Question

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP=BQ (see the given figure) Show that:
$\Delta APD\cong \Delta CQB$

Answer 1

In $\Delta APDand\Delta CQB$,
$\angle ADP=\angle CBQ$ (Alternate interior angles)
AD =CB (Opposite sides of parallelogram ABCD)
DP=BQ (Given)
$\therefore \Delta APD\cong \Delta CQB$ ( using SAS congruence rule)