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)