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

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