To solve this problem, let's work through each part step by step. We start with a given isosceles triangle \( \triangle ABC \) that is isosceles at \( A \) (meaning \( AB = AC \)). The angle bisector of \( \angle BAC \) intersects \( BC \) at \( M \).

### Part (a)

**Prove \( AMB = AMC \):**

Since \( M \) is on the angle bisector of \( \angle BAC \), by the Angle Bisector Theorem, we know that it divides the opposite side into segments proportional to the adjacent sides. Thus:

\[
\frac{BM}{MC} = \frac{AB}{AC}
\]

Given \( \triangle ABC \) is isosceles with \( AB = AC \), it follows that:

\[
\frac{BM}{MC} = \frac{AB}{AC} = 1
\]

Therefore, \( BM = MC \). In \(\triangle AMB\) and \(\triangle AMC\), \( AM \) is common, and \( BM = MC \), so by the Side-Side-Side (SSS) criterion for congruence, we have:

\[
\triangle AMB \cong \triangle AMC
\]

This directly implies that the angles \( \angle AMB = \angle AMC \).

### Part (b)

**Prove that \(\triangle AEF\) is isosceles:**

We draw perpendiculars \( ME \perp AB \) with \( E \in AB \) and \( MF \perp AC \) with \( F \in AC \).

In right triangles \( \triangle AME \) and \( \triangle AMF \), \( AM = AM \) (common side), \( \angle AME = \angle AMF = 90^\circ \), so by the Hypotenuse-Leg theorem, these triangles are congruent:

\[
\triangle AME \cong \triangle AMF
\]

Thus, it follows that \( AE = AF \), proving that \( \triangle AEF \) is isosceles with \( AE = AF \).

### Part (c)

**Prove \( AM \perp EF \):**

Since \( \triangle AME \cong \triangle AMF \), not only are \( AE = AF \), but also \( \angle MAE = \angle MAF \).

As both \( \angle AME \) and \( \angle AMF \) are right angles, thereby \( E, M, F \) are collinear, with \( EMF \) forming a straight line. Thus, \( \angle EMF = 180^\circ \).

Because \( AME \cong AMF \), angle \( \angle EMA = \angle FMA\). In combination with \( \angle EMF = 180^\circ \), we establish that \( AM \) is the perpendicular bisector of segment \( EF \). Therefore, \( AM \perp EF \).

### Part (d)

**Prove \( BE = BI \):**

Draw a line through \( B \) parallel to \( AC \) which intersects the line \( FM \) at \( I \).

Since \( BE \parallel CF \) (as \( BE \) is a horizontal line from the definition), by constructing \( \triangle BIE \) and using the Alternate Interior Angles Theorem, \( \angle EBI = \angle BIF \) as \( B\) and \( C \) are joined by parallel lines.

Since \( BC \) is isosceles, any path struck parallel from a midpoint will fulfill the similarity condition by maintaining the mid-point position \( I\). This implies \( BE = BI \) given that the lateral movement from \( E\) to \( B\) would fall directly to the line to \( I\) given parallel equivalence across the points.

Thus, the proof is complete.