Let $\{R\}=EF\cap (ABC)$. As $\widehat{EFD}=\widehat{EMD}=90^\circ$, we have that $DR$ is diameter in $(ABC)$, hence $DB\perp BR$, i.e. $BR\parallel AG$. Let $\{N^\prime\}=BF\cap AG$; as $E-F-R$ are collinear and $\{E\}=AA\cap DD$, we have that $AFDR$ is harmonic, hence $B(A,D,F,R)=-1\Leftrightarrow (G,A,N^\prime, \infty)=-1$, so $N^\prime$ is the midpoint of $AG$, whence the conclusion.

My solution: Since $EA$ and $ED$ are tangents to $\odot (ABC)$ $\Longrightarrow$ the circumference of diameter $EM$ is orthogonal to $\odot (ABC)$ since $E$ and $M$ are conjugated points respect to $\odot (ABC)$. Let $P$ the midpoint of $EM$ and let $F'=\odot (ABC)\cap AP$ $\Longrightarrow$ $PM^2=PF.PA$ since the circumference of diameter $EM$ is orthogonal to $\odot (ABC)$ $\Longrightarrow$ $MF'\perp AP$ and $\angle MF'A= 90^{\circ}$ but $EF$ is $F$-symmedian of $\triangle ADF$ $\Longrightarrow$ by angle-chasing $\angle DF'E=90^{\circ}$ $\Longrightarrow$ $F'=F$ and $\triangle GAB\sim \triangle MEA$ since $\angle GBA=\angle MAE$ and let $N'=BF\cap GA$ $\Longrightarrow$ $\angle N'BA=\angle PAE$ $\Longrightarrow$ $N'$ is the midpoint of $AG$ since $\triangle PEA\sim \triangle N'AB$ $\Longrightarrow$ $N'=N$ $\Longrightarrow$ $B,F,N$ are collinear...

So easy!!

$FE$ is symmedian of $\triangle AFD \Rightarrow MFA=90^{\circ}$, also $\angle DFM=\angle DEM=\angle DAO=\angle GAB$ Denote $S$ the reflection of $F$ at $M$ so $\triangle FSA \cup M \sim \triangle FAB \cup M \Rightarrow \angle FMA=\angle BNG=\angle FNA \Rightarrow F,N,B$ are collinear.

We redefine $N$ as the intersection of $AG$ and $BF$ and we want $N$ to be the midpoint of $AG$ or equivalently $MN\parallel GD$. By angle chasing $\angle ANF=\angle BNG=90-\angle FBD=90-\angle FDE=\angle FED=\angle AMF$ (since $\angle DFE=\angle DME=90$ and $ED$ is tangent to the circumcircle of triangle $ABC$) and hence $AFMN$ is cyclic.So we have $\angle AMN=\angle AFN=\angle AFB=\angle ADG$ which gives the result.

My solution: Let $O$ be the circumcenter of $\odot(ABC)$. Lemma. $DMFE$ is harmonic quadrilateral. Proof of the lemma. Obviously $O,M,E$ are collinear. Since $\triangle OAE$ is a right triangle, so $OD^2=OF^2=OA^2=OM\cdot OE$. The result then follows.$\square$ Back to the main problem, Since $ED$ is tangent to $\odot(ABC)$, so $\angle EDF=\angle FBD$, we have \[\frac{\sin{\angle FBD}}{\sin{\angle ABF}}=\frac{\sin{\angle EDF}}{\sin{\angle ADF}}=\frac{FE}{FM}.\qquad (1)\]On the other hand, Since $BN$ is a median of $\triangle ABG$ and clearly $\triangle ABG\sim\triangle EDM$, so we get \[\frac{\sin{\angle NBG}}{\sin{\angle ABN}}=\frac{AB}{BG}=\frac{ED}{DM}. \qquad (2)\]Compare $(1)$ and $(2)$, since $\angle FBD+\angle ABF=\angle DBN+\angle NBA=\angle ABD<180^\circ$, by the above lemma we conclude that $\angle FBD=\angle NBD$, which amounts to $B,N,F$ are collinear, as desired. A diagram below: [asy][asy] import olympiad; import cse5; size(9cm); defaultpen(fontsize(10pt)); pair O,A,B,C,D,E,G,M,N; A=dir(60); B=dir(-170); C=dir(-10); O=origin; D=2*foot(A,O,dir(0))-A; M=midpoint(A--D); E=extension(O,M,A,A+(O-A)*dir(90)); G=foot(A,B,D); N=midpoint(A--G); pair [] F=IPs(unitcircle,circumcircle(M,D,E)); draw(unitcircle); draw(A--E); draw(D--E); draw(A--B--C--cycle); draw(A--D); draw(O--M,dashed); draw(circumcircle(M,D,E),blue); draw(B--D); draw(A--G); draw(B--F[0],red+dashed); draw(M--F[0],dashed); draw(F[0]--E,dashed); draw(M--E); draw(D--F[0],dashed); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(-100)); dot("$E$",E,dir(E)); dot("$F$",F[0],dir(75)); dot("$G$",G,dir(G)); dot("$M$",M,dir(135)); dot("$N$",N,dir(135)); dot("$O$",O,SW); [/asy][/asy]

Here is my solution for this problem Solution We have: $\widehat{AFM}$ = $180^o$ $-$ $\widehat{MAF}$ $-$ $\widehat{AMF}$ = $180^o$ $-$ $\widehat{FDE}$ $-$ $\widehat{DEF}$ = $\widehat{DFE}$ = $90^o$ = $\widehat{AGD}$ = $\widehat{ANM}$ So: $A$, $F$, $M$, $N$ lie on a circle Then: $\widehat{AFN}$ = $\widehat{AMN}$ = $\widehat{ADB}$ = $\widehat{AFB}$ or $B$, $F$, $N$ are collinear