Question

$ABC$ is an isosceles triangle, right angled at $C.$ Prove that $A{B}^2=2A{C}^2$.

Answer 1

In $△ABC$,

By Pythagoras Theorem,

$(AB{)}^2=(AC{)}^2+(BC{)}^2.....(1)$

Since, $△ABC$ is an isosceles triangle,

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

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

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