Question

$ABC$ is an isosceles triangle with $AC=BC$. If $A{B}^2=2A{C}^2$, prove that $ABC$ is a right angled triangle.

Answer 1

Given, $△ABC$ is an isosceles triangle,

$\therefore$ $AC=BC....\left(1\right)$

Also, given that,

$(AB{)}^2=2(AC{)}^2$

$\therefore$ $(AB{)}^2=(AC{)}^2+(AC{)}^2.....(2)$

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

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

Hence, By converse of Pythagoras theorem,

$△ABC$ is an isosceles right-angled triangle. $\left[henceproved\right]$