Is there a possible geometric method to find length of this equilateral triangle? 
Problem Given that $AD \parallel BC$, $|AB| = |AD|$, $\angle A=120^{\circ}$, $E$ is the midpoint of $AD$, point $F$ lies on $BD$, $\triangle EFC$ is a equilateral triangle and $|AB|=4$, find the length $|EF|$.


Attempt At first glance, I thought it could be solved using a geometric method. I considered the law of sines/cosines, similar triangles, Pythagorean theorem, even Menelaus' theorem, however, got properties which contributed nothing to calculate $|EF|$.
What I've got after draw a line perpendicular to $BC$ through $E$


*

*$\triangle ABH$ and $\triangle AHD$ are both equilateral triangles of length 4.

*$\triangle EFD \sim \triangle GEH$

*$|EH|=2\sqrt{3}$

Algebraic method Eventually, I've changed my mind to embrace algebra. I found it is easy to coordinate $E,A,B,D$ and $C$ is related to $F$ (rotation) and $B$ (same horizontal line). Make $E$ as the origin, $AD$ points to $x$-axis, $HE$ points to $y$-axis, we got


*

*$E = (0,0)$

*$A = (-2,0)$

*$B = (-4,-2\sqrt{3})$

*$D = (2,0)$
Point $(x, y)$ in line $BD$ has $y=\frac{1}{\sqrt{3}}(x-2)$. Assume $F=(x_0,y_0)$, $C=(x_1,y_1)$, we can obtain $C$ by rotating $F$ around pivot $E$ $60^{\circ}$ counter-clockwise
$$
\begin{bmatrix}
x_1 \\ y_1
\end{bmatrix} 
= 
\begin{bmatrix}
\cos{\theta} & -\sin{\theta} \\
\sin{\theta} & \cos{\theta} 
\end{bmatrix}
\begin{bmatrix}
x_0 \\ y_0
\end{bmatrix} 
$$
, also we know that $BC$ is parallel to $x$-axis, then
$$
\begin{align*}
y_1 
 & = \sin{60^{\circ}} x_0 + \cos{60^{\circ}} y_0 \\
 & = \sin{60^{\circ}} x_0 + \cos{60^{\circ}} \frac{1}{\sqrt{3}}(x_0-2) \\
 & = -2\sqrt{3}
\end{align*}
$$
, thus $F=(-\frac{5}{2}, -\frac{3\sqrt{3}}{2})$, and finally $|EF|=\sqrt{13}$
Thoughts afterwords I noticed that $F$ (through its coordinate) is actually the midpoint of $BK$. It may be a key point in geometric method, but I cannot prove it either.
Graph I made it in GeoGebra and it is shared. Please go and edit it to save your time if you have any idea. 
Link: https://www.geogebra.org/graphing/yqhbzdem
 A: I like the following way.
Let $\vec{AB}=\vec{a}$, $\vec{AD}=\vec{b}$, $\vec{BF}=p\vec{BD}$ and $\vec{BC}=k\vec{AD}.$
Thus, $$\vec{FE}=-p(-\vec{a}+\vec{b})-\vec{a}+\frac{1}{2}\vec{b}=(p-1)\vec{a}+\left(\frac{1}{2}-p\right)\vec{b}$$ and
$$\vec{FC}=-p(-\vec{a}+\vec{b})+k\vec{b}=p\vec{a}+(k-p)\vec{b}.$$
Now, we obtain the following system:
$$|\vec{FE}|=|\vec{FC}|$$ and
$$\frac{\vec{FE}\cdot \vec{FC}}{|\vec{FE}||\vec{FC}|}=\frac{1}{2}$$
with variables $p$ and $k$.
We can solve this system and the rest is smooth.
A: Let $\alpha=\angle DEC$. We can apply the sine law to triangle $FED$:
$$
{ED\over\sin(90°-\alpha)}={EF\over\sin30°}={FD\over\sin(\alpha+60°)},
$$
that is:
$$
EF={1\over\cos\alpha}\quad\text{and}\quad FD={2\over\cos\alpha}\sin(\alpha+60°).
$$
Applying then the sine law to triangle $BFC$ one gets:
$$
FB={2\over\cos\alpha}\sin(\alpha-60°)=4\sqrt3-FD=4\sqrt3-{2\over\cos\alpha}\sin(\alpha+60°).
$$
From this it follows $\tan\alpha=2\sqrt3$ and $EF^2=1/\cos^2\alpha=1+\tan^2\alpha=13$.
A: Since $$\angle EDF = {1\over 2}\angle FCE $$ we see that $D$ is on a circle with center at $C$ and radius $CE =CF$ so $CD=CE$.  

If $M$ is midpoint of $ED$ we have $$CE^2 = ME^2+CM^2 = 1+AG^2 = 13$$
so $CE = \sqrt{13}$.
A: Let $P$ be the perpendicular foot of $E$ to $BD$. We find that $|EP|=\sin(\angle EDP )\cdot|ED|=1$.

 We also find that $\triangle EPF$ is congruent to $\triangle CHE\ $ implying that
$$
|EP|=|CH|=1.
$$ By Pythagorean theorem, it follows
$$
|EC|^2 =|EH|^2 +|CH|^2 =13,
$$i.e. $$
|EF|=|EC| =\sqrt{13}.
$$
A: This can be solved in your imagination.  It takes a lot of words to describe it, but you don't need these words when you imagine it.
Imagine moving $F$ back and forth along $BD$, while holding $E$ fixed, so $C$ (defined as the third point of the equilateral triangle) moves around.  $C$ is always a 60° counter-clockwise rotation of $F$ (rotating around $E$), so the set of points visited by $C$ is a 60° counter-clockwise rotation of $BD$ (around $E$).  So $C$ moves vertically.  When $F$ is at $D$, then the equilateral triangle is small and $C$ is above the midpoint of $DE$.  So we see that $C$ is always on the perpendicular bisector of $DE$.
Now we return to the diagram as shown.  The distance between $AD$ and $BC$ is $\sqrt{4^2-2^2=12}$, and since half of $ED$ is 1, we have EF$\;=\;$EC$\;=\sqrt{12+1^2=13}$.
