Question

In the figure, PSDA is a parallelogram in which PQ = QR = RS and AP || BQ || CR.
Prove that ar $\left(△PQE\right)=ar\left(△CFD\right)$

Answer 1

Given : In the figure, PSDA is a ||gm in which QP = QR = Rs and AP || BQ || CR || DS
To Prove : $ar\left(△PQE\right)=ar\left(△CFD\right)$
Construction : Join PD
Proof : $\because$ PA || BQ || CR || DS and PQ = QR = RS
$\therefore$ AB = BC = CD
$\therefore$ PQ = CD
Now in $△BED,$ F is mid point of ED
$\therefore EF=FD$
Similarly, EF = PE
$\Rightarrow$ PE = FD
In $△PQE$ and $△CFD,$
$\therefore \angle EPQ=\angle FDCPQ=CDPE=FD\therefore △PQE\cong △CFD\therefore ar\left(△PQE\right)=ar\left(△CFD\right)$