Question

In a parallelogram ABCD, P is any point on the side AB. A line AQ drawn parallel to PC, intersects DC at Q. BD intersects AQ at S and PC at R. Prove that $ar\left(\Delta PRB\right)=ar\left(\Delta DSQ\right)$

Answer 1

AQ || PC and AP || CQ
APCQ is a parallelogram.
$\Rightarrow$ AP = CQ [Opposite side of parallelogram are equal]
$\Rightarrow$ AB - AP = CD - CQ
$\Rightarrow$ BP = DQ
Also in APCQ, we have
$\Rightarrow$ $\angle$ APC = $\angle$AQC
$\Rightarrow$ ${180}^{\circ }$ - $\angle$ APC = ${180}^{\circ }$ - $\angle$AQC
$\Rightarrow$ $\angle$ BPC = $\angle$ DQA
Now, in $\Delta$ BPR and $\Delta$ DSQ we have,
$\angle$ BPR = $\angle$ DQS
PB = DQ
$\angle$ RBP = $\angle$ SDQ
So by ASA critreria triangles are congruent
Hence, $ar\left(\Delta PRB\right)=ar\left(\Delta DSQ\right)$