In Fig., PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR = RS and PA || QB || RC. Prove that ar (PQE) = ar (CFD).

In Fig., PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR = RS and $PA \parallel QB \parallel RC$ . Prove that ar (PQE) = ar (CFD).

Answers (1)

Solution.

Given: a parallelogram PSDA with points Q and R taken on PS such that PQ = RS = QR
and $PA\parallel QB\parallel RC$
To prove: ar(PQE) = ar(CFD)
Proof:
PQ = QR = RS
PS = PQ + QR + RS = 3 PQ
Also, $PA \parallel QB \parallel RC$ and $PS \parallel AD$
So PABQ, QBCR, RCDS – all are parallelograms.
Hence,   $AB=BC=CD=\frac{1}{3}AD$
We know that opposite sides of a parallelogram are equal, so in $\parallel gm$ APSD:
PS = AD
$\Rightarrow \frac{1}{3}PS=\frac{1}{3}AD$
$\Rightarrow PQ=CD$ …(1)
Consider $\triangle PEQ$ and $\triangle CFD$
$PS\parallel AD$ and PD is transversal
\ $\angle QPE = \angle CDF$   [Alternate interior angles] …(2)
Now, $QB \parallel RC$ and AD is a transversal
\ $\angle QBD = \angle RCD$    [Corresponding angles] …(3)
$PS \parallel AD$ and PD is transversal
$\therefore \angle PQB = \angle QBC$ [Alternate interior angles] …(4)
From (3) and (4)
$\angle PQE=\angle FCD$ $\angle PQE = \angle FCD$ $???????\angle PQE = \angle FCD$ …(5)
From (1), (2), (5)
$\triangle PEQ\cong \triangle CFD$ [ASA congruence rule]
Hence, $ar (\triangle PQE) = ar(\triangle CED)$ [Congruent figures have equal area]
Hence proved