Question

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

Answer 1

Answer:

Hope its helpful.....!!!!

Answer 2

Answer:

• In ∆ABL, we have

∠ BAL + ∠ ALB + ∠B = 180°

↦ ∠BAL + 90° + ∠B = 180°

↦ ∠BAL + ∠B = 90°

↦ ∠BAL = 90° - ∠B [ eqⁿ 1 ]

• In ∆ABC, we have

∠A + ∠B + ∠C = 180°

↦ 90° + ∠B + ∠C = 180°

↦ ∠B + ∠C = 180° - 90°

↦ ∠B + ∠C = 90°

↦ ∠C = 90° - ∠B [ eqⁿ 2 ]

From 1 and 2, we get

∠BAL = ∠ACB