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

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

To prove:

Let BO be the external angle bisector of ∠ABC, i.e., BO is angle bisector of ∠CBD, and CO be the external angle bisector of ∠ACB, i.e., CO is the angle bisector of ∠BCE. The figure of the above question is as shown below,

Now in ΔABC,

We know in a triangle the measure of any exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.

So in this case,

∠CBD = ∠ACB + ∠BAC………(i)

And also,

∠BCE = ∠ABC + ∠BAC……..(ii)

But it is also given BO is angle bisector of ∠CBD, so

∠DBO = ∠CBO

⇒ ∠CBD = 2∠CBO

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

2∠CBO = ∠ACB + ∠BAC

And it is also given CO is angle bisector of ∠BCE, so

∠ECO = ∠BCO

⇒ ∠BCE = 2∠BCO

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

2∠BCO = ∠ABC + ∠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 + ∠BCO + ∠CBO = 180°

Substituting equation (iii) and (iv) in above equation, we get

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°

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

Hence proved