Really nice problem. The given problem is actually quite trivial. Let $M'$ be the second intersection of $(DFG)$ and $\overline{AB}$. Then, \[\measuredangle GDM' = \measuredangle GFB = \measuredangle CFB = \measuredangle AFD = \measuredangle M'FD = \measuredangle M'GD \]So, $M'D=M'G$ which implies that in fact $M'=M$ which finishes the problem (the fact that $\measuredangle CFB = \measuredangle AFD$ follows from the fact that the orthocenter is the incenter of the orthic triangle). Remark : The is a LOT going on in this picture. For example, the second intersection of $\overline{EF}$ and $(DFG)$ in fact lies on the perpendicular bisector of $DG$ (proof entirely the same). Further, if this second intersection point is $N$ , $\overline{NC}$ passes through the second intersection $R$ of $(AB)$ and $(DGF)$. Further, if the perpendicular bisector of $DG$ hits $AC$ at $X$ , $X$ in fact lies on both the circles $(DNE)$ and $(DAM)$.

Clearly, $CEFB$ are concyclic. Hence, $BG=\frac{BC\sin GCB}{BC\sin GCB+CE\sin ECG}BE=\frac{BC\cdot BF}{BC\cdot BF+CE\cdot EF}BE$. Then, we also have $BF=BE\cos EBF=BE\cdot \frac{BD}{BA}$. Finally, $BM=\frac{BD+BG}{2\cos DBA}=\frac{BA(BD+BG)}{2BD}$. Thus, $\frac{BD\cdot BG}{BF\cdot BM}=\frac{BD\cdot \frac{BC\cdot BD}{BC\cdot BD+CE\cdot DA}\cdot BE}{BE\cdot \frac{BD}{BA}\cdot BA\cdot (BD+BG)/(2BD)}$, which is then equivalent to $\frac{2BD\cdot \frac{BC\cdot BD}{BC\cdot BD+CE\cdot DA}}{BD+BG}$, which is $\frac{2BD\cdot \frac{BC\cdot BD}{BC\cdot BD+CE\cdot DA}}{BD+\frac{BC\cdot BD}{BC\cdot BD+CE\cdot DA}\cdot BE}=\frac{2BD\cdot \frac{BD}{BD+DE}}{BD+\frac{BD}{BD+DE}\cdot BE}=\frac{2BD^2}{BD^2+BD\cdot DE+BD\cdot BE}=1$ as wanted.