In a Δ ABC, ABC =ACB and the bisectors of ABC and ACB intersect at O such that BOC = 120°. Shoe that A =B =C = 60°.

Given,


ABC = ACB


Divide both sides by 2, we get


ABC = ACB


OBC = OCB [Therefore, OB, OC bisects B and C]


Now,


BOC = 90o + A


120o – 90o = A


30o * 2 = A


A = 60o


Now in


A + ABC + ACB = 180o [Sum of all angles of a triangle]


60o + 2ABC = 180o [Therefore, ABC = ACB]


2ABC = 180o – 60o


2ABC = 120o


ABC = 60o


Therefore, ABC = ACB = 60o


Hence, proved


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