Question

∆ABC is right-angled at A. If AL ⊥ BC, prove that ∠BAL = ∠ACB.

Answer 1

We know that the sum of two acute angles of a right angled triangle is $90°$
From the right $∆ABL$, we have:
$$\begin{gathered} \therefore \angle BAL+\angle ABL=90° \\ \Rightarrow \angle BAL=90°-\angle ABL \\ \Rightarrow \angle BAL=90°-\angle ABC...\left(1\right) \end{gathered}$$

Also, from the right $∆ABC$, we have:
$$\begin{gathered} \angle ABC+\angle ACB=90° \\ \Rightarrow \angle ACB=90°-\angle ABC...\left(2\right) \\ \mathrm{From}\left(1\right)\mathrm{and}\left(2\right),\text{we get:} \\ \angle ACB=\angle BAL\left[\because \angle BAL=90°-\angle ABC\right] \\ \therefore \angle BAL=\angle ACB \end{gathered}$$