imc

Triangle $AFG$ is isosceles, so $AG=AF=1$. $FJ=FG$ since $△FGJ$ is also isosceles. Using the symmetry of pentagon $FGHIJ$, notice that $△JHG\cong △AFG$. Therefore, $JH=AF=1$. Since $△AJH\sim △AFG$, $$\frac{JH}{AF+FJ}=\frac{FG}{FA}$$ $$\frac{1}{1+FG}=\frac{FG}{1}$$ $$1=F{G}^2+FG$$ $$F{G}^2+FG-1=0$$ $$FG=\frac{-1\pm \sqrt{5}}{2}$$ 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, $FG+JH+CD=\frac{-1+\sqrt{5}}{2}+1+\frac{1+\sqrt{5}}{2}=\boxed{1+\sqrt{5}}$ 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.