The internal bisectors of ∠ABC and ∠ACB of ΔABC meet at O. Let’s prove that ∠ BOC = 90° + 1/2 ∠ BAC

Given: ΔABC, internal angle bisectors of ∠ABC and ∠ACB meet at O

To prove:

Let BD be the internal angle bisector of ∠ABC and CE be the internal angle bisector of ∠ACB. The figure of the above question is as shown below,

Now in ΔABC,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠BAC + ∠ACB + ∠ABC = 180°………(i)

But from above figure,

∠ABC = ∠ABO + ∠OBC……(ii)

And it is also given BD is internal angle bisector of ∠ABC, so

∠ABO = ∠OBC

Substituting the above value in equation (ii), we get

⇒ ∠ABC = 2∠OBC………(iii)

And also from figure,

∠ACB = ∠ACO + ∠OCB……….(iv)

And it is also given CE is internal angle bisector of ∠ACB, so

∠ACO = ∠OCB

Substituting the above value in equation (iv), we get

⇒ ∠ACB = 2∠OCB………(v)

Substituting equation (iii) and (v) in equation (i), we get

∠BAC + (2∠OCB ) + ( 2∠OBC ) = 180°

⇒ 2(∠OBC + ∠OCB) = 180° - ∠BAC

Now consider ΔBOC,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠BOC + ∠OCB + ∠OBC = 180°

Substituting equation (vi) in above equation, we get

Hence proved