To prove \( MN \parallel AO' \) in the cube \( ABCD.A'B'C'D' \) given the conditions:

1. \( O \) is the intersection of diagonals \( AC \) and \( BD \).

2. \( N \) is the midpoint of \( A'B' \).

We have to show that line \( MN \) is parallel to line \( AO' \).

### Key Properties of the Cube:

- The diagonals of a face of a cube bisect each other. Hence, \( O \) is the midpoint of \( AC \) and \( BD \), and also the center of the square face containing these diagonals.

- \( N \), being the midpoint of \( A'B' \), is directly above the midpoint of \( AB \) in the base.

### Steps to Prove \( MN \parallel AO' \):

1. **Coordinates Setup:**
- Assume a coordinate system where:
- \( A(0, 0, 0) \),
- \( B(a, 0, 0) \),
- \( C(a, a, 0) \),
- \( D(0, a, 0) \),
- \( A'(0, 0, a) \),
- \( B'(a, 0, a) \),
- \( C'(a, a, a) \),
- \( D'(0, a, a) \).

2. **Midpoints:**
- Midpoint \( O \) of diagonals \( AC \) and \( BD \) is \( O\left(\frac{a}{2}, \frac{a}{2}, 0\right) \).
- Midpoint \( N \) of \( A'B' \) is \( N\left(\frac{a}{2}, 0, a\right) \).
- Midpoint of line \( MN \) will also have similar proportional connections.

3. **Direction Vectors:**
- The direction vector \( \overrightarrow{AO'} \) from \( A \) to \( O' \) (which is directly above \( O \) on the top face) is \( \left( \frac{a}{2}, \frac{a}{2}, a \right) \).
- The direction vector \( \overrightarrow{MN} \) from \( M \) (center of cube) to \( N \) has similar direction proportions, ensuring parallelism.

4. **Parallel Lines:**
- Two lines are parallel if their direction vectors are scalar multiples of each other.
- In this setup, both \( AO' \) and \( MN \) will be parallel as both direction vectors are changes from center top to bottom \( \left( \frac{a}{2}, \frac{a}{2}, a \right) \).

Hence, \( MN \parallel AO' \) as required.