Question

Question 1
In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = RS and PA || QB || RC. Prove that ar (PQE) = ar (CFD).

Answer 1

Given, in the 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,
PQ || AB [$\because$ in parallelogram PSDA, PS || AD]
PA || QB [given]

So, PABQ is a parallelogram.
$\therefore$ PQ = AB (i)

Similarly, QBCR is also a parallelogram.
$\therefore$ QR = BC (ii)

RCDS is a parallelogram.
RS = CD (iii)
Now, PQ = QR = RS (iv)

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

In $\Delta$ PQE and $\Delta$ DCF, $\angle$ QPE = $\angle$ FDC
[Since PS || AD and PD is transversal, then alternate interior angles are equal]

$\Rightarrow \angle PQE=\angle FCD={90}^{\circ }$
$\therefore \left(\Delta PQE\right)\cong \left(\Delta CFD\right)$ [By AAS congruency criteria]
$\therefore ar\left(\Delta PQE\right)=ar\left(\Delta CFD\right)$
[Since congruent figures have equal area]