Triangle $ABC$ is equilateral with $AB=12$. Points $E$ and $G$ are on $\overline{AC}$ and points $D$ and $F$ are on $\overline{AB}$ such that both $\overline{DE}$ and $\overline{FG}$ are parallel to $\overline{BC}$. Triangle $ADE$ and trapezoid $DFGE$ have the same perimeter, and trapezoid $FBCG$ has perimeter $28$. What is $DE+FG$?

Let $AD=x$ and $AF=y$. Since $DE\parallel FG\parallel BC$, triangles $ADE$ and $AFG$ are equilateral, so $DE=AD=AE=x$ and $FG=AF=AG=y$.

1. The perimeter of triangle $ADE$ is $3x$.
2. In trapezoid $DFGE$, $DF=EG=y-x$, so its perimeter is $2(y-x)+x+y=3y-x$.
3. In trapezoid $FBCG$, $FB=GC=12-y$, so its perimeter is $2(12-y)+12+y=36-y$.
4. From $3x=3y-x$ we get $y=\frac43x$. From $36-y=28$ we get $y=8$, hence $x=6$.
5. $DE+FG=x+y=6+8=\boxed{14}$.