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)`.

Explanation

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,
`PQ || AB` [`because` in parallelogram `PSDA, PS || AD]`
and `PA || QB` [given]
So, PABQ is a parallelogram.
`therefore` PQ = AB ...(i)
Similalry, QBCR is also a parallelogram.
`therefore` QR = BC ...(ii)
and RCDS is a parallelogram.
`therefore` RS = CD ...(iii)
Now, PQ = QR = RS ...(iv)
From Eqs. (i), (ii) (iii) and (iv),
PQ = QR = RS =AB = BC = CD ...(v)
In `DeltaPQE` and `DeltaDCF`, `angleQPE = angleFDC`
[since, `PS || AD` and PD is transversal, then alternate interior angles are equal]
PQ = CD [from Eq. (v)]
and `anglePQE = angleFCD`
[`because anglePQE = anglePRC` corresponding angles and `anglePRC = angleFCD` alternate interior angles]
`therefore` `DeltaPQE ~= DeltaDCF` [by ASA congruence rule]
`therefore` `ar (DeltaPQE) = ar (DeltaCFD)` [ since, congruent figures have equal area]
Hence proved.