# 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?

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?
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Step 1
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 ${84}^{2}-\left(42-x{\right)}^{2}={112}^{2}-\left(56+x{\right)}^{2}$.
Step 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.