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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).


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Solution

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 [ in parallelogram PSDA, PS || AD]
PA || QB [given]

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

Similarly, QBCR is also a parallelogram.
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 Δ PQE and Δ DCF, QPE = FDC
[Since PS || AD and PD is transversal, then alternate interior angles are equal]

PQE=FCD=90
(ΔPQE)(ΔCFD) [By AAS congruency criteria]
ar(ΔPQE)=ar(ΔCFD)
[Since congruent figures have equal area]


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