Question

A triangle ABC is right -angled at A. AL is drawn perpendicular to BC. Prove that $\angle BAL=\angle ACB.$

Answer 1

In $△BAL,\angle B+\angle 1+90={180}^0$

$\Rightarrow \angle B+\angle 1={90}^0$

In $△ABC,\angle A+\angle B+\angle C={180}^0$

$\Rightarrow \angle B+\angle C={90}^0$

$\Rightarrow \angle B=\angle C\Rightarrow \angle BAL=\angle BCA$