Question

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

Answer 1

In $△ABL$,

$\angle BAL+\angle ALB+\angle B={180}^o$

$\angle BAL+{90}^o+\angle B={180}^o$

$\angle BAL={90}^o-\angle B$ ......(1)

In $△ABC$,

$\angle A+\angle B+\angle C={180}^o$

$\angle B+\angle C={90}^o$

$\angle C={90}^o-\angle B$

$\angle ACB={90}^o-\angle B$ .......(2)

From 1 and 2, we get,

$\angle BAL=\angle ACB$