It is given to us –

ΔABC is a right - angled triangle.

∠A = 90°

L is a point on BC

AL ⊥ BC

We have to prove ∠BAL = ∠ACB.

We know that the sum of the angles of a triangle is equal to 180°. Thus, in ΔABC,

∠BAC + ∠B + ∠C = 180° - - - - (i)

Now, in ΔABL,

AL ⊥ BC, i.e., ∠ALB = 90°

Since, the sum of the angles of a triangle is equal to 180°,

∠BAL + ∠ALB + ∠B = 180° - - - - (ii)

From equation (i) and equation (ii), we can say that

∠BAC + ∠B + ∠C = ∠BAL + ∠ALB + ∠B

⇒ ∠C + ∠BAC = ∠BAL + ∠ALB

⇒ ∠C = ∠BAL (Since, ∠A = 90° and ∠ALB = 90°, so they are equal)

⇒ ∠BAL = ∠ACB

Hence, proved.