Triangle $ABC$ is equilateral with $AB=1$. 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 $\overline{BC}$. Let $P_1$ be the perimeter of triangle $ADE$, $P_2$ be the perimeter of trapezoid $DFGE$, and $P_3$ be the perimeter of trapezoid $FBCG$. If $P_1:P_2:P_3=1:2:3$, what is $DE+FG$?