In the figure, if $a r △ R A S = a r △ R B S$ and $[a r (△ Q R B)] = a r (△ P A S)$ then show that both the quadrilaterals PQSR and RSBA are trapeziums.

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Solution

In given figure $a r e a (Δ R A S) = a r e a (Δ R B S)$ and $a r e a (Δ Q R B) = a r e a (Δ P A S)$
Given $a r e a (Δ R A S) = a r e a (Δ R B S)$ .....................(1)
This possible when those triangles in (1) are equal on same base RS and two line AB and RS
Then line RS and AB are parallel each other
So RSAB is a trapezium
Given $a r e a (Δ Q R B) = a r e a (Δ P A S)$
Less $a r e a (Δ R A S)$ and $a r e a (Δ R B S)$ we get
$\Rightarrow a r e a (Δ Q R B) - a r e a (Δ R A S) = a r e a (Δ P A S) - a r e a (Δ R B S)$
$\Rightarrow a r e a (Δ Q R S) = a r e a (Δ P S R)$ ..........(2)
This possible when those triangle are in (2) On same base RS and two line PQ and RS
Then line RS and PQ are parallel each other
So RSAB is a trapezium
So PQRS and ABRS are trapezium

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