Solution :
Since ABC is an isosceles triangle with AC = BC and \({AB}^2\) = \(2{AC}^2\), therefore,
\({AB}^2\) = \({AC}^2\) + \({AC}^2\)
\(\implies\) \({AB}^2\) = \({AC}^2\) + \({BC}^2\) (because AC = BC, given)
\(\therefore\) \(\triangle\) ABC is right angled at C.