In `DeltaABC,AB=AC` and the bisectors of angles B and C intersect at point O. Prove that
`BO=OC`

Answer

To prove that \( BO = OC \) in triangle \( ABC \) where \( AB = AC \) and the angle bisectors of angles \( B \) and \( C \) intersect at point \( O \), we can follow these steps:

### Step-by-Step Solution:

1. **Draw Triangle ABC**:
- Construct triangle \( ABC \) such that \( AB = AC \). This makes triangle \( ABC \) isosceles.

2. **Label Angles**:
- Let the angles at vertices \( B \) and \( C \) be \( \angle B \) and \( \angle C \) respectively. Since \( AB = AC \), we know that \( \angle B = \angle C \) (angles opposite to equal sides are equal).

3. **Draw Angle Bisectors**:
- Draw the angle bisector of \( \angle B \) which intersects \( AC \) at point \( O \).
- Similarly, draw the angle bisector of \( \angle C \) which intersects \( AB \) at point \( O \).

4. **Identify Angles**:
- Let \( \angle XOB \) be the angle formed by the angle bisector of \( \angle B \) and \( \angle YOC \) be the angle formed by the angle bisector of \( \angle C \).
- Since \( O \) is the intersection of the angle bisectors, we have \( \angle XOB = \angle YOC \).

5. **Use Vertically Opposite Angles**:
- The angles \( \angle BOX \) and \( \angle COY \) are vertically opposite angles, hence \( \angle BOX = \angle COY \).

6. **Use the Angle Bisector Property**:
- The angles \( \angle XBO \) and \( \angle YCO \) are equal because they are half of \( \angle B \) and \( \angle C \) respectively (as they are angle bisectors). Therefore, \( \angle XBO = \angle YCO \).

7. **Establish Equality of Angles**:
- Since \( \angle B = \angle C \), we can say that:
\[ \frac{1}{2} \angle B = \frac{1}{2} \angle C \implies \angle XBO = \angle YCO \]

8. **Consider Triangles BOX and COY**:
- In triangles \( BOX \) and \( COY \):
- \( \angle XOB = \angle YOC \) (from step 4)
- \( \angle BOX = \angle COY \) (from step 5)
- \( \angle XBO = \angle YCO \) (from step 6)

9. **Apply the Angle-Angle (AA) Criterion**:
- Since two angles in triangle \( BOX \) are equal to two angles in triangle \( COY \), by the Angle-Angle (AA) criterion, triangles \( BOX \) and \( COY \) are similar.

10. **Conclude with Side Ratios**:
- From the similarity of triangles \( BOX \) and \( COY \), the sides opposite to equal angles are equal. Therefore, we conclude that:
\[ BO = OC \]