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, we have

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

$\Rightarrow$ $\angle$BAL = ${90}^{\circ }$ - $\angle$B .....(i)

In $△$ABC, we have

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

$\Rightarrow$ $\angle$C = ${90}^{\circ }$ - $\angle$B .....(ii)

From (i) and (ii), we get

$\angle$ BAL = $\angle$ACB.