Question

ABC is an isosceles triangle with AC = BC. If ${AB}^2=2{AC}^2$, prove that $\Delta ABC$ is right triangle. [1 MARK]

Answer 1

Concept: 0.5 Mark
Application : 0.5 Mark

We have,

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

Consider,

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

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

$\Rightarrow {AB}^2={AC}^2+{BC}^2$ [ $\because$ AC = BC (Given)]

$\Rightarrow \Delta ABC$ is a right triangle right angled at C.