Question

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

Answer 1

Given: PSDA is a parallelogram,

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

To prove: ar (∆PQE) = ar (∆CFD)

Proof: In quadrilateral PABQ,

PQ || AB (∵ PS || AD)

and PA || QB (given)

So, PABQ is parallelogram

PQ = AB …(i)

Similarly, QBCR is also a parallelogram

QR = BC …(ii)

And RCDS is also a parallelogram

RS = CD …(iii)

But PQ = QR = RS (given) …(iv)

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

PQ = QR = RS = AB = BC = CD …(v)

Now, in ∆PQE and ∆DCF

∠QPE = ∠FDC (alternate interior angle)

PQ = CD [from equation (v)]

∠PQE = ∠FCD

∠PQE = ∠QBC (alternate interior angles)

and ∠QBC = ∠FCD (corresponding angles)

∆PQE = ∆DCF (by ASA congruencey rule)

Then, ar (∆PQE) = ar (∆DCF) (since, congruent figures have equal area)

Hence proved.