Question

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

Answer 1

Given that $\Delta ABC$ is an isosceles triangle having AC = BC and $A{B}^2=2A{C}^2$
In $\Delta ABC$,
AC = BC (Given)
$\Rightarrow$ $A{B}^2=2A{C}^2$ (Given)
$\Rightarrow$ $A{B}^2=A{C}^2+A{C}^2$
$\Rightarrow$ $A{B}^2=A{C}^2+B{C}^2$ [Since, AC = BC]
Hence, by using Pythagoras theorem $\Delta ABC$ is right angled triangle.