In  right angled at  A, AL is drawn perpendicular to BC. Prove that .

The correct answer is: Hence proved

In    ABC ,

∠BAC  + ∠ABC + ∠ACB = 180°(sum of angles in the triangle)

∠ABC  + ∠ACB= 180°- 90°(given, ∠ BAC = 90° )

∠ABC  +  ∠ ACB = 90°

∠ABC = 90°- ∠ ACB
In   ABL ,

∠BAL + ∠ABC + ∠ ALB = 180°(sum of angles in the triangle)

∠BAL + ∠ABC= 180°- 90°(given, ∠ ALB = 90° )

∠BAL + ∠ABC = 90°

∠BAL =  90°- ∠ ABC

Substitute ∠ABC = 90°- ∠ACB in the above equation

∠BAL = 90°- (90°- ∠ ACB)

∴ ∠ BAL =  ∠ ACB
Hence proved