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In Below Fig., Psda is a Parallelogram in Which Pq = Qr = Rs and Ap || Bq || Cr. Prove that Ar (δ Pqe) = Ar (δCfd). - Mathematics

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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APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 14 Areas of Parallelograms and Triangles
Exercise 14.3 | Q 22 | Page 47
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