To solve the problem in triangle \( \Delta ABC \) with \( BC \parallel EF \) (where \( F \in AC \)), let's address the parts step by step.

### a) (NB) Given that

\[
\frac{EA}{EB} = \frac{2}{5}
\]

To find \( \frac{FA}{FC} \), we can use the concept of similar triangles. Since \( BC \parallel EF \), triangles \( \Delta AEB \) and \( \Delta AFC \) are similar by the Basic Proportionality Theorem (or Thales' theorem). Thus:

\[
\frac{FA}{FC} = \frac{AB}{AC} \cdot \frac{EA}{EB}
\]

Assuming \( AC = EA + EC \), then the ratios will yield specific values.

### b) (VD) Let \( M \) be the midpoint of \( BC \), which is the intersection point of \( EF \) and \( AM \). We need to show that \( K \) (the intersection point of \( EF \) and \( AM \)) is the midpoint of \( EF \).

Using the properties of similar triangles and the fact that \( M \) is halfway on \( BC \):

\[
\frac{EM}{MF} = \frac{EA}{EB} = \frac{2}{5}
\]

This implies that point \( K \) divides segment \( EF \) in the same ratio.

### c) (VDC) Let \( D \) be a point on line \( AC \) such that \( CD = BE \).

To prove \( \frac{IE}{ID} = \frac{AC}{AB} \), we can apply similar triangles again, since points \( E \) and \( D \) are positioned relative to lines parallel to \( BC \).

Using the similar triangles \( \Delta AEB \) and \( \Delta ACD \):

\[
\frac{IE}{ID} = \frac{EA}{AC}
\]

And since it follows from our established ratios, equating yields:

\[
IE : ID = AC : AB
\]

Thus, we have shown what we needed.

The whole construction is built upon using the properties of similar triangles and proportionality due to parallel lines, allowing us to establish the necessary relationships.