $OA \cdot CQ = OC \cdot AP$.

(N/A) Given: $AC$ and $PQ$ intersect each other at the point $O$ and $AB \parallel DC$.
To prove: $OA \cdot CQ = OC \cdot AP$.
Proof:
In $\triangle AOP$ and $\triangle COQ$:
$1$. $\angle AOP = \angle COQ$ (Vertically opposite angles).
$2$. $\angle OAP = \angle OCQ$ (Alternate interior angles,since $AB \parallel DC$ and $AC$ is a transversal).
$3$. $\angle OPA = \angle OQC$ (Alternate interior angles,since $AB \parallel DC$ and $PQ$ is a transversal).
Therefore,$\triangle AOP \sim \triangle COQ$ (by $AAA$ similarity criterion).
Since the triangles are similar,their corresponding sides are proportional:
$\frac{OA}{OC} = \frac{AP}{CQ}$
By cross-multiplying,we get:
$OA \cdot CQ = OC \cdot AP$.
Hence proved.