Let's analyze the problem step by step.

**Given:**
Triangle \( \Delta ABC \) is isosceles at \( A \), and \( M \) is the midpoint of side \( BC \).

### Part a: Prove that \( \Delta AMB = \Delta AMC \)

1. **Sides:**
- \( AM = AM \) (common side)
- \( MB = MC \) (since \( M \) is the midpoint of \( BC \))
- \( AB = AC \) (given that \( \Delta ABC \) is isosceles)

2. **Angles:**
- \( \angle AMB = \angle AMC \) (they are vertical angles)

By the criteria of congruence (SAS), we have:
\[
\Delta AMB \cong \Delta AMC
\]

### Part b: Prove \( EA = FA \)

1. **Construction:**
- Draw perpendiculars \( ME \perp AB \) and \( MF \perp AC \).
- Since \( M \) is a point on the perpendiculars to sides \( AB \) and \( AC \), and \( E \) and \( F \) are the foot of the altitudes from \( M \) to \( AB \) and \( AC \), respectively.

2. **Triangles:**
- In \( \Delta AEM \) and \( \Delta AFM \):
- \( AM = AM \) (common side)
- \( ME = MF \) (both are heights dropped from the same point \( M \) to the respective sides)
- \( \angle AEM = \angle AFM = 90^\circ \) (right angles)

By the criteria of congruence (SAS), we conclude \( \Delta AEM \cong \Delta AFM \), which gives:
\[
EA = FA
\]

### Part c: Prove \( EF \parallel BC \)

Since \( ME \) is perpendicular to \( AB \) and \( MF \) is perpendicular to \( AC \), we know that:

- \( \angle EMF = 90^\circ \)
- \( AB \) and \( AC \) are equal angles because \( \Delta ABC \) is isosceles.

Thus, by the alternate interior angle theorem, since \( EA = FA \) and both angles \( \angle AME \) and \( \angle AMF \) (formed with line \( EF \)) are equal to the angles at \( M \) due to alternate angles, it follows that:
\[
EF \parallel BC
\]

### Conclusion:
- We have demonstrated that \( \Delta AMB = \Delta AMC \), \( EA = FA \), and \( EF \parallel BC \).