Maths-

General

Easy

Question

# In , the bisectors of and intersect each other at O. Prove that

Hint:

### As we know OB and OC are angular bisectors we find ∠OBC and ∠OCB in terms of ∠A and ∠B .We find the value of ∠BOC by using the sum of angles in the triangle is 180°.

## The correct answer is: Hence proved

### Aim :- Prove

Step 1:- Find ∠OBC and ∠OCB

∠OBC = ∠B(OB is the angular bisector)

∠OCB = ∠C (OC is the angular bisector)

∠A+ ∠B+ ∠C =180° (sum of angles in triangle ABC)

∠B+ ∠C =180°-∠A

Step 2:- Find ∠BOC

∠BOC+ ∠OBC + ∠OCB =180° (sum of angles in triangle OBC)

∠B+ ∠C =180°-∠A

Step 2:- Find ∠BOC

∠BOC+ ∠OBC + ∠OCB =180° (sum of angles in triangle OBC)

∠BOC+ ∠B + ∠C = 180°

∠BOC= 180° -( ∠B +∠C )

∠BOC= 180°-( ∠B +∠C)

∠BOC= 180°-( 180°-∠A)

Hence proved

Step 1:- Find ∠OBC and ∠OCB

∠OBC = ∠B(OB is the angular bisector)

∠OCB = ∠C (OC is the angular bisector)

∠A+ ∠B+ ∠C =180° (sum of angles in triangle ABC)

∠B+ ∠C =180°-∠A

Step 2:- Find ∠BOC

∠BOC+ ∠OBC + ∠OCB =180° (sum of angles in triangle OBC)

∠B+ ∠C =180°-∠A

Step 2:- Find ∠BOC

∠BOC+ ∠OBC + ∠OCB =180° (sum of angles in triangle OBC)

∠BOC+ ∠B + ∠C = 180°

∠BOC= 180° -( ∠B +∠C )

∠BOC= 180°-( ∠B +∠C)

∠BOC= 180°-( 180°-∠A)

Hence proved

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