Question

$\Delta ABC$ is an isosceles triangle with $\angle A={90}^{\circ }$. Show that $B{C}^2=2A{B}^2$

Answer 1

In $△ABC,\angle A={90}^{\circ }$

$△ABC$ is an isosceles triangle

$⟹\angle B=\angle C={45}^{\circ }$

$\sin C=\frac{AB}{BC}⟹\sin {45}^{\circ }=\frac{AB}{BC}$

$⟹\frac{AB}{BC}=\frac{1}{\sqrt{2}}⟹\frac{A{B}^2}{B{C}^2}=\frac{1}{2}⟹B{C}^2=2A{B}^2$