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 $|EF|$.

Figure 1

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.
  • $\triangle EFD \sim \triangle GEH$
  • $|EH|=2\sqrt{3}$

Figure 2

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 $CD=CE$.

enter image description here

If $M$ is midpoint of $ED$ we have $$CE^2 = ME^2+CM^2 = 1+AG^2 = 13$$

so $CE = \sqrt{13}$.

  • $\begingroup$ Extremely beautiful! +1. $\endgroup$ – Michael Rozenberg Jan 27 at 5:27
  • 1
    $\begingroup$ 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. $\endgroup$ – Mick Jan 28 at 3:54
  • $\begingroup$ 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. $\endgroup$ – C Perkins Jan 28 at 5:06
  • $\begingroup$ Yes, just that it is not. But if you read further you see $CF=CE$! @Mick @C Perkins $\endgroup$ – Aqua Jan 28 at 9:03
  • 1
    $\begingroup$ 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. $\endgroup$ – Mick Jan 28 at 17:28

Let $P$ be the perpendicular foot of $E$ to $BD$. We find that $|EP|=\sin(\angle EDP )\cdot|ED|=1$. enter image description here
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. $$ |EF|=|EC| =\sqrt{13}. $$

  • $\begingroup$ Can you post a diagram? I cannot visually follow this at all. $\endgroup$ – The Great Duck Jan 26 at 17:47
  • 1
    $\begingroup$ @TheGreatDuck I've added a figure. I hope this will help. $\endgroup$ – Song Jan 26 at 18:19
  • 1
    $\begingroup$ 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. $\endgroup$ – Henning Makholm Jan 26 at 19:57
  • $\begingroup$ @HenningMakholm Thank you for suggesting, sir. I've updated the figure. I hope this is better .. $\endgroup$ – Song Jan 26 at 22:38
  • $\begingroup$ Much nicer this way. $\endgroup$ – 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 $EF^2=1/\cos^2\alpha=1+\tan^2\alpha=13$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.