Let $Z$ be the second intersection point of $NF$ and circle with diameter $AB$. From the cyclic quadrilaterals we get that: $\angle NZE = \angle FZE = \angle FAE = \angle FAM = \angle EMN$, where the last equality holds because $AC \parallel MN$, and hence $ZNEM$ is cyclic. Also, $D,N,F,M$ are concyclic in the Euler circle. By radical axis theorem on $(AFEDBZ), (DMFN), (ZNEM)$ we get that $MN, DF, EZ$ are concurrent. Let $X$ be the concurrecy point and $T = NF \cap DE.$ By Brocard we have that: $ZF \cap DE= T$ lies on the polar of $ZE \cap DF = X$ with respect to $(AFEDBZ)$. Also, we have that: $\angle ZMX = \angle NMZ = \angle NEZ = \angle NZE$, so $NZ$ is tangent to $(ZXM)$, i.e $NZ^2=NX \times NM => NX \times NM = R^2$, where $R$ is the radius of $(AFEDBZ)$. From this equality, and using that $X$ lies on $MN$, we conclude that $M$ also lies on the polar of $X$, with respect to $(AFEDBZ)$. Hence, $MT$ is the polar of $X$ with respect to $(AFEDBZ)$, so $NM \perp MT => MT \perp AC$, as $AC \parallel MN$. Also, $MG \parallel BF => MG \perp AC => M,G,T$ are collinear, as needed.

Solution using Pascal Theorem Define point $Z$ as above Consider the cyclic hexagon $FZBDEA$, from Pascal, $ZF \cap DE = T$,$ZB \cap EA = X$,$BD \cap AF = C$ are collinear. Notice that $ZB \parallel AF$, and $AM$ passes through the middle of $BC$, by similar triangles we get $ABCX$ is a parallelogram, so $TC$ is parallel to $AB$. Now consider $L$ to be the intersection of $NF$ and $MG$. we will prove that $LC$ is parallel to $AB$ which will force $NF$,$DE$,$MG$ to be concurrent. (We cannot have two different parallel lines to $AB$ passing through $C$). It is clear that $MG$ is a medial line in right angle triangle $BFC$, from $ MG \perp FC$ and $FG$=$GC$, we get that $MFLC$ is a kite. Now by angle chasing: $\angle FCL = \angle CFL = \angle NFA = \angle NAF$, so we have shown $LC$ is parallel to $AB$, as desired.

Nice exercise for constructing a parallelogram!

Let $T$ be such that $AFBT$ is a rectangle. Clearly, $T \in \odot(ABF)$. Moreover, define $X=TD \cap GM$. Notice that $BF \parallel GM \parallel AT$, therefore by the converse of Reim's theorem we have that quadrilateral $DEMX$ is cyclic. Now we do simple angle chasing to see that $ \angle FTD = \angle DAF = \angle FBD = \angle GMC = \angle FMG$. This means that quadrilateral $TFMX$ is cyclic. Now by radical axis theorem on $\odot(TFXM)$, $\odot(DEMX)$ and $\odot(TFDE)$ we have that lines $TF, DE, XM$ are concurrent, which is the same as lines $NF,DE,GM$ being concurrent.

Quick bash from contest: We are going to use standart symbols: Let $NF\cap MG=T$. We will prove that $D,E$ and $T$ are collinear. $\bigtriangleup AMB:$ Cotangent technique $\Rightarrow$ $ ctg(<BAM)=\frac{\frac{BA}{AM}-cos(<ABM)}{sin(<ABM)}=\frac{\frac{2sin(\gamma)}{sin(\alpha)}-cos(\beta)}{sin(\beta)}=$$\frac{2sin(\alpha+\beta)}{sin(\alpha)sin(\beta)}-ctg(\beta)=2ctg(\alpha)+ctg(\beta)$ $\bigtriangleup BFC: MG\parallel BF \Rightarrow MG \perp AC$ $\bigtriangleup NAF: NA=NF \Rightarrow <NFA=<NAF=\alpha$ $\bigtriangleup FTC: TG$ - median and altitude $\Rightarrow FT=TC \Rightarrow <FCT=\alpha$ $\bigtriangleup GTC:TC=\frac{GC}{cos(\alpha)}=R.cos(\gamma)tg(\alpha)$ $\bigtriangleup DCT:$ Cotangent technique $\Rightarrow$ $ctg(<CDT)=\frac{\frac{DC}{CT}-cos(<DCT)}{sin(<DCT)}=\frac{\frac{2cos(\gamma)sin(\beta)}{cos(\gamma)tg(\alpha)}+cos(\beta)}{sin(\beta)}=2ctg(\alpha)+ctg(\beta)$ This means that $ctg(<EDC)=ctg(<BAM)=ctg(<TDC)$ $\Rightarrow <EDC=<TDC \Rightarrow$ $E, D$ and $T$ are collinear $\Rightarrow NF, MG$ and $DE$ are concurrent.

