# ABC is an isosceles triangle right angled at C. Prove that $${AB}^2$$ = $$2{AC}^2$$.

## 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,

$$\therefore$$  BC = AC

$$\implies$$ $${AB}^2$$ = $${AC}^2$$ + $${AC}^2$$

So, $${AB}^2$$ = $$2{AC}^2$$