Question

In a triangle, if the square of one side is equal to the sum of the squares of other two sides, then the angle opposite to the first side is a right angle. Prove it

Answer 1

Given:- $ABC$ is a triangle

${AC}^2={AB}^2+{BC}^2$

To prove:- $\angle B=90°$

Construction:- Construct a triangle $PQR$ right angled at $Q$ such that, $PQ=AB$ and $QR=BC$

Proof:-

In $△PQR$

${PR}^2={PQ}^2+{QR}^2\left(\text{By pythagoras theorem}\right)$

$\Rightarrow {PR}^2={AB}^2+{BC}^2.....\left(1\right)(\because AB=PQ\text{and}QR=BC)$

${AC}^2={AB}^2+{BC}^2.....\left(2\right)\left(\text{Given}\right)$

From equation $\left(1\right)\&\left(2\right)$, we have

${AC}^2={PR}^2$

$\Rightarrow AC=PR.....\left(3\right)$

Now, in $△ABC$ and $△PQR$

$AB=PQ$

$BC=QR$

$AC=PR\left(\text{From}\left(3\right)\right)$

$\therefore △ABC\cong △PQR\left(\text{By SSS congruency}\right)$

Therefore, by C.P.C.T.,

$\angle B=\angle Q$

$\because \angle Q=90°$

$\therefore \angle B=90°$

Hence proved.