Angle quadrisection in a triangle

In triangle ABC, AB=84, BC=112, and AC=98. Angle B is bisected by line segment BE, with point E on AC. Angles ABE and CBE are similarly bisected by line segments BD and BF, respectively. What is the length of FC?

I solved this problem with the angle bisector theorem and Stewart's theorem, but I was wondering if there was a more elegant approach?

Once you've used the angle bisector theorem you know that the center bisector breaks up 98 into 42 and 56. Now drop a height to the 98 side. You know that it has to be closer to the 84 side than that center bisector. Why? Now break up the base into 42-x and 56+x according to where you put the height. Using the Pythagorean theorem, we know $84^2 - (42-x)^2 = 112^2 - (56+x)^2$. Solve and you get $x=21$. This means that the altitude (height)of the triangle formed by the 84 side, the 42 side, and the center bisector is also a median. Therefore, that triangle is isosceles and the middle bisector has length 84 and the result follows from another application of the angle bisector theorem.