In the figure shown below, $ABCDE$ is a regular pentagon and $AG=1$. What is $FG + JH + CD$?

$\textbf{(A) } 3 \qquad\textbf{(B) } 12-4\sqrt5 \qquad\textbf{(C) } \dfrac{5+2\sqrt5}{3} \qquad\textbf{(D) } 1+\sqrt5 \qquad\textbf{(E) } \dfrac{11+11\sqrt5}{10}$
Triangle $AFG$ is isosceles, so $AG=AF=1$. $FJ = FG$ since $\triangle FGJ$ is also isosceles. Using the symmetry of pentagon $FGHIJ$, notice that $\triangle JHG \cong \triangle AFG$. Therefore, $JH=AF=1$.
Since $\triangle AJH \sim \triangle AFG$, 
\[\frac{JH}{AF+FJ}=\frac{FG}{FA}\]
\[\frac{1}{1+FG} = \frac{FG}1\]
\[1 = FG^2 + FG\]
\[FG^2+FG-1 = 0\]
\[FG = \frac{-1 \pm \sqrt{5} }{2}\]
So, $FG=\frac{-1 + \sqrt{5}}{2}$ since $FG$ must be greater than 0.
Notice that $CD = AE = AJ = AF + FJ = 1 + \frac{-1 + \sqrt{5}}{2} = \frac{1 + \sqrt{5}}{2}$.
Therefore, $FG+JH+CD=\frac{-1+\sqrt5}2+1+\frac{1+\sqrt5}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }$

Note by Fasolinka: Alternatively, extend $FI$ and call its intersection with $DC$ $K$. It is not hard to see that quadrilaterals $FGCK$ and $JHKD$ are parallelograms, so $DC=DK+KC=JH+FG=1+\frac{-1+\sqrt{5}}{2}$, and the same result is achieved.