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Question
In below fig., PSDA is a parallelogram in which PQ = QR = RS and AP || BQ || CR. Prove
that ar (Δ PQE) = ar (ΔCFD).
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Solution
Given that PSDA is a parallelogram
Since, AP || BQ ||CR || DS and AD || PS
∴PQ = CD .....(1)
In ΔBED,C is the midpoint of BD and CF || BE
∴ F is the midpoint of ED
⇒ EF = PE
similarly
EF = FD
∴ PE = FD ........ (2)
In Δ SPQE and CFD , WE have
PE = FD
∠EDQ = ∠FDC, [Alternative angles]
And PQ = CD
So by SAS congruence criterion, we have . ΔPQE ≅ ΔDCF .
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