Cute and easy. Let $(CFE)$ meets $BC$ at $T$. Then $\angle FTC= \angle FEC = \angle ABC$, so $TF||AB$. We want $B$ to lie on the radical axis of $(DNF)$ and $(EFC)$, so we shall prove $BT.BC=BD.BN \iff \frac {AF} {AD}=\frac {BT} {BD}=\frac {BN} {BC} \iff \frac {DF} {AF}= \frac {CN} {BN}= \frac {DN} {BN}$ which is trivial by Menelaus for $\triangle ABD$ and the line $MN$, done.

By defining $\mathcal P(X)$ to be the difference of powers from $X$ to $(DFN)$ and $(EFC)$, respectively, we have that \[ \frac{DC}{BC}\mathcal P(B) +\frac{BD}{BC}\mathcal P(C)=\mathcal P(D) \]by linearity of PoP. We just need to prove that $\mathcal P(B) = 0$, which is equivalent to \begin{align*} BD\cdot \mathcal P(C) = BC\cdot \mathcal P(D) &\iff BD\cdot CD\cdot CN = BC\cdot DE\cdot DF\\ &\iff BD\cdot CD\cdot CN = BC\cdot DF\cdot \frac{DB\cdot DC}{DA}\\ &\iff CN\cdot DA = BC\cdot DF. \end{align*}But overserve that if $L$ is the midpoint of $AC$, then $\triangle MNL \sim \triangle NDF$, hence \[ \frac{DN}{DF} = \frac{ML}{LN} = \frac{2\cdot BC}{2\cdot DA}=\frac{BC}{DA}, \]thus we're done.

Projective cookie. Let $(CGFE) \cap BC=L$ then by Reim's theorem $AB \parallel FL$ and now by projecting $$-1=(B, A; \infty_{AB}, M) \overset{F}{=} (B, D; L, N)$$And from McLaurin $BL \cdot BC=BD \cdot BN$ which means that $B$ lies in radax of $(DFGN),(CGFE)$ which gives that $B,G,F$ are colinear thus we are done