Cho hình chữ nhật ABCD (AB > AD), AC cắt BD tại I, tia phân giác của góc BAD cắt BD tại M

To solve this geometry problem related to rectangle \(ABCD\):

### a) Prove that \( M \) belongs to segment \( ID \)

1. Since \( I \) is the intersection of diagonals \( AC \) and \( BD \), we know \( IA = IB = IC = ID \) (the diagonals of a rectangle are equal).
2. Triangle \( ABD \) is significant as it has the angle bisector \( AM \) which divides the angle \( BAD \).
3. In triangle \( ABD \):
- Given \( AB > AD \), we observe that \( MB > MD \).
4. Therefore, we can conclude:
- \( MB \) and \( MD \) imply \( MB > MD \) because \( AB > AD \).
5. Hence, \( M \) lies on the line segment \( ID \), which proves that \( M \in ID \).

### b) Prove \( MN \parallel AD \)

1. Utilize the ratio in \( \triangle ABD \):
- \(\frac{MB}{MD} = \frac{NC}{NA}\) since \( AM \) is the angle bisector.
2. Therefore, we conclude:
- \( MB \cdot NA = MD \cdot NC \).
3. Since \( M \) and \( N \) on segments \( ID \) and \( AC \) respectively maintain this ratio from triangle similarity, it follows that line \( MN \) is parallel to \( AD \).

### Final Conclusion

- **Thus, we have proven:**
- (a) \( M \in ID \)
- (b) \( MN \parallel AD \)