As I read it, your parallelogram is not valid. This is because ABF should form a right isosceles triangle, yet BAE is a 5-4-3 triangle (so the angle BAE is less than BAF). Thus, F cannot lie on the line segment CD.

Perhaps you mean that F lies on the line determined by CD. In this case, one can find EF by simply using the pythagorean theorem and scaling relations. Here is what I imagine your description to be:

I don’t quite know exactly what you mean by no trigonometry, but the rest is simply solving and applying 5-4-3 triangles (or more generally, it is a simple scaling relationship to the triangle BAE). First, we see BC is the hypotenuse of BCF with short leg BF, so BC=5/3*BF (thus BC=25/3=AD).

Then, notice that HBE has short leg EH and long leg BE, so EH=3/4*EB (in fact, EH=9/4). This gives us HD=AD-AE-EH (so HD=25/12).

HD is the hypotenuse of HFD with short leg HF, so HF=3/5*HD (HF=5/4).

The last triangle is HFD with hypotenuse HF, so the two legs are HG=3/5*HF (HG=3/4) and FG=4/5*HF (FG=1). Now we can get EG=EH+HG (EG=9).

Finally, using the Pythagorean theorem, EF^2=EG^2+FG^2 ( )