Question

In $\Delta PQR$, if $P{R}^2=P{Q}^2+Q{R}^2$, prove that $\angle Q$ is right angle.

Answer 1

This is converse of Pythagoras theorem

We can prove this contradiction sum $q^2=p^2+r^2$ in $\Delta PQR$ while triangle is not a rightangle

Now consider another triangle $\Delta ABC$ we construct $\Delta ABC$ $AB=qCB=b$ and C is a Right angle

By the Pythagorean theorem $(AC{)}^2=p^2+r^2$

But we know $p^2+r^2=q^2$ and $q=PR$

So $(AB{)}^2=p^2+r^2=(SR{)}^2$

Since $PQ$ and $AB$ are length of sides we can take positive square roots

$AC=PQ$

All the these sides $\Delta ABC$ are congruent to $\Delta PQR$

So they are congruent by sss theorem