O is an interior point of ΔABC; Lets prove that ∠BOC > ∠BAC

Given: ΔABC, O is an interior point of ΔABC

To prove that ∠BOC > ∠BAC

The figure for the given 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 the above figure,

∠ABC = ∠ABO + ∠OBC………(ii)

And also,

∠ACB = ∠ACO + ∠OCB…….(iii)

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

∠BAC + (∠ACO + ∠OCB ) + ( ∠ABO + ∠OBC ) = 180°

⇒ ∠OBC + ∠OCB = 180° - ∠BAC - ∠ACO - ∠ABO………(iv)

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 (iv) in the above equation, we get

∠BOC + (180° - ∠BAC - ∠ACO - ∠ABO ) = 180°

⇒ ∠BOC = 180° - (180° - ∠BAC - ∠ACO - ∠ABO )

⇒ ∠BOC = ∠BAC + ∠ACO + ∠ABO

⇒ ∠BOC > ∠BAC

Hence proved