To solve the problems related to the right triangle ABC and its midpoints D, E, F, we start by addressing each part:

1. **Finding EF parallel to AB**: Since D and E are midpoints of sides AB and BC respectively, a line drawn through these points (EF) is parallel to side AB. This follows from the midpoint theorem, which states that a line segment connecting the midpoints of two sides of a triangle is parallel to the third side.

2. **Finding point K**: To show that points B, K, F are collinear and that segments KA and KF are equal, we can argue that since D and E are midpoints, and by symmetry in a right triangle, the median from A to the hypotenuse BC will bisect EF at some point K (the midpoint of EF). By properties of isosceles triangles, we have KA = KF.

3. **Distance between points B and E**: Given the positions of points A, F, and C, where AF = 40m, FC = 20m, and EC = 30m, we can place these points on a coordinate system. Let:
- A = (0, 0)
- F = (40, 0)
- C = (40, 20)
- E = (x, 20) where x is determined from EC = 30.

Using the distance formula and the given lengths, we can find the coordinates of E and then compute the distance between B (which lies vertically above A) and E.

4. **Calculating distance BE**: After establishing the coordinates, we can apply the distance formula:

\[
Distance = \sqrt{(x_B - x_E)^2 + (y_B - y_E)^2}
\]

Here, \(x_B\) and \(y_B\) represent the coordinates of point B, while \(x_E\) and \(y_E\) correspond to point E.

By closely following these steps and employing trigonometric principles and coordinate geometry, we can accurately gauge the distances and relationships between points B and E in this triangle scenario.