In parallelogram $ABCD$,two points $P$ and $Q$ are taken on diagonal $BD$ such that $DP = BQ$ (see Fig). Show that: $\Delta APD \cong \Delta CQB$.

(N/A) We have parallelogram $ABCD$. $BD$ is a diagonal and $P$ and $Q$ are points on $BD$ such that:
$DP = BQ$ [Given]
To prove that $\Delta APD \cong \Delta CQB$:
Since $AD \parallel BC$ and $BD$ is a transversal,$[\because ABCD$ is a parallelogram $]$
$\therefore \angle ADB = \angle CBD$ [Interior alternate angles]
$\Rightarrow \angle ADP = \angle CBQ$
Now,in $\Delta APD$ and $\Delta CQB$,we have:
$AD = CB$ [Opposite sides of a parallelogram are equal]
$DP = BQ$ [Given]
$\angle ADP = \angle CBQ$ [Proved above]
$\therefore$ By $SAS$ congruence criterion,we have:
$\Delta APD \cong \Delta CQB$.