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 $\overline{BC}$. Furthermore, triangle $ADE$ and trapezoids $DFGE$ and $FBCG$ all have the same perimeter. What is $DE+FG$?

$\textbf{(A) }1\qquad \textbf{(B) }\dfrac{3}{2}\qquad \textbf{(C) }\dfrac{21}{13}\qquad \textbf{(D) }\dfrac{13}{8}\qquad \textbf{(E) }\dfrac{5}{3}\qquad$
Let $AD = x$, and $AG = y$. We want to find $DE + FG$, which is nothing but $x+y$.
Based on the fact that $ADE$, $DEFG$, and $BCFG$ have the same perimeters, we can say the following:
$3x = x + 2(y-x) + y = y + 2(1-y) + 1$
Simplifying, we can find that
$3x = 3y-x = 3-y$
Since $3-y = 3x$, $y = 3-3x$.
After substitution, we find that $9-10x = 3x$, and $x$ = $\frac{9}{13}$.
Again substituting, we find $y$ = $\frac{12}{13}$.
Therefore, $x+y$ = $\frac{21}{13}$, which is $C$