Question

The bisectors BO and CO of the angles at B and C of ABC meet at O. Prove that BOC = 90° + A/2.​

Answer 1

The prove that BOC = 90° + A/2. is given below:

What is the bisector about?

Note that, bisectors of angles B and C and that of a triangle ABC is one that pass across each other at the point O.

Hence: one need to to prove that ∠BOC = 90° + A/2

Since:

BO is said to be the bisector of angle B e.g. ∠OBC = (∠1)

CO is said to be the bisector of angle C e.g.  ∠OCB = (∠2)

Looking at the triangle BOC, to interpret  by the use of angle sum property, then:

∠OBC + ∠BOC + ∠OCB = 180°

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

Looking at triangle ABC and by the use of angle sum property, So;

∠A + ∠B + ∠C = 180°

∠A + 2(∠1) + 2(∠2) = 180°

Then we divide by 2 on the two sides, and it will be:

∠A/2 + ∠1 + ∠2 = 180°/2

∠A/2 + ∠1 + ∠2 = 90°

∠1 + ∠2 = 90° - ∠A/2 ---(equation 2)

Then one need to Substitute (2) in (1), So:

∠BOC + 90° - ∠A/2 = 180°

∠BOC - ∠A/2 = 180° - 90°

∠BOC - ∠A/2 = 90°

Hence , ∠BOC = 90° + ∠A/2

Therefore, from the above, we care able to prove that BOC = 90° + A/2.

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