Is there a possible geometric method to find length of this equilateral triangle?

Question

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

Answer 1

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 $CE = \sqrt{13}$.

Answer 2

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.

Answer 3

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

Answer 4

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 $EF^2=1/\cos^2\alpha=1+\tan^2\alpha=13$.

Answer 5

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$\;=\sqrt{12+1^2=13}$.