Question

In the figure $AB||DC$ and $AC$ and $PQ$ intersect each other at the point $O$. Prove that $OA.CQ=OC.AP$.

Answer 1

In $△APO$ and $△CQO$

$\angle AOP=\angle COQ$ (vertically opposite angle)

$\angle APO=\angle CQO$ (alternative angle)

$\angle PAO=\angle QCO$ (alternative angle)

Thus, $\frac{OA}{AP}=\frac{OC}{CQ}$

So, $OA.CQ=OC.AP$

Hence proved