It is not true.
If the line (which looks like altitute) is not perpendicular as it appears but rather slanated it fails.
I've attached a drawing of a triangle with the given conditions.
(This sign < means angle)
<(EFD) = 90° (it is the half of 180° and therefore constant)
The intersection of AF ∩ DE = P
Now reflect P over FD to P'. Then PP' is parallel to FE because PP' is perpendicular to FD and FD is perpendicular to FE. P' lies on CB because the triangle PFP' is an isosceles triangle. Therefore P'F must be parallel to PE because PEFP' is a parallelogramm.
Do the same procedure with P, P'', F and D.
You get DE is parallel to P'P'' and therefore DE is parallel to BC.