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

Given: ABC is right triangle right angled at A and AL ⊥ BC

Now, in ΔBAL and ΔBCA

∠ABL = ∠CBA  (Common)

∠BLA = ∠BAC = 90°

So, ΔBAL ∼ ΔBCA  (by AA similarity criteria)

⇒ ∠BAL = ∠ACB  (C.P.C.T)