Question

Question 9 (iv)
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP=BQ (see the given figure) Show that:

AQ = CP

Answer 1

In $\Delta AQB\cong \Delta CPD$,
$\angle ABQ=\angle CDP$ (Alternate interior angles for AB||CD)
AB=CD (Opposite sides of parallelogram ABCD)
BQ=DP( Given)
$\Delta AQB\cong \Delta CPD$ (Using SAS congruence rule)
As $\Delta AQB\cong \Delta CPD$,
$\Rightarrow$AQ=CP (CPCT)