In the adjoining figure,ABCD and BQSC are two parallelograms.Prove that $a r (△ R S C) = a r (△ P Q B)$ .

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Solution

In ∆RSC and ∆PQB
∠∠CRS = ∠∠BPQ (CD || AB) so, corresponding angles are equal)
∠∠CSR = ∠∠BQP ( SC || QB so, corresponding angles are equal)
SC = QB (BQSC is a parallelogram)
So, ∆RSC ≅≅ ∆PQB (AAS congruency)
Thus, ar(∆RSC) = ar(∆PQB

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