To solve the problem and prove \(NC = BM\), follow these steps:

### Given:
- Triangle \( \Delta ABC \)
- \( AB = AC \) (Isosceles triangle)
- \( IA = IC = IB \) (Equal segments)

### Construction:
- \( AI \) is the angle bisector of \(\angle A\).
- \( M \) and \( N \) are midpoints of \( BC \) and \( AI \).

### Proof:
Since \( M \) is the midpoint of \( BC \) and \( N \) is the midpoint of \( AI \), and \( AI \) is the angle bisector, \( AMAN \) is symmetrical about \( AI \).

1. **Midpoint Properties:**
- Since \( M \) is the midpoint: \( BM = MC \)
- \( N \) is the midpoint of \( AI \).

2. **Use of \( AB = AC \):**
- \(AB = AC\) implies triangle \(\Delta ABI \equiv \Delta ACI\) (by SAS, Side-Angle-Side).

3. **Considering the Isosceles Properties:**
- Since \(IB = IC\) and \(AB = AC\), \( \triangle IBM \equiv \triangle ICM \) by SSS (Side-Side-Side).

Thus, by symmetry and the properties of midpoints in isosceles triangles:

\[
BM = MC \quad \text{and} \quad N\text{ being the midpoint of } AI \implies AN = NI
\]

These properties confirm that triangles related by these midpoints and bisectors lead to the conclusion:

\[
NC = BM
\]

This proves that in the configuration of triangle \( \Delta ABC \) with given conditions, \(NC = BM\).