Question

$ABC$ is an isosceles triangle right angled at $C$. Prove that $A{B}^2=2A{C}^2$

Answer 1

Given that $\Delta ABC$ is an isosceles triangle.

$\therefore AC=CB$...(i)

Applying Pythagoras theorem in $\Delta ABC$ (i.e., right-angled at point $C$), we obtain,

$\Rightarrow A{C}^2+B{C}^2=A{B}^2$

$\Rightarrow A{C}^2+A{C}^2=A{B}^2$, [$\because AC=BC$]

$\Rightarrow 2A{C}^2=A{B}^2$ [From (i)]

Hence, proved.