i think my solution is too much boring
let's apply Menelaus's and Menelaus's ;
let , $CF\cap AB=N ; BF\cap AC=P ; BF\cap CM=R ; CF\cap BD=S ; CM\cap BD=T$
Now
$\frac{BD}{DT}\frac{TF}{FA}\frac{AM}{MB}=1\Leftrightarrow \frac{AF}{FT}=\frac{BD}{DT}$
$\frac{AC}{CD}\frac{DS}{ST}\frac{TF}{FA}=1 \Leftrightarrow \frac{DS}{ST}=\frac{1}{3}\frac{AF}{FT}=\frac{1}{3}\frac{BD}{DT}$
$\frac{DB}{BT}\frac{TE}{EA}\frac{AC}{CD}=1\Leftrightarrow \frac{DB}{BT}=\frac{1}{3}\frac{EA}{ET}$
so, we have $\frac{DS}{ST}=\frac{DB}{BT}$
$\therefore (D,S,T,B) harmonic$
now, $\frac{AC}{CD}\frac{DF}{FM}\frac{MN}{NA}=1 \Leftrightarrow \frac{MN}{NA}=\frac{CD}{AC}\frac{FM}{DF}=
\frac{1}{3}\frac{3}{2}=\frac{1}{2}$
and, ceva for first time $\frac{BN}{NA}\frac{AP}{PC}\frac{CE}{EB}=1\Leftrightarrow \frac{AP}{PC}=1$
lastly, $\frac{AB}{BM}\frac{MR}{RC}\frac{CP}{PA}=1\Leftrightarrow \frac{MR}{RC}=\frac{BM}{AB}=\frac{1}{2}$
so, $\frac{MN}{NA}=\frac{MR}{RC}=\frac{CE}{EB}=\frac{1}{2}$
$\Delta NCM\sim \Delta RBM\sim \Delta EAC \Leftrightarrow \angle NCM=\angle RBM=\angle EAC$
$\therefore MFCB $ concyclic.
$\therefore \angle SFB=\angle CMB=90$
finally $FS$ bisects $\angle DFT$
so, $FC$ bisects $\angle DFE$