Question
In the adjoining figure, ABCD and BQSC are two parallelograms. Prove that $\text{ar}(\triangle\text{RSC})=\text{ar}(\triangle\text{PQB}).$
✓
Answer
In $\triangle\text{RSC}$ and $\triangle\text{PQB},$$\angle\text{CRS}=\angle\text{BPQ}$ [RC || PB, corresponding angles]
$\angle\text{RSC}=\angle\text{PQB}$ [RC || PB, corresponding angles]
$\text{SC}=\text{QB}$ [Opposite sides of a parallelogram BQSC]
$\therefore\ \triangle\text{RSC}\cong\triangle\text{PQB}$ [by AAS congruence criterion]
$\Rightarrow\ \text{ar}(\triangle\text{RSC})=\text{ar}(\triangle\text{PQB})$
$\angle\text{RSC}=\angle\text{PQB}$ [RC || PB, corresponding angles]
$\text{SC}=\text{QB}$ [Opposite sides of a parallelogram BQSC]
$\therefore\ \triangle\text{RSC}\cong\triangle\text{PQB}$ [by AAS congruence criterion]
$\Rightarrow\ \text{ar}(\triangle\text{RSC})=\text{ar}(\triangle\text{PQB})$
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