Mediterranean Mathematical Competition

a) If $M,N,F$ are collinear, we must have $DE\parallel BC$, hence the distances from $D$ and $E$ to $BC$ be equal, and this is equivalent to $b=c$.

b) If $A,M,F$ are collinear, $\tan \overrightarrow{BAM}=\frac{b\sin A}{c-b\cos A}$, $\tan \overrightarrow{DAN}=\frac{b\cos A}{c-b\sin A}$, and $\overrightarrow{DAN}=45^{\circ }+\overrightarrow{BAM}\iff b=c$.

c) If $A,M,N$ are collinear, we call $H=CD\cap AB$, $G=BE\cap AC$ and computing we have $$\frac{AG}{GC}=\frac{c\sin (A+45^{\circ })}{a\sin (B+45^{\circ })}$$ and similarly $\frac{AH}{HB}$. By Ceva the condition becomes $b=c$.

d) if $A,F,N$ are collinear, to use Ceva in $DEF$ we consider $X=CD\cap AE$, $Y=BG\cap AD$ (eventually improper). We get $$\overrightarrow{HCG}=\overrightarrow{HBG},\text{ hence HG is antiparallel to }BC,i.e.AH\cdot AB=AG\cdot AC\iff b=c.$$

e) If $b=c$, all four considered points are collinear.