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In the figure, PSDA is a parallelogram. ...

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)`.

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Given : PSDA is a parallelogram. Points Q and R are taken on PS such that `PQ = QR = RS and PA || QB || RC`.
To prove : `ar(DeltaPQE) = ar(DeltaCFD)`
Proof: Now, `PS = AD` (opposite side of a parallelogram)
`:. (1)/(3) PS = (1)/(3) AD rArr PQ = CD`...(1)
Again, `PS ||AD` and QB cut them.
`:. anglePQE = angleCBE` (alternate angles)...(2)
Now, `QB||RC` and AD cuts them
`:. angleQBD = angle RCD` (corresponding angles)...(3)
So, `anglePQE = angleFCD`...(4)
[from (2) and (3), `angleCBE and angleQBD` are same and `angleRCD and angleFCD` are same] ,brgt Now, in `DeltaPQE and DeltaCFD`
`:' {(angleQPE = angleCDF," (alternate angles)"),(" "PQ = CD," [from(1)]"),(anglePQE = angleFCD," [from (4)]"):}`
`:. DeltaPQE ~= DeltaCFD` (by ASA congruence rule)
Hence, `ar(Delta PQE) = ar(Delta CFD)` (congruent triangles are equal in area)
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