$\min(DE+EF)=\ ?$

The shortest path is one traveled by light . So we can treat $BC$ as a mirror, if $DE$ and $EF$ are the incident and reflected beams respectively, then the angle of incident is equal to the angle of refection. We note that $D'E$ , the image of $DE$ , and $EF$ is a straight line. Then the minimum $DE+EF=D'E+EF = \sqrt{FG^2+D'G^2}$ .

Since $\triangle ABC$ is equilateral, we can work out it has a side length of $9$ , and $HI=FG=4$ , $DH=3\sqrt 3$ , and $FI=2\sqrt 3$ . We also note that $FG = DH+FI = 5\sqrt 3$ . Then $\min(DE+EF) = \sqrt{4^2 + (5\sqrt 3)^2} \approx \boxed{9.54}$ .

Let $|\overline {BE}|=a$ , position coordinates of $A, B, C, E$ be $\left (\dfrac{9}{2}, \dfrac{9\sqrt 3}{2}\right), (0,0), (9,0), (a, 0)$ respectively. Then those of $D$ and $F$ are $(3,3\sqrt 3)$ and $(7,2\sqrt 3)$ respectively. So $|\overline {DE}|=\sqrt {(3-a)^2+27}$ and $|\overline {EF}|=\sqrt {(7-a)^2+12}$ .

$|\overline {DE}|+|\overline {EF}|=\sqrt {(3-a)^2+27}+\sqrt {(7-a)^2+12}$ is optimum when it's first derivative with respect to $a$ is zero, that is, $5a^2-102a+405=0\implies a=15$ and $a=5.4$ . For $a=15$ , the sum is $\approx 21.7945$ . For $a=5.4$ , the sum is $\approx 9.53939$ . So the minimum value of the sum is $\approx \boxed {9.53939}$ .