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

Name: A triangle ABC is right angled at A. L is a point on BC such that AL ⊥ BC. Prove that ∠ BAL = ∠ ACB.
Uploaded: Fri, 2021-01-15 23:59
Description: A triangle ABC is right angled at A. L is a point on BC such that AL ⊥ BC. Prove that ∠ BAL = ∠ ACB.

#Exemplar Maths for Class 9
#Lines and Angles
#Maths
#NCERT

#1.3
#6.2
#6.3
#6.4
#6.1
#9.1

1.0
1.0
1.0
1.0
2.0
2.0
2.0
2.0
3.0
3.0
3.0
3.0
4.0
4.0
4.0
4.0
5.0
5.0
5.0
5.0
6.0
6.0
6.0
6.0
7.0
7.0
7.0
7.0
8.0
8.0
8.0
8.0
9.0
9.0
10.0
10.0

A triangle ABC is right angled at A. L is a point on BC such that $AL\perp BC$ . Prove that $\angle BAL = \angle ACB$ .

Answers (1)

Solution: $AL\perp BC$ $\angle BAL = \angle ACB$

Proof: In $\triangle ABC$ and $\triangle LAB$ ,
$\angle BAC = \angle ALB$ (each $90^{\circ}$ ) (i)
And $\angle ABC = \angle ABL$ (Common angle) (ii)
In $\triangle ABC$ ,
$\angle BAC + \angle ABC + \angle ACB = 180^{\circ}$ (angle sum property)
$\angle ABC + \angle ACB = 180^{\circ} -90^{\circ}$ (from i)
$\angle ACB = 90^{\circ} - \angle ABC$
In $\triangle ABL$ ,
$\angle BAL + \angle ALB + \angle ABL = 180^{\circ}$ (angle sum property)
$\angle BAL + \angle ABL = 180^{\circ} - 90^{\circ}$ (from i)
$\angle BAL = 90^{\circ} - \angle ABL$
$\angle BAL = 90^{\circ} - \angle ABC$ (from ii)
Hence, $\angle ACB = 90^{\circ} -\angle ABC = \angle BAL$