In △ A B C, ∠ B < ∠ C, and B D and C...

In $\mathrm{△}ABC$, $\mathrm{\angle }B<\mathrm{\angle }C$, and $BD$ and $CE$ are angle bisectors. $D$ is on $AC$ and $E$ is on $AB$. Prove that $CE<BD$.

Using the fact that $\mathrm{\angle }B<\mathrm{\angle }C$, I got that $AB>AC$ and $BF>CF$ (where $F$ is the intersection of the two angle bisectors).
I then expressed $CE=CF+FE$ and $BD=BF+DF$. Since $CF<BF$, we only have to prove that $FE<DF$. I tried to use the angle bisector theorem to get some relations that might help me prove that, but here's where I got stuck. Could someone please help me out?