To prove that \( EF \parallel AB \) and \( HF = 2FK \) in the given geometric configuration, we can apply the properties of parallel lines and triangles. Here’s a structured approach based on the information provided:

### Step 1: Proving \( EF \parallel AB \)

1. **Given:** The planes \( ABCD \) and the lines \( AM \), \( BC \) intersecting \( EF \) at points \( E \) and \( F \) respectively.
2. **Use the transversal theorem:** Since \( EF \) intersects the lines \( AM \) and \( BC \), we can check the angles formed. If the corresponding angles ∠AEF and ∠ABM are equal, then \( EF \parallel AB \) by the Corresponding Angles Postulate.

### Step 2: Proving \( HF = 2FK \)

1. **Establish triangles:** Consider triangles \( HFE \) and \( KFE \). Since \( EF \parallel AB \), triangles \( HFE \) and \( KFE \) are similar.
2. **Ratio of sides in similar triangles:** The sides opposite to equal angles of similar triangles maintain a constant ratio. Therefore, if you can establish \( HF \) being double the length of \( FK \) due to the ratios, you can conclude that:
\[
HF = 2FK
\]

### Additional Calculation

If you know the lengths of the segments or if additional conditions are given (like \( AB \) and the specific coordinates of points), you might apply the Segment Division Theorem or similar triangles' properties to find explicit lengths or ratios.

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