Question

$\Delta$ACB is an isosceles triangle such that AC$=$BC. If $A{B}^2=2A{C}^2$, prove that $\Delta$ACB is a right angled triangle.

Answer 1

$△ACB$ is an isosceles triangle ...Given

In order to prove $△ACB$ is right angled, we must prove $\angle C={90}^o$

And $A{B}^2=2A{C}^2$

$\Rightarrow A{B}^2=A{C}^2+A{C}^2$
$\Rightarrow A{B}^2=A{C}^2+B{C}^2$ ...$[\because AC=BC]$
$\Rightarrow \angle ACB={90}^o$ ....[Converse of Pythagoras theorem]

$\therefore \Delta$ACB is a right angled triangle.