ABC is an isosceles triangle right angled at C.

Question.

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

Solution:

In $\triangle \mathrm{ABC}, \angle \mathrm{ACB}=90^{\circ}$. We are given that

$\triangle \mathrm{ABC}$ is an isosceles triangle.

$\Rightarrow \angle \mathrm{A}=\angle \mathrm{B}=45^{\circ}$

$\Rightarrow \mathrm{AC}=\mathrm{BC}$ ... (1)

By pythagoras theorem, we have

$\mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{BC}^{2}$

$=\mathrm{AC}^{2}+\mathrm{AC}^{2} \quad\{\because \mathrm{BC}=\mathrm{AC}$ by $(1)]$

$=2 \mathrm{AC}^{2}$