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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.


• $triangle EFD sim triangle GEH$


• $|EH|=2sqrt3$

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

• $E = (0,0)$


• $A = (-2,0)$


• $B = (-4,-2sqrt3)$


• $D = (2,0)$

Point $(x, y)$ in line $BD$ has $y=frac1sqrt3(x-2)$. 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

$$
beginbmatrix
x_1 \ y_1
endbmatrix
=
beginbmatrix
costheta & -sintheta \
sintheta & costheta
endbmatrix
beginbmatrix
x_0 \ y_0
endbmatrix
$$
, also we know that $BC$ is parallel to $x$-axis, then
$$
beginalign*
y_1
& = sin60^circ x_0 + cos60^circ y_0 \
& = sin60^circ x_0 + cos60^circ frac1sqrt3(x_0-2) \
& = -2sqrt3
endalign*
$$
, thus $F=(-frac52, -frac3sqrt32)$, and finally $|EF|=sqrt13$

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

Since $$angle EDF = 1over 2angle FCE $$ 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 $CE = sqrt13$.

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. $$
|EF|=|EC| =sqrt13.
$$

I like the following way.

Let $vecAB=veca$, $vecAD=vecb$, $vecBF=pvecBD$ and $vecBC=kvecAD.$

Thus, $$vecFE=-p(-veca+vecb)-veca+frac12vecb=(p-1)veca+left(frac12-pright)vecb$$ and
$$vecFC=-p(-veca+vecb)+kvecb=pveca+(k-p)vecb.$$

Now, we obtain the following system:
$$|vecFE|=|vecFC|$$ and
$$fracvecFEcdot vecFC=frac12$$
with variables $p$ and $k$.

We can solve this system and the rest is smooth.

Let $alpha=angle DEC$. We can apply the sine law to triangle $FED$:
$$
EDoversin(90°-alpha)=EFoversin30°=FDoversin(alpha+60°),
$$
that is:
$$
EF=1overcosalphaquadtextandquad FD=2overcosalphasin(alpha+60°).
$$
Applying then the sine law to triangle $BFC$ one gets:
$$
FB=2overcosalphasin(alpha-60°)=4sqrt3-FD=4sqrt3-2overcosalphasin(alpha+60°).
$$
From this it follows $tanalpha=2sqrt3$ and $EF^2=1/cos^2alpha=1+tan^2alpha=13$.