smc

Let the points be labeled as in the new diagram above, with $BF=6$ and $CG=4$ (from the problem). Because $\overline{BC}\parallel \overline{FG}$ and $\frac{BC}{FG}=\frac{1}{3}$, $AB=\frac{AF}{3}=\frac{AB+6}{3}$ and $AC=\frac{AG}{3}=\frac{AC+4}{3}$. Solving these equations for $AB$ and $AC$, respectively yields $AB=3$ and $AC=2$. Let $DE=x$. Thus, because $\overline{DE}\parallel \overline{BC}$, $\frac{BC}{DE}=\frac{3}{x}=\frac{AB}{AD}=\frac{3}{3+BD}$. Solving this equation for $BD$ yields $BD=x-3$. Similarly, $\frac{3}{x}=\frac{AC}{AE}=\frac{2}{2+CE}$, and solving this equation for $CE$ yields $CE=\frac{2x}{3}-2$. Now, we can set the perimeters of $BCED$ and $DEGF$ equal to each other to solve for $x$: $$\begin{array}{rl}3+(x-3)+x+(\frac{2x}{3}-2)& =x+[6-(x-3)]+9+[4-(\frac{2x}{3}-2)]\\ \frac{8x}{3}-2& =9+9+4-\frac{2x}{3}+2\\ \frac{10x}{3}& =26\\ \frac{5x}{3}& =13\\ 5x& =39\\ x& =\frac{39}{5}\end{array}$$ To find the ratio $\frac{BD}{DF}=\frac{x-3}{6-(x-3)}$, we substitute $x=\frac{39}{5}$ into this expression to find our answer: $$\begin{array}{rl}\frac{x-3}{6-(x-3)}& =\frac{x-3}{9-x}\\ & =\frac{\frac{39-15}{5}}{\frac{45-39}{5}}\\ & =\frac{24}{6}\\ & =\frac{4}{1}\end{array}$$ Thus, our answer is $\boxed{4:1}$.