In a ` A B C ,\ /_A B C=\ /_A C B` and the bisectors of `/_A B C\ a n d\ /_A C B` intersect at `O` such that `/_B O C=120^0dot` Show that `/_A=/_B=/_C=60^0` .

Explanation

To solve the problem, we need to show that in triangle \( ABC \), where \( \angle ABC = \angle ACB \) and the angle bisectors of these angles intersect at point \( O \) such that \( \angle BOC = 120^\circ \), the angles \( \angle A, \angle B, \) and \( \angle C \) are all \( 60^\circ \).

### Step-by-Step Solution:

1. **Identify the Given Information:**
- \( \angle ABC = \angle ACB \) (let's denote this common angle as \( \angle 1 \)).
- The bisectors of \( \angle ABC \) and \( \angle ACB \) intersect at point \( O \).
- \( \angle BOC = 120^\circ \).
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