Question

A triangle ABC is right angled at A. L is a point on BC such that AL ⊥ BC. Prove that ∠BAL = ∠ACB.

Answer 1

Given In ΔABC, ∠A = 90° and AL ⊥ BC

To prove ∠BAL = ∠ACB

Proof In ΔABC and ΔLAC, ∠BAC = ∠ALC [Each 90°]  ...(i)

And ∠ABC = ∠ABL [Common angle]  ...(ii)

On adding equation (i) and (ii), we get

∠BAC + ∠ABC = ∠ALC + ∠ABL  ...(iii)

Again, in ΔABC,

∠BAC + ∠ACB + ∠ABC = 180°  ...[Sum of all angles of a triangle is 180°]

⇒ ∠BAC + ∠ABC = 180° – ∠ACB   ...(iv)

In ΔABL,

∠ABL + ∠ALB + ∠BAL = 180°   ...[Sum of all angles of a triangle is 180°]

⇒ ∠ABL + ∠ALC = 180° – ∠BAL  ...[∴ ∠ALC = ∠ALB = 90°] ...(v)

On substituting the value from equations (iv) and (v) in equation (iii), we get

180° – ∠ACS = 180° – ∠SAL

⇒ ∠ACB = ∠BAL

Hence proved.