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In the 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).
- Firstly, use the formula, area of parallelogram = Base x Altitude
- Further, prove that A PQE = AD CF, by ASA congruent rule.
- At the end use the property that congruent figures have same area.
Given In a parallelogram PSDA, points 0 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 Eqs. (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 Eq. (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]