To prove that triangles \( \Delta AOM \) and \( \Delta BOM \) are congruent and that \( MB = MA \), we can follow these steps:

### Proof for \( \Delta AOM = \Delta BOM \)

1. **Common Side (\( OM \))**:
- The segment \( OM \) is common to both triangles.

2. **Angle \( AOB \)**:
- Given that \( \angle D = 60^\circ \) and \( \angle E = 60^\circ \), we can infer that \( \angle AOB = \angle D + \angle E = 60^\circ + 60^\circ = 120^\circ \).

3. **Given Angles**:
- We also know that \( \angle AOM + \angle BOM = \angle AOB = 120^\circ \). Each triangle shares the common vertex at \( O \), creating two angles that sum to \( 120^\circ \).

4. **Using the criteria for triangle congruence**:
- We can conclude that \( \Delta AOM \cong \Delta BOM \) by the Angle-Side-Angle (ASA) criterion, since:
- \( OM \) is common.
- \( \angle AOM = \angle BOM \) (since \( OA = OB \)).
- \( \angle AOB = 120^\circ \).

### Conclusion for \( MB = MA \)

5. **Consequences of Congruence**:
- Since \( \Delta AOM \cong \Delta BOM \), the corresponding sides are equal.
- Therefore, from the congruence, we can deduce that:
\[
MB = MA
\]

This completes the proof that \( \Delta AOM = \Delta BOM \) and \( MB = MA \).