Answer :

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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