Solution :
Since ABC is an isosceles right triangle, right angled at C, therefore
\({AB}^2\) = \({AC}^2\) + \({BC}^2\)
Since given that the triangle is isosceles,
\(\therefore\)ย BC = AC
\(\implies\) \({AB}^2\) = \({AC}^2\) + \({AC}^2\)
So, \({AB}^2\) = \(2{AC}^2\)