Question

Hey guys, I am just asking this question for fun and i will continue to ask questions every week. Today's question:
In triangle abc, lines bo and co bisect angle b and angle c respectively. Prove:
angle boc = (angle a)/2

Answer 1

Step-by-step explanation:

Given:- ΔABC, where BO bisects ∠B and CO bisects ∠C.

To prove:- ∠BOC = 90° + (∠A)/2

Proof:-

In ΔABC,

∠A + ∠B + ∠C = 180° [Angle Sum Property of Triangles]

∴ ∠B + ∠C = 180° - ∠A ------ 1

Now,

In ΔBOC,

∠OBC + ∠OCB + ∠BOC = 180° [Angle Sum Property of Triangles]

Also,

We know that,

BO bisects ∠B

∴ ∠OBC = (1/2)∠B

Similarly,

CO bisects ∠C

∴ ∠OCB = (1/2)∠C

then,

∠OBC + ∠OCB + ∠BOC = 180°

∠BOC + (1/2)∠B + (1/2)∠C = 180°

(1/2) is common so factoring it out we get

∠BOC + (1/2)(∠B + ∠C) = 180°

Thus,

∠BOC = 180° - (1/2)(∠B + ∠C)

From eq.1 we get,

∠BOC = 180° - (1/2)(180° - ∠A)

∠BOC = 180° - ((1/2)(180°) - (1/2)(∠A))

∠BOC = 180° - (90° - ∠A/2)

Opening brackets we get,

∠BOC = 180° - 90° + (∠A)/2

∴ ∠BOC = 90° + (∠A)/2

Hence Proved

Hope it helped and you understood it........All the best