Bisectors of angles B and C of a triangle ABC intersect each other at the point O. Prove that ∠BOC = 90° + 1/2 ∠A

Solution:

Given, bisectors of angles B and C of a triangle ABC intersect each other at the point O.

We have to prove that ∠BOC = 90° + 1/2 ∠A.

BO is the bisector of angle B such that ∠OBC = (∠1)

CO is the bisector of angle C such that ∠OCB = (∠2)

Considering the triangle BOC,

By angle sum property,

∠OBC + ∠BOC + ∠OCB = 180°

∠1 + ∠BOC + ∠2 = 180° -------------------------- (1)

Considering triangle ABC,

By angle sum property,

∠A + ∠B + ∠C = 180°

∠A + 2(∠1) + 2(∠2) = 180°

Dividing by 2 on both sides,

∠A/2 + ∠1 + ∠2 = 180°/2

∠A/2 + ∠1 + ∠2 = 90°

∠1 + ∠2 = 90° - ∠A/2 ------------------------------- (2)

Substituting (2) in (1),

∠BOC + 90° - ∠A/2 = 180°

∠BOC - ∠A/2 = 180° - 90°

∠BOC - ∠A/2 = 90°

Therefore, ∠BOC = 90° + ∠A/2

✦ Try This: In fig., if l∥m, find the value of a and b.

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 6

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NCERT Exemplar Class 9 Maths Exercise 6.4 Sample Problem 3

Bisectors of angles B and C of a triangle ABC intersect each other at the point O. Prove that ∠BOC = 90° + 1/2 ∠A

Summary:

An angle bisector or the bisector of an angle is a ray that divides an angle into two equal parts. Bisectors of angles B and C of a triangle ABC intersect each other at the point O. It is proven that ∠BOC = 90° + 1/2 ∠A

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