In parallelogram $ABCD$,two points $P$ and $Q$ are taken on diagonal $BD$ such that $DP = BQ$ (see Fig). Show that: $AP = CQ$.

(N/A) Given: $ABCD$ is a parallelogram. $P$ and $Q$ are points on diagonal $BD$ such that $DP = BQ$.
To prove: $AP = CQ$.
Proof:
In $\Delta APD$ and $\Delta CQB$:
$1$. $AD = CB$ (Opposite sides of a parallelogram are equal)
$2$. $\angle ADP = \angle CBQ$ (Alternate interior angles as $AD \parallel BC$ and $BD$ is a transversal)
$3$. $DP = BQ$ (Given)
Therefore,by $SAS$ congruence rule,$\Delta APD \cong \Delta CQB$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$AP = CQ$.