Here is a solution that i found during the contest. We will denote the intersection of $NF$ and $MG$ with $O$ and also: $$\angle A=\alpha, \angle B=\beta, \angle MAC=\alpha_1, \angle BAM= \alpha_2, \angle ABF =\beta_1, \angle FBC = \beta_2$$We clearly see that $\alpha=\alpha_1+\alpha_2$ , $\beta=\beta_1+\beta_2$ and by construction we have $ABDEF-$ cyclic (1) therefore $\angle AEF =\angle ABF =\beta_1$ and : $$\angle ABD+\angle DEA=180^{\circ}\Rightarrow \angle DEA=180^{\circ}-\beta$$Now, notice that $MG$ is the midline in $\triangle BCF$ (2) thus $MG\parallel FB$ which implies that $\angle FBC = \angle GMC = \beta_2$ but also $FB\perp AC$ because $AB$ is diameter,so by (2) it follows that $MG\perp FC$ .But $MG$ is also a median in $\triangle FMC$ therefore this triangle is isosceles which implies that: MG is also the angle bisector of $\angle FMC \Rightarrow \angle FMG=\angle CMG=\beta_2$ Claim: $FEMO-$ cyclic Proof: From $NB=NF \Rightarrow \angle NFB=\angle NBF=\beta_1$ , from (1) $\Rightarrow \angle BFE=\angle BAE = \alpha_2$ so $\angle EFO= 180^{\circ}-\angle NBF -\angle BFE=180^{\circ}-\beta_1-\alpha_2 $ . On the other side by (1) $\angle EDM= \angle BAE = \alpha_2$ and $$\angle DEM=180^{\circ}-\angle DEA=\beta \Rightarrow \angle EMC = \angle EDM + \angle DEM=\alpha_2+\beta $$But $\angle EMC = \angle EMG +\angle GMC=\angle EMO+\beta_2$ so $\angle EMO=\beta_1+\alpha_2$. Therefore $\angle EMO + \angle EFO=180^{\circ}$ and we get the desired claim. By our previous claim it follows that: $\angle FEO=\angle FMO=\beta_2$ Finally we know that: $$\angle DEO=\angle DEA +\angle AEF +\angle FEO=180^{\circ}-\beta+\beta_1+\beta_2=180^{\circ}$$Thus proving that $D$, $E$ and $O$ are collinear and so we are done .

Different approach and sketch of my sol. Let $NF \cap MG$ at $X$. Let $(FXM) \cap AM$ at $E'$.Let $XE \cap BC$ at $D'$.From angle chasing $E'$ and $D'$ coincide with $E$ and $D$ respectively.

We can also apply trigonometric Ceva to $\triangle DMN$ and lines $DE$, $MG$, and $NF$.

Always projective on top!, solved in a couple minutes, really nice problem. Let $NF \cap BC=J$ and $D'$ a point such that $D'ADB$ is a rectangle, now just note: \[ -1=(F, C; G, \infty_{AC}) \overset{M}{=} (F, J; MG \cap NF, N)=(J, F; MG \cap NF, N) \]\[-1=(B, C; M, \infty_{BC}) \overset{A}{=} (B, F; E, D') \overset{D}{=} (J, F; DE \cap NF; N) \]Both are enough to imply that these lines are concurrent as desired thus done .

Didn't use projective but still solved in a couple of minutes! Cool problem. Let $R = \overline{NF} \cap \overline{MG}$. First of all, it is well known that $D$ and $F$ are the feet of the altitudes from $A$ and $B$ , and that $F$ also lies on the with diameter $(BC)$. We also know that the circles $(AB)$ and $(BC)$ have centers $N$ and $M$ respectively. The entirety of the solution is the following key observation. Claim : Points $R$ , $F$ , $E$ and $M$ lie on the same circle. Proof : Now, since $MF=MC$ $M$ lies on the perpendicular bisector of segment $FC$ so $MG \perp FC$. Simply note that, \[\measuredangle EMR = \measuredangle AMG = \frac{\pi}{2} + \measuredangle EAF = \measuredangle EFN\]which implies the claim. Next, also note that since $R$ lies on the perpendicular bisector of segment $FC$ , $FR=CR$ and $\triangle RFC$ and $\triangle NFA$ (which is also isoceles clearly) share a vertically opposite angle. Thus, $\triangle RFC \sim \triangle NFA$ and in particular, \[\measuredangle ACR = \measuredangle FCR = \measuredangle FAN = \measuredangle CAB\]which implies that $CR \parallel AB$. Now we are essentially done since, \[\measuredangle MED = \measuredangle AED = \measuredangle AFD = \measuredangle ABC = \measuredangle RCB = \measuredangle RCM = \measuredangle REM\]which implies that points $D$ , $E$ and $R$ are collinear, which implies that these lines are concurrent, and we are done.