Question

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

• A. $\Delta CBQ$
• B. $\Delta PDQ$
• C. $\Delta QBC$
• D. $\Delta PQB$

Answer 1

The correct option is A $\Delta CBQ$

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)