Question

Prove that in a $△PQR$ if $Q{R}^2=P{Q}^2+P{R}^2$ then $\angle P$ is a right angle.

Answer 1

Construction:

Draw a $\Delta ABC$ right angled at $A$ such that

$AB=PQ$

$CA=RP$

Now

In $\Delta ABC$, $\angle A={90}^{\circ }$

By Pythagoras Theorem, $(BC{)}^2=(CA{)}^2+(AB{)}^2$

As $AB=PQ$ and $CA=RP$

$(BC{)}^2=(RP{)}^2+(PQ{)}^2$ (1)

Given: $(QR{)}^2=(RP{)}^2+(PQ{)}^2$ (2)

Hence from (1) and(2) $BC=QR$

In $\Delta ABC$ and $\Delta PQR$

$AB=PQ$ (from construction)

$CA=RP$ (from construction)

$BC=QR$

$\Delta ABC\cong \Delta PQR$ (By SSS congruence)

So $\angle A=\angle P={90}^{\circ }$

Hence proved.