An isoceles trapezoid has the following properties:
it is a quadrilateral
it has congruent diagonals
its diagonals bisect its internal angles
the lower base angles are congruent
the upper base angles are congruent
the pair of non-parallel edges are congruent
the pair of parallel edges are not congruent
Thus, if ADFB can be proven to be an isoceles trapezoid, ADFB has all these properties. That would show that DF is parallel to the base segment, that the base angles are equal, and that my solution holds.
From the information given, we know that ADFB has congruent diagonals which bisect the lower base angles. I have taken this to prove that ADFB is an isoceles trapezoid; you are questioning whether that is proof enough.
In other words, you are asking:
Can't there exist a quad which has congruent diagonals that bisect the internal angles, but which is NOT an isoceles quadrilateral?
I've tried looking that up, but so far haven't managed to find anything about it. From testing it so far, it seems the answer to the question is no - but if anyone can show otherwise, I'm open to persuasion.
...
So - I will accept, for the moment, that may proof might not hold (though it isn't ready to be tossed out the window yet!).
Keep in mind that there may be more than one solution.
Meanwhile, I'll see what I can do with the law of sines...