Dtyjlui

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.


• $triangle EFD sim triangle GEH$


• $|EH|=2sqrt{3}$

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

• $E = (0,0)$


• $A = (-2,0)$


• $B = (-4,-2sqrt{3})$


• $D = (2,0)$

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
$$
begin{align*}
y_1
& = sin{60^{circ}} x_0 + cos{60^{circ}} y_0 \
& = sin{60^{circ}} x_0 + cos{60^{circ}} frac{1}{sqrt{3}}(x_0-2) \
& = -2sqrt{3}
end{align*}
$$
, thus $F=(-frac{5}{2}, -frac{3sqrt{3}}{2})$, 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

Since $$angle EDF = {1over 2}angle FCE $$ 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 $CE = sqrt{13}$.

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. $$
|EF|=|EC| =sqrt{13}.
$$

I like the following way.

Let $vec{AB}=vec{a}$, $vec{AD}=vec{b}$, $vec{BF}=pvec{BD}$ and $vec{BC}=kvec{AD}.$

Thus, $$vec{FE}=-p(-vec{a}+vec{b})-vec{a}+frac{1}{2}vec{b}=(p-1)vec{a}+left(frac{1}{2}-pright)vec{b}$$ and
$$vec{FC}=-p(-vec{a}+vec{b})+kvec{b}=pvec{a}+(k-p)vec{b}.$$

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.

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