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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).
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Solution
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.
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