Lemma: Let $ABC$ be a scalene triangle. The perpendicular bisector of $BC$ intersects $(ABC)$ at $M$ and $N$ respectively. $D$ lies on $AC$ such that $ND \perp AC$. $K$ is the midpoint of $AM$. Prove that $A$, $K$, $D$, $B$ are concyclic. Proof: Let $T$ be the second intersection of $(N,NC)$ and $AC$. It can be easily seen that $D$ is the midpoint of $TC$. We have $\angle BTC = 180^{\circ} - \frac{1}{2} \angle BNC = 180^{\circ} - \angle BNM = \angle BAM$. Also, $\angle BMA = \angle BCT$, so $\triangle BAM \cup \{ K \} \sim \triangle BTC \cup \{ D \}$. Therefore, $\angle AKB = \angle TDB$, so quadrilateral $AKDB$ is cyclic. Back to the problem: Using the lemma, we can see that $D$ and $F$ lies on $(AN)$. Let $I$ be the midpoint of $BC$, $X$ be a point on $MN$ such that $AX \perp MN$. Then $A(DF, IX) = A(CB, IX) = -1 = (DF, NA) = X(DF, IA)$, so $\overline{D,I,F}$. Similarly, $\overline{E,I,G}$, and because $I$ lies on $MN$, we can conclude that $DF$, $EG$ and $MN$ are concurrent.

Attachments:

Another easy way to see $D - I - F$ is simply noticing that it is the Simson line with respect to $N$ in $\triangle ABC$.

Oh, somehow the Simson line escaped my mind. Tks hakN for improving my solution

mrleo_of_alf wrote: Lemma: Let $ABC$ be a scalene triangle. The perpendicular bisector of $BC$ intersects $(ABC)$ at $M$ and $N$ respectively. $D$ lies on $AC$ such that $ND \perp AC$. $K$ is the midpoint of $AM$. Prove that $A$, $K$, $D$, $B$ are concyclic. It can be easily seen that by angle chasing, triangles $ABT$ and $MBC$ are positive similar, so using spiral symmetry wrt.B (and some complex numbers), we obtain $ABT \sim KBD$, which finishes the proof from the uniqueness of point $D$

Another (lengthy) way to prove the lemma of mrleo_of_alf: $MN$ cuts $BC$ at $Z$. Wlog $AC>BC$. $\textbf{Lemma:} \frac{MA}{MB}=\frac{|AB-AC|}{BC}$. Proof: Trigonometry . $\textbf{Lemma:} BKM \sim BDC$. Proof: It is enough to prove $\frac{MK}{MB}=\frac{CD}{CB}$ because $\angle KMB=\angle DCB$. $\frac{MK}{MB}=\frac{CD}{CB} \leftrightarrow \frac{AC-AB}{2BC}=\frac{CD}{CB} \leftrightarrow \frac{AC-AB}{2}=CD$, $NDC \sim NAM \implies CD=\frac{AM\cdot CN}{MN}$. $\frac{AC-AB}{2}=CD \leftrightarrow \frac{AC-AB}{2}=\frac{AM\cdot CN}{MN}$. Because $AC$ and $MN$ intersects inside the circumcircle Ptolemy's Theorem on $ACMN$ gives $\frac{AM\cdot CN}{MN}=AC-\frac{AN\cdot MC}{MN}$. $\frac{AC-AB}{2}=\frac{AM\cdot CN}{MN} \leftrightarrow \frac{AC-AB}{2}=AC-\frac{AN\cdot MC}{MN} \leftrightarrow AB+AC=2\frac{AN\cdot MC}{MN}$. Ptolemy's Theorem on $ABCN$ gives $AB+AC=\frac{AN \cdot BC}{CN}$. $AB+AC=2\frac{AN\cdot MC}{MN} \leftrightarrow \frac{BC}{2CN}=\frac{MC}{MN} \leftrightarrow \frac{CZ}{CN}=\frac{MC}{MN}$. Last one is true because we have $NMC \sim NCZ$ From the last lemma we have $\angle AKB=\angle ADB \implies AKDB$ is cyclic. $\textbf{Note:}$ $BKM \sim BDC$ can be proved easily by proving $ABM \sim TBC$. Last can be proved by what mrleo_of_alf did in his proof.

by Menelaus theorem $D,F,\frac{BC}{2}$ collinear. Similarity $E,G,\frac{BC}{2}$ collinear.

Here is another solution. Let $N$ lie on the minor arc $\widehat{BC}$ and $M$ lie on major arc $\widehat{BC}$. It suffices to prove that $F$ is the foot from $N$ to $AB$, since then analogous results will prove the result by Simson lines. Let $O$ be the center of $(ABC)$. For this, we define $F$ to be the foot from $N$ to $AB$ and let $(ACF)$ meet $AM$ at $K$, then we want to show $K$ is the midpoint of $AM$. Let $NF$ meet $(ABC)$ again at $P$. By easy angle chasing we can prove $AM \parallel PC$. Let $A'$ be the antipode of $A$ in $(ABC)$ and let $J = NF \cap (ACF) \neq F$. Since $90^\circ = \angle ACA' = \angle ACJ$, we have $J,A',C$ are collinear. Also, clearly $ANMA'$ is a rectangle. Thus, we have $PC \parallel AM \parallel NA'$, so $PCNA'$ is an isosceles trapezoid and hence $J$ lies on the perpendicular bisectors of $AM,PC,NA'$. Since $O$ also lies on the perpendicular bisector of $AM$, we get that $JO$ is the perpendicular bisector of $AM$. Now let $JO \cap AM = K'$. We have $K'$ is the midpoint of $AM$ and $\angle AK'J = 90^\circ = \angle AFJ = \angle ACJ$, so $K'$ lies on the circle $(ACFJ)$, and hence $K' = K$, we are done. $\square$

Here is a rather quick solution First notice that we must have that $BK=KD$ and by spiral similarity we must have that $KF=KC$. Let $D'$ be a point on $AC$ such that $ND' \perp AC$ nd let $F'$ be a point on $AB$ such that $DF' \perp AB$. Throw the configuration on the complex plane such that $(ABC)$ is the unit circle and such that $c=\frac{1}{b}$. Then we must have that $m=-1$ and that $n=1$, $k=\frac{1}{2}(a-1)$ and $d'=\frac{1}{2}(1+a+\frac{1}{b}-\frac{a}{b})$. Then notice that $ABKD'$ must be cyclic, since we have that: $$\frac{\frac{k-d'}{k-b}}{\frac{a-d'}{a-b}} \in \mathbb{R}$$thus we have that $D \equiv D'$, similarly we must have that $F \equiv F'$. Let $H$ be the midpoint of $BC$, we easily see that $D,F,H$ must be colinear, since we have that: $$\frac{f-h}{\overline{f-h}} = \frac{f-d}{\overline{f-d}}$$where $h=\frac{1}{2}(b+c)$. Similarly $EG$ must pass through the midpoint of $BC$. Since $MN$ passes through the midpoint of $BC$ we must have that $EG,DF$ and $MN$ are concurrent lines.

Actually, the solutions above are quick and elegant. We can also prove some lenght equalities by using the perpendecular things which are proven above. After that, menelaus theorem handles this property with ease.

Using Brazil EGMO TST 2021 P3 we get that $ND\perp CA$ and analogous results for $E$, $F$ and $G$. Then using Simson's Line we find that $DF$, $EG$ and $MN$ concur at the midpoint of $BC$.