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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 $|EF|$.

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

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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 $CE = \sqrt{13}$.

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  • $\begingroup$ Extremely beautiful! +1. $\endgroup$ – Michael Rozenberg Jan 27 at 5:27
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    $\begingroup$ 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
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    $\begingroup$ 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
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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. $$ |EF|=|EC| =\sqrt{13}. $$

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  • $\begingroup$ Can you post a diagram? I cannot visually follow this at all. $\endgroup$ – The Great Duck Jan 26 at 17:47
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    $\begingroup$ @TheGreatDuck I've added a figure. I hope this will help. $\endgroup$ – Song Jan 26 at 18:19
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    $\begingroup$ 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
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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.

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

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