Question:
In the adjoining figure, ABCD and BQSC are two parallelograms. Prove that ar(∆RSC) = ar(∆PQB).
Solution:
In $\triangle \mathrm{RSC}$ and $\triangle \mathrm{PQB}$
$\angle \mathrm{CRS}=\angle \mathrm{BPQ}$ (CD || AB) so, corresponding angles are equal)
$\angle \mathrm{CSR}=\angle \mathrm{BQP}$ ( SC || QB so, corresponding angles are equal)
SC = QB (BQSC is a parallelogram)
So, $\Delta \mathrm{RSC} \cong \Delta \mathrm{PQB}$ (AAS congruency)
Thus, $\operatorname{ar}(\Delta \mathrm{RSC})=\operatorname{ar}(\Delta \mathrm{PQB})$