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

(N/A) Given: $ABCD$ is a parallelogram. $P$ and $Q$ are points on diagonal $BD$ such that $DP = BQ$.
To prove: $AQ = CP$.
Proof:
In $\Delta AQB$ and $\Delta CPD$:
$1$. $AB = CD$ (Opposite sides of a parallelogram are equal)
$2$. $\angle ABQ = \angle CDP$ (Alternate interior angles as $AB \parallel CD$ and $BD$ is a transversal)
$3$. $BQ = DP$ (Given)
Therefore,by $SAS$ congruence criterion,$\Delta AQB \cong \Delta CPD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$AQ = CP$.