A beautiful construction

We first note that triangle $AED$ is isosceles with vertex angle $120^{\circ}$ , so $\angle EDA = 30^{\circ}$ , which then makes $\angle FDG = 30^{\circ}$ and $\angle CDF = 90^{\circ}$ .

Since triangle $CDF$ is right, we know that $\sin \theta = \dfrac{1}{x}$ . Also, applying Sine Law to triangle $DFG$ , we get

$\begin{aligned} \dfrac{\sin 30^{\circ}}{1} &= \dfrac{\sin \alpha}{y} \\ \\ \dfrac{1}{2} &= \dfrac{\sin \theta}{y} \qquad \small \text{[ Because } \theta = 180^{\circ} - \alpha \ ] \\ \\ \dfrac{1}{2} &= \dfrac{1}{x y} \qquad \small \text{[ Using } \sin \theta = \dfrac{1}{x} \ ] \\ \\ y &= \dfrac{2}{x} \end{aligned}$

Now we use Cosine Law on triangle $CDG$ :

$\begin{aligned} (x + 1)^2 &= 1^2 + y^2 - \ 2 \ (1) (y) \cos 120^{\circ} \\ \\ x^2 + 2x + 1 &= 1 + \dfrac{4}{x^2} + \dfrac{2}{x} \qquad \small \text{[ Now multiply by } x^2 \ ] \\ \\ x^4 + 2x^3 + x^2 &= x^2 + 2x + 4 \\ \\ x^4 + 2x^3 &= 2x + 4 \\ \\ x^3 (x + 2) &= 2 (x + 2) \qquad \small \text{[ Since } x \text{ cannot be -} 2 \text{, we can divide by } (x + 2) \ ] \\ \\ x^3 &= 2 \\ \\ x &= \sqrt[3]{2} \approx \boxed{1.259} \end{aligned}$

By Menelaus' Theorem considering the transversal $ADG$ of the triangle $CEF$ , $\frac{CA}{AE} \times \frac{ED}{DF} \times \frac{FG}{GC} \; = \; -1$ (here lengths are signed). Thus $1 \times \frac{\sqrt{3}}{\sqrt{CF^2-1}} \times \frac{1}{CF+1} \; = \; 1$ and hence $\sqrt{CF^2-1}(CF+1) = \sqrt{3}$ , and so $CF$ is a positive real root of the equation $\begin{aligned} (x^2-1)(x+1)^2 - 3 & = \; 0 \\ x^4 + 2x^3 - 2x - 4 & = \; 0 \\ (x+2)(x^3 - 2) & = \; 0 \end{aligned}$ and so $CF= \boxed{\sqrt[3]{2}}$ .

$Let~DF=X.\\ ~~~~\\ From~the~give~data,\\ \Delta~CAD~is~~1-1-1, ~\therefore~~\angle s~~ CAD=ADC=60^o.\\ So~\angle~DAE=120^o, ~~but~AE=AD,\\ \therefore~in~\Delta~AED,~~\angle~FDG=EDA=30^o,~~and~~\angle~CDF=90^o.\\ ~~~~\\ rt.~\angle~\Delta~CDF,~~\angle~FCD=atanX.\\ ~~~~\\ Using~Sine~Law~in~\Delta~FDG, ~1/sin30=X/sinDGF,\\ \therefore~\angle~DGF=asin(X/2).\\ ~~~~\\ In~\Delta~CDG,~~\angle s ~DGF+FCD=180-120=60^o\\ Solving~asin(X/2)+atanX=60^o~ by~ trial~ and ~error,\\ we~get~X=.766.\\ ~~~~\\ \implies~CF=\sqrt{1^2+.766^2}=\huge \color{#D61F06}{1.25966}.$

Letting $CF=x$ , and using the Law of Sines, we solve for $x$

$\dfrac { Sin( ArcCsc(x) -\frac{\pi}{6} ) } { \sqrt{x^2-1} } = \dfrac { Sin(\frac{\pi}{6}) } {1}$

which delivers $x=1.25992$

See the diagram.

Considering that $\angle CDF=\dfrac{\pi}{2}$ , $\sin \angle DFC=\dfrac{CD}{CF}=\dfrac{1}{x}$

Let $DG=y$ . Employing the sine law, we see that for triangle $DFG$ :

$\dfrac{\sin\dfrac{\pi}{6}}{1}=\dfrac{\sin \angle DFG}{y}$

$\dfrac{1}{2}=\dfrac{\sin (\pi-\angle DFC}{y}$

$\dfrac{1}{2}=\dfrac{\sin \angle DFC}{y}$

$\dfrac{1}{2}=\dfrac{\dfrac{1}{x}}{y}$

$\dfrac{1}{2}=\dfrac{1}{xy}$

$xy=2$

Employing the cosine law, we see that for triangle $CDG$ :

$(x+1)^2=1^2+y^2-2(1)(y) \cos \dfrac{2\pi}{3}$

$x^2+2x+1=1+y^2-2y\left(-\dfrac{1}{2}\right)$

$x^2+2x=y^2+y$

$x(x+2)=y(y+1)$

$x(x+xy)=y(y+1)$

$xx\cancel{(1+y)}=y\cancel{(y+1)}$

$x^2=y$

$xx^2=xy$

$x^{1+2}=2$

$x^3=2$

$x=\sqrt[3]{2}$

With some angle chasing we can find out that $\angle BFG = 30 °$ and $\angle CDG = 120 °$ Let $DF = y$ and $CF = x$

Observing that $\angle CDF = 90 °$ we know from pytaghorean theorem that

$1^2 + y^2 = x^2$ and $x>y>0$

Now, from law of sines in $\triangle FDG$

$\frac{y}{\sin(\angle FGD)} = \frac{1} {\sin(30°)}$ $\therefore$ $y = 2 \times \sin(\angle FGD)$

Then, law of sines in $\triangle CDG$

$\frac{1}{\sin(\angle FGD)} = \frac{1+x}{\sin(120° )} \therefore x = \frac{1}{2} \sqrt{3} \csc(\angle FGD) - 1$

putting al together and solving those equations we get the answer $\sqrt[3]{2}$