The Answer $\ne$ 1!

Let the points and angles be labelled as shown in the diagram below:

We first note that triangle $ABD$ is isosceles with vertex angle $120^{\circ}$ , so $\angle BDA = 30^{\circ}$ , which then makes $\angle EDF = 30^{\circ}$ and $\angle CDE = 90^{\circ}$ .

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

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

In the picture above, all of the possible angle measurements that can be found via angle hunting are expressed, $\angle\phi$ and $\angle\theta$ are also expressed as shown above, the line segment, $?$ , is labelled as $x$ , and the line segment opposite of $\angle\phi$ and $\angle\theta$ is labelled as $a$ . Also, there are many ways to solve this problem, this is just the way I found the answer.

To begin solving for $x$ , we need to find an expression for it. By observation, you can find such an expression using the law of sines as such:

The law of sines states: $\frac{a}{\sin A}=\frac{b}{\sin B}$

$\therefore \frac{x}{\sin \frac{\pi}{2}}=\frac{a}{\sin \phi}$ $\Rightarrow$ $\boxed{x=\frac{a}{\sin \phi}}$

Now, again using the law of sines and observation, we can express $a$ utilizing the $\triangle$ with $\angle\theta$ and $\angle30$ as such:

$\frac{1}{\sin \frac{\pi}{6}}=\frac{a}{\sin \theta}$ $\Rightarrow$ $2=\frac{a}{\sin \theta}$ $\Rightarrow$ $\boxed{a=2\sin \theta}$

Now that we have an expression for $a$ , we can substitute $2\sin \theta$ for $a$ as such:

$x=\frac{(2\sin \theta)}{\sin \phi}$

At this moment, it is necessary to find $\theta$ in terms of $\phi$ to find a value for $x$ .

The $\triangle$ in the picture above is sufficient for this necessity because it contains both $\angle\theta$ and $\angle\phi$ . Using the property that all inner $\angle 's$ of a $\triangle$ add up to $180°$ , we can find $\theta$ in terms of $\phi$ as such:

$180°=\theta+60°+60°+\phi$ $\Rightarrow$ $180°=\theta+120°+\phi$ $\Rightarrow$ $60°=\theta+\phi$ $\Rightarrow$ $\boxed{\theta=60°-\phi}$

Now that we have an expression for \theta in terms of /phi, we can substitute $60°(or \frac{\pi}{3})-\phi$ for $\theta$ as such:

$x=\frac{2\sin (\frac{\pi}{3}-\phi)}{\sin \phi}$

At this point, the equation above has a trigonometric identity , $\sin (a-b)$ . This identity states the following:

$\sin (a-b)=\sin (a)\cos (b)-\sin (b)\cos (a)$

$\therefore x=\frac{2(\sin (\frac{\pi}{3})cos (\phi)-\sin (\phi)cos (\frac{\pi}{3})}{\sin (\phi)}$ $\Rightarrow$ $x=\frac{2(\frac{\sqrt{3}}{2})\cos (\phi)-\sin(\phi)(\frac{1}{2})}{\sin (\phi)}$ $\Rightarrow$ $x=\sqrt{3} \times {\frac{\cos (\phi)-\sin (\phi)}{\sin (\phi)}}$ $\Rightarrow$ $x=\sqrt{3} \times {\frac {\cos (\phi)}{\sin (\phi)}-1}$ $\Rightarrow$ $\boxed{x=\sqrt{3}\cot (\phi)-1}$

Now, because of the fact that we have $\cot (\phi)$ , we need to find another expression that satisfies $\cot (\phi)$ .

Luckily, in the right $\triangle$ above, $\cot (\phi)$ clearly can be expressed with side lengths as such:

$\cot (n)=\frac{adj}{opp}$

$\therefore \cot (\phi)=\frac{1}{a}$

Even though we have this expression for $\cot (\phi)$ , to solve for $x$ , we need the expression for $\cot (\phi)$ to be in terms of $x$ . This can be accomplished by utilizing the Pythagorean Theorem as such:

$a^{2}=x^{2}-1^{2}$ $\Rightarrow$ $\boxed{a=\sqrt{x^{2}-1}}$

Now that we have $\cot (\phi)$ in terms of $x$ , we can substitute $\sqrt{x^{2}-1}$ for $a$ as such:

$\cot (\phi)=\frac{1}{x^{2}-1}$

$\therefore x=\sqrt{3} \times {\frac{1}{\sqrt{x^{2}-1}}}-1$ $\Rightarrow$ $x+1=\frac{\sqrt{3}}{\sqrt{x^{2}-1}}$ $\Rightarrow$ $(x+1)(\sqrt{x^{2}-1})=\sqrt3$ $\Rightarrow$ $(x+1)^{2}(x^{2}-1)=3$ $\Rightarrow$ $(x^{2}+2x+1)(x^{2}-1)=3$ $\Rightarrow$ $x^{4}+2x{3}+x^{2}-x^{2}-2x-4=0$ $\Rightarrow$ $\boxed{x^{4}+2x^{3}-2x-4=0}$

At this point, we have a quartic polynomial , so factoring is necessary to solve for the value of $x$ . If you observe carefully, you'll notice that you can factor by grouping in this case by factoring $x^{3}$ in the first group and $-2$ in the second group as such:

$x^{3}(x+2)-2(x+2)=0$

$\therefore \boxed{(x+2)(x^{3}-2)=0}$

Now, all we have to do is set both of the factors to $0$ to solve for $x$ as such:

$x+2=0$ $\Rightarrow$ $\boxed{x=-2}$

$x^{3}-2=0$ $\Rightarrow$ $x^{3}=2$ $\Rightarrow$ $\boxed{x=\sqrt[3]{2}}$

Now we have two values for $x$ , but since $x$ represents a side length, $x\ne-2$ .

$\therefore x=\sqrt[3]{2}\approx\boxed{1.259}$

Using Zico Quintina's diagram and applying Menelaus's Theorem to triangle FCB with ADE as the transversal we can say: $\frac{CA}{AB}$ x $\frac{BD}{DF}$ x $\frac{FE}{EC}$ =1 or $\frac{2}{1}$ x $\frac{1}{y}$ x $\frac{1}{x}$ =1 or y= $\frac{2}{x}$ . Now apply Stewart's Theorem in triangle FCD to obtain: 1+x* $\frac{4}{x²}$ =(1+x)(x+x²-1) which yields, x=2^(1/3)~=1.2599