Question

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

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Solution

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

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