In the given parallelogram ABCD, DP = BQ and ∠ADP=∠CBQ. To which triangle is ΔADP congruent to?
ΔPDQ
ΔCBQ
ΔQBC
ΔPQB
In ΔADP and ΔCBQ ∠ADP=∠CBQ (given) AD = BC (∵opposite sides of parallelogram are equal) DP = QB (given) ∴ By S.A.S. ΔADP≅ΔCBQ.