$A$ triangle $ABC$ is right-angled at $A$. $L$ is a point on $BC$ such that $AL \perp BC$. Prove that $\angle BAL = \angle ACB$.

(N/A) Given: In $\triangle ABC$,$\angle BAC = 90^{\circ}$ and $AL \perp BC$.
To prove: $\angle BAL = \angle ACB$.
Proof:
In $\triangle ABC$,the sum of angles is $180^{\circ}$.
$\angle BAC + \angle ABC + \angle ACB = 180^{\circ}$
$90^{\circ} + \angle ABC + \angle ACB = 180^{\circ}$
$\angle ABC + \angle ACB = 90^{\circ} \quad \dots(1)$
In $\triangle BAL$,$\angle ALB = 90^{\circ}$ (since $AL \perp BC$).
So,$\angle BAL + \angle ABL + \angle ALB = 180^{\circ}$
$\angle BAL + \angle ABC + 90^{\circ} = 180^{\circ}$
$\angle BAL + \angle ABC = 90^{\circ} \quad \dots(2)$
From equations $(1)$ and $(2)$:
$\angle BAL + \angle ABC = \angle ACB + \angle ABC$
Subtracting $\angle ABC$ from both sides,we get:
$\angle BAL = \angle ACB$.
Hence,proved.