Ceva's Theorem implies
$$\frac{BF}{FA} \cdot \frac{AE}{EC} \cdot \frac{CD}{DB} = 1,$$
hence $\frac{BF}{FA} = \frac{BD}{DC}$. The converse of Thales' Theorem yields $DF \parallel AC$. From $\triangle BFD \sim \triangle BAC$ follows $\frac{BF}{BA} = \frac{FD}{AC}$, whence $BF = \frac{AB}{AC} \cdot FD$. Now $\triangle BDM \sim \triangle CDF$ implies $\frac{DM}{FD} = \frac{BD}{DC}$. On the other hand, $BD$ being a bisector, $\frac{BD}{DC} = \frac{AB}{AC}$, so $DM = \frac{AB}{AC} \cdot FD$. Finally, $DM = BF$.