# 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

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 $$CE = \sqrt{13}$$.

• Extremely beautiful! +1. – Michael Rozenberg Jan 27 at 5:27
• I think the condition "$\angle EDF = {1\over 2}\angle FCE$" is NOT strong enough to guarantee D lies on the same circle that is centered at C with radius = CF. – Mick Jan 28 at 3:54
• At least explain why the first assertion is true. The explanation takes care to list the obvious radius equality, but doesn't even explain the logic of the circle. – C Perkins Jan 28 at 5:06
• Yes, just that it is not. But if you read further you see $CF=CE$! @Mick @C Perkins – Aqua Jan 28 at 9:03
• That clarifies one thing. Another requirement we need to say is both C and D should be on the same side of the line EF, but the original diagram clearly showed they are. – Mick Jan 28 at 17:28

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}.$$

• Can you post a diagram? I cannot visually follow this at all. – The Great Duck Jan 26 at 17:47
• @TheGreatDuck I've added a figure. I hope this will help. – Song Jan 26 at 18:19
• This makes sense (+1), but it would help if the diagram didn't show quite so many unneeded points and lines. CD is useless-but-distracting even though mentioned in the problem, and the intersection points I, J, K, L, M, N, O are completely irrelevant. – Henning Makholm Jan 26 at 19:57
• @HenningMakholm Thank you for suggesting, sir. I've updated the figure. I hope this is better .. – Song Jan 26 at 22:38
• Much nicer this way. – Henning Makholm Jan 26 at 23:49

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.

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