This is Megh Parikh

Order of $F,N,M,C$

case 1: $F-M-N-C$

Then N is mid-point of CF & E is midpoint AC => EN should be parallel to AF i.e AB. But that cant be true. (I have not twisted the question to become like this. This is how the question originally appeared to me.)

case 2: $F-N-M-C$

diagram

Applying meneleaus theorem in $\triangle ACF$ with $BE$ transversal $\frac{AE}{EC}\cdot \frac{CN}{NF}\cdot \frac{FB}{AB}=1$

Evaluating this we get $2a=b$ .

Applying meneleaus theorem in $\triangle BCF$ with $AD$ transversal $\frac{BD}{DC}\cdot \frac{CM}{MF}\cdot \frac{FA}{AB}=1$

Evaluating this we get $5c^2=21a^2$ .

Also we know that length of angle bisector is 4. $a\cdot b\cdot (1- \frac{c^2}{(a+b)^2} ) =4$

Evaluating this we get $a=\sqrt{15}$ $b=2\sqrt{15}$ $c= 3\sqrt{7}$