Given,
In ΔABC
∠ ABC = ∠ ACB
Divide both sides by 2, we get
\(\frac{1}{2}\)∠ABC = \(\frac{1}{2}\)∠ACB
∠OBC = ∠OCB [Therefore, OB, OC bisects ∠B and ∠C]
Now,
∠BOC = 90° + \(\frac{1}{2}\)∠A
120° – 90° = \(\frac{1}{2}\)∠A
30° x 2 = ∠A
∠A = 60°
Now in ΔABC
∠A + ∠ABC + ∠ACB = 180°[Sum of all angles of a triangle]
60° + 2∠ABC = 180°[Therefore, ∠ABC = ∠ACB]
2∠ABC = 180° – 60°
2∠ABC = 120°
∠ABC = 60°
Therefore,
∠ABC = ∠ACB = 60°
Hence, proved