To prove the statements given in the triangle \( ABC \) with specific points \( M \) and \( N \) on sides \( AC \) and \( AB \), respectively, where \( BM \) and \( CN \) are the angle bisectors that intersect at point \( G \), we can follow these steps:

### Given:
- Triangle \( ABC \) is isosceles with \( AB = AC \).
- \( BM \) and \( CN \) are the angle bisectors.

### Proof:
1. **Show that \( BM = CN \):**
- Since triangle \( ABC \) is isosceles with \( AB = AC \), the angles \( \angle ABC = \angle ACB \).
- By the properties of angle bisectors in an isosceles triangle, the segments \( BM \) and \( CN \) will also be equal. Thus:

\[
BM = CN
\]

2. **Show that triangle \( GBC \) is isosceles:**
- Since \( BM \) and \( CN \) are angle bisectors, and they are equal, triangles \( GBC \) will have \( GB = GC \) (because the sides opposite equal angles are equal):

\[
GB = GC
\]

Hence, triangle \( GBC \) is isosceles.

3. **Show that \( AG \) is perpendicular to \( BC \):**
- The property of the altitudes and angle bisectors in an isosceles triangle implies that \( AG \), being drawn from the vertex \( A \) to the segment \( BC \), will also be perpendicular to \( BC \):

\[
AG \perp BC
\]

### Conclusion:
Thus, we have proved:
- \( BM = CN \)
- Triangle \( GBC \) is isosceles.
- \( AG \) is perpendicular to \( BC \).

This concludes the proof for the statements in the problem.