Question

In the following figure, 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).

Answer 1

Given: In a parallelogram PSDA, points Q and R are on PS such that

PQ = QR = RS and PA || QB || RC.

To prove: ar (PQE) = ar (CFD)

Proof: In parallelogram PABQ,

And PA || QB  ...[Given]

So, PABQ is a parallelogram.

PQ = AB   ...(i)

Similarly, QBCR is also a parallelogram.

QR = BC  ...(ii)

And RCDS is a parallelogram.

RS = CD   ...(iii)

Now, PQ = QR = RS  ...(iv)

From equations (i), (ii), (iii) and (iv),

PQ || AB  ...[∴ In parallelogram PSDA, PS || AD]

In ΔPQE and ΔDCF,

∠QPE = ∠FDC  ...[Since, PS || AD and PD is transversal, then alternate interior angles are equal]

PQ = CD  ...[From equation (v)]

And ∠PQE = ∠FCD   ...[∴ ∠PQE = ∠PRC corresponding angles and ∠PRC = ∠FCD alternate interior angles]

ΔPQE = ΔDCF  ...[By ASA congruence rule]

∴ ar (ΔPQE) = ar (ΔCFD)  ...[Since, congruent figures have equal area]

Hence proved.