# ABC is an isosceles triangle with AC = BC. If $${AB}^2$$ = $$2{AC}^2$$, prove that ABC is a right triangle.

## 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.