This is a beautiful problem Let $\Delta M_AM_BM_C$ be the medial triangle WRT $\Delta ABC$, Let $D'$ be reflection of $D$ over $M_A$, hence, $G$ is the centroid WRT $\Delta ADD'$ $\implies$ $DG$ bisects $AD'$, Let a parallel to $BC$ through $A$ $\cap$ $\odot (ABC)$ $=$ $T$, then, trivial to see, $T \in \odot (ADD')$ $\implies$ $ADD'T$ is a rectangle $\implies$ $T \in NG$ $$\angle ATN=\angle AMN=\angle TDC \implies DM_AMN \text{ cyclic }$$Hence $EF,BC,MN$ concur at, suppose, $X$, then, $XN \cdot XM =XD \cdot XM_A= XF \cdot XE \implies EFNM$ cyclic

AlastorMoody wrote: This is a beautiful problem ......................... Hence $EF,BC,MN$ concur at, suppose, ......................... Very nice ! This point should perhaps have been discussed a little more ! Radical center gives a not clear answer. How could you explain this point ?

Apply Radical Axes Theorem WRT $\odot (ABC), \odot (FEDM_A), \odot (DM_AMN)$ and use the fact that the orthic axis WRT $\Delta ABC$ is the radical axis WRT $\odot (ABC)$ and $\odot (FEDM_A)$ and hence suppose orthic axis $\cap BC$ at $X$ then obviously $EF$ passes through $X$

AlastorMoody wrote: Apply Radical Axes Theorem WRT $\odot (ABC), \odot (FEDM_A), \odot (DM_AMN)$ and use the fact that the orthic axis WRT $\Delta ABC$ is the radical axis WRT $\odot (ABC)$ and $\odot (FEDM_A)$ and hence suppose orthic axis $\cap BC$ at $X$ then obviously $EF$ passes through $X$ Thank you ! This is also aproximately my explanation .Any way, the greek students have given and some other information after the test.See : https://www.mathematica.gr/forum/viewtopic.php?f=58&t=62039&p=308036&hilit=%CF%80%CF%81%CE%BF%CE%BA%CF%81%CE%B9%CE%BC%CE%B1%CF%84%CE%B9%CE%BA%CF%8C%CF%82+2018#p308036 I put the figure(in greek letters and some different notation, but the content is clear .

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Similar to Alastoormoody Let the second intersection of $(ABC)$ with $GD$ be $O$. It is well known that that $BC \parallel AO$.Let $M_A$ and $M_B$ be the midpoints of $BC$ and $CA$. Note that $\angle ANM =\angle AON =\angle ODC \implies M,N,D,M_A$ conclic. $$(I,D;B,C)=-1 \implies ID.IM_A=IB.IC=IN.IM$$and so through Radical axis theorem $E,FM,N$ all lie on a circle.

Let $I$ $\equiv$ $GD$ $\cap$ $(ABC)$; $J$ $\equiv$ $EF$ $\cap$ $BC$ It's easy to see that: $AJ$ $\parallel$ $BC$ We have: $\dfrac{NB}{NC} . \dfrac{MB}{MC} = \dfrac{DB}{DC} . \dfrac{IC}{IB} . \dfrac{AC}{AB} = \dfrac{DB}{DC} . \dfrac{AB}{AC} . \dfrac{AC}{AB} = \dfrac{DB}{DC} = \dfrac{JB}{JC}$ Then: $M$, $N$, $J$ are collinear So: $\overline{JM} . \overline{JN} = \overline{JB} . \overline{JC} = \overline{JE} . \overline{JF}$ or $E$, $F$, $M$, $N$

A bit different finishing... Let $DG\cap\odot(ABC)=J$ for the second time. So from this we get $AJ\| BC$. So, $\angle AJN=\angle JDC=\angle AMN\implies MNDM_a$ is a cyclic quadrilateral, where $M_a$ is the midpoint of side $BC$. Let $EF\cap BC=K'$ and $MN\cap BC=K$. Hence, $K'F.K'E=K'D'.K'M_a=K'B.K'C$ as $E,F,D,M_a$ lie on the Nine Point Circle of $\triangle ABC$. Also, $KD.KM_a=KN.KM=KB.KC$. So, $K'\equiv K$. Hence, $EF,BC,MN$ concur at point $K$. Hence, $KF.KE=KD.KM_a=KN.KM\implies F,E,M,N$ are concyclic. $\blacksquare$.

Let $D'$ be such on $\odot(ABC)$ such that $AD' \parallel BC$. It is well-known that points $D',G,D$ are collinear. Let $K$ be midpoint of $BC$. Note that $(B,C;K,\infty) =-1$. Now observe that: $$ -1 = (B,C;K, \infty) \overset{A}{= }(B,C;M,D') \overset{N}{=} (B,C; NM \cap BC, D)$$Since $(B,C; EF \cap BC, D)=-1$, we conclude that lines $EF, BC, MN$ are concurrent at point $X_A$. Now by PoP we have that: $$ X_AE \cdot X_AF = X_AB \cdot X_AC = X_M \cdot X_AN $$This implies that quadrilateral $EFMN$ is cyclic as desired.

Let $A'$ be the second intersection of $\overline{GD}$ with $(ABC)$. It is well-known that $\overline{AA'} \parallel \overline{BC}$. Then $$GD \cdot GN = \frac 12 GA' \cdot GN = \frac 12 GA \cdot GM = GK \cdot GM,$$so $(DKMN)$ is cyclic. Furthermore, $(EFDK)$ is concyclic along the nine-point circle. Let $\overline{MN} \cap \overline{BC} = P_1$ and $\overline{EF} \cap \overline{BC} = P_2$. Then we can check that $$P_1B \cdot P_1C = P_1D \cdot P_1K \text{ and } P_2B \cdot P_2C = P_2D \cdot P_2K.$$It can easily be checked that such a point is unique (say, by setting variables for the side lengths $BD, DK$ and solving for $BP$, so $P_1 = P_2$). Thus we are done by radical axis.