Selection Examination

Let $K$ be the midpoint of $BC$ and $P$ the second intersection of $GD$ with $c$. Then $c$ and the Euler's circle are homothetic with center $G$ and ratio $-2$, so $$\frac{GP}{GD}=\frac{GA}{GK}=2\Rightarrow GP=2GD.$$ From the power of a point theorem we have $$GM\cdot GA=GN\cdot GP\Rightarrow GM\cdot (2GK)=GN\cdot (2GD)\Rightarrow GM\cdot GK=GN\cdot GD,\text{ so the quadrilateral }DKMN\text{ is inscribed in a circle, let it be }{c}_1.$$ Moreover $F,D,K,E$ belong to a circle (the Euler circle), let it be $c_2$. Therefore the lines $FE,DK,MN$ are concurrent to the radical center, let it be $T$, of the circles $c,c_1,c_2$. To this end, $TF\cdot TE=TD\cdot TK=TN\cdot TM$, so the points $F,E,M,N$ are cocyclic.