Show that $CD\parallel AB$ with square $FEBD$ As the figure shows, $FEBD$ is a squre. $AE=GE$, $FA=FB$ and $CD=BA$. Show that $CD\parallel AB$.

I have found that $\triangle BGD$ must be an equilateral triangle, but I have no proof yet. Please help. It's better to offer a synthetical solution without much computation. Thanks in advance. 

Let $FEBD$ be the square with vertices $F=(-1,0)$, $E=(0,1)$, $B=(1,0)$, $D=(0,-1)$ and $A=(a,b)$ a point such that $AF = BF$.
$AF=BF$
By the formula of distance between two points
$\begin{array}{} (a+1)^{2}+b^2=4 & ⇒ & b^{2}=4-(a+1)^{2} & (I) \end{array}$
$\begin{array}{} CD∥AB & ⇒ & m_{CD}=m_{AB} \end{array}$
$\frac{y_{D}-y_{C}}{x_{D}-x_{C}}=\frac{y_{B}-y_{A}}{x_{B}-x_{A}}$
$\begin{array}{} x_{C}=\frac{a-1}{b} & (II)  \end{array}$
$\begin{array}{} CD=BA & C∈BF &⇒ & y_{C}=0 \end{array}$
$\begin{array}{} x_{C}^{2}=3-4·a & (III)  \end{array}$
$\begin{array}{} \text{From (I) (II) (III)} & \frac{(a-1)^{2}}{b^{2}}=3-4·a  \end{array}$
$\frac{(a-1)^{2}}{4-(a+1)^{2}}=3-4·a$
$\begin{array}{} \text{solutions: } & a:\left( \begin{array}{} -1+\sqrt{3} \\ -1-\sqrt{3} \end{array}   \right)   \end{array}$
$AB=CD=\sqrt{6}-\sqrt{2}$
$\begin{array}{} x_{C}=-2+\sqrt{3} & y_{C}=0 \\ x_{G}=\frac{1-\sqrt{3}}{2} & y_{G}=\frac{\sqrt{3}-1}{2} \\ \end{array}$
$med(\angle{AFB})=\frac{ π }{6}$
$\begin{array}{} BG=DG=BD=\sqrt{2} & ⇒ & \triangle{BDG} & \text{ is an equilateral triangle} \end{array}$

