Question

$△ABC$ is an isosceles right triangle, right angled at $\mathrm{C}.$ Prove that $\mathrm{AB}{}^2=2{\mathrm{AC}}^2$

Answer 1

Proving $\mathrm{AB}{}^2=2{\mathrm{AC}}^2$:

In $△ABC$ using Pythagoras theorem,

${\left(AB\right)}^2={\left(AC\right)}^2+{\left(BC\right)}^2\dots ..\left(1\right)$

Since, $△ABC$is an isosceles triangle,

$AC=BC\dots .\left(2\right)$

∴ From $\left(1\right)$ and$\left(2\right)$,

${\left(AB\right)}^2=2{\left(AC\right)}^2$

Hence we proved ${\left(AB\right)}^2=2{\left(AC\right)}^2$