Question

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

Answer 1

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

$\mathrm{\Delta ABC}$is an isosceles triangle right angled at $C$.

In $\mathrm{\Delta ACB},\angle \mathrm{C}=90°$

$\mathrm{AC}=\mathrm{BC}$ [By isosceles triangle property]

${\mathrm{AB}}^2={\mathrm{AC}}^2+{\mathrm{BC}}^2$ [By Pythagoras theorem]

${\mathrm{AB}}^2={\mathrm{AC}}^2+{\mathrm{AC}}^2$ [Since, $\mathrm{AC}=\mathrm{BC}$]

$\Rightarrow$${\mathrm{AB}}^2=2{\mathrm{AC}}^2$

Hence we proved that ${\mathrm{AB}}^2=2{\mathrm{AC}}^2$