Question

In the given figure, $PQRS$ is a square and $\angle BAC={90}^o$. Prove that $:R{S}^2=BRxSC$

Answer 1

Solution:-

Given that:- ABC is a $△$ with $\angle BAC=90°$ and PQRS is a square.

To prove:- ${RS}^2=BR\times SC$

Proof:-

In $△APQ$ and $△RBP$,

$\angle APQ=\angle PBR[\because \text{corresponding angles}]$

$\angle PAQ=\angle BRP[\because \text{each}90°]$

$\therefore △APQ\sim △RBP\left[\text{AA criteria}\right].....\left(1\right)$

Similarly, in $△AQP$ and $△SCQ$

$\angle AQP=\angle QCS[\because \text{corresponding angles}]$

$\angle PAQ=\angle CSQ[\because \text{each}90°]$

$△AQP\sim △SCQ\left[\text{AA criteria}\right].....\left(2\right)$

From ${eq}^n\left(1\right)\&\left(2\right)$, we get

$△RBP\sim △SCQ$

As we know that corresponding sides of similar triangles are proportional.

$\therefore \frac{BR}{SQ}=\frac{RP}{SC}$

$\Rightarrow BR\times SC=RP\times SQ.....\left(3\right)$

As given that PQRS is a square.

$\therefore PQ=QS=RS=PR.....\left(4\right)$

From ${eq}^n\left(3\right)\&\left(4\right)$, we get

$RS\times RS=RP\times SQ$

$\Rightarrow {RS}^2=RP\times PQ$

Hence proved that ${RS}^2=RP\times PQ$.