In parallelogram $ABCD$,two points $P$ and $Q$ are taken on diagonal $BD$ such that $DP = BQ$ (see figure). Show that: $\Delta AQB \cong \Delta CPD$.
(N/A) We have parallelogram $ABCD$. $BD$ is a diagonal and $P$ and $Q$ are such that:
$PD = QB$ [Given]
To prove that $\Delta AQB \cong \Delta CPD$.
Since $AB \parallel CD$ and $BD$ is a transversal,$[\because ABCD$ is a parallelogram$]$
$\therefore \angle ABD = \angle CDB$
$\Rightarrow \angle ABQ = \angle CDP$
Now,in $\Delta AQB$ and $\Delta CPD$,we have:
$QB = PD$ [Given]
$\angle ABQ = \angle CDP$ [Proved above]
$AB = CD$ [Opposite sides of parallelogram $ABCD$]
$\therefore \Delta AQB \cong \Delta CPD$ [By $SAS$ congruence criterion]
$PD = QB$ [Given]
To prove that $\Delta AQB \cong \Delta CPD$.
Since $AB \parallel CD$ and $BD$ is a transversal,$[\because ABCD$ is a parallelogram$]$
$\therefore \angle ABD = \angle CDB$
$\Rightarrow \angle ABQ = \angle CDP$
Now,in $\Delta AQB$ and $\Delta CPD$,we have:
$QB = PD$ [Given]
$\angle ABQ = \angle CDP$ [Proved above]
$AB = CD$ [Opposite sides of parallelogram $ABCD$]
$\therefore \Delta AQB \cong \Delta CPD$ [By $SAS$ congruence criterion]
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