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]
$\Rightarrow\ \text{ar}(\triangle\text{RSC})=\text{ar}(\triangle\text{PQB})$
$\angle\text{CRS}=\angle\text{BPQ}$ [$RC \| PB$, 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]
$\Rightarrow\ \text{ar}(\triangle\text{RSC})=\text{ar}(\triangle\text{PQB})$
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