# O is an interior

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

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