Question

In the given parallelogram ABCD, DP = BQ and $\angle ADP=\angle CBQ$. Which triangle is $\Delta ADP$ congruent to?

Answer 1

In $\Delta ADP$ and $\Delta CBQ$
$\angle ADP=\angle CBQ$ (given)
AD = BC (opp. sides of parallelogram)
DP = QB (given)
$\therefore \Delta ADP\cong \Delta CBQ$ (by S.A.S. rule)