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Can you provide a proof for the following claim:

Claim. Given an arbitrary equilateral triangles $\triangle ABC$ and $\triangle BDE$ with common vertex $B$ . The points $H$ and $I$ divide line segments $CE$ and $AD$ respectively in the ratio $2 : 1$ . Let $G_1$ be the centroid of triangle $\triangle ABC$ . Then triangle $\triangle G_1IH$ is an equilateral triangle as well.

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GeoGebra applet that demonstrates this claim can be found here.

So far I have managed to prove that $|AE|=|CD|$ and that line segments $AE$ and $CD$ intersect each other at angle of $60^{\circ}$ , but I dont know if this can be of any use. Also, can we apply the fundamental theorem of similarity in some way?

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  • $\begingroup$ What you've proven is enough! All that needs to be done is a simple modification and you will have your result. I'll elaborate. $\endgroup$ – Orange Mushroom Jan 20 at 6:49
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Use complex numbers. Put $G_1=0$, $B=1$ and $BD=z$ (so $BE=ze^{i\pi/3}$). Then, letting $w=e^{i\pi/3}$, $$H=\frac13(w^2+2(1+zw))=\frac13(w^2+2zw+2)$$ $$I=\frac13(w^4+2(1+z))=\frac13(w^4+2z+2)$$ But $w^2+2=w(w^4+2)$, so $$H=w\cdot\frac13(w^4+2z+2)=wI$$ Since $|w|=1$ and $\arg w=60^\circ$, $G_1H=G_1I$ and $\angle HG_1I=60^\circ$, so $\triangle HG_1I$ is equilateral.

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  • $\begingroup$ A somewhat similar issue was posted here shortly after this one and I had, like you, the idea to use complex numbers... $\endgroup$ – Jean Marie Jan 21 at 23:06
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Here's a synthetic solution. You have already proven the required facts.

Let $X, Y$ be the midpoints of $\overline{BC}, \overline{BA}$, respectively. Then $\triangle XBY$ is equilateral. Apply your results to $\triangle XBY$ and $\triangle EBD$ to obtain

  1. $\angle(\overline{XD}, \overline{YE}) = 60^{\circ}$.
  2. $XD = YE$.

Now here's the critical point of the argument: $\triangle CYE \sim \triangle CG_1H$ and $\triangle AXD \sim \triangle AG_1I$. Thus $\overline{YE} \parallel \overline{G_1H}$ and $\overline{XD} \parallel \overline{G_1I}$. Furthermore, $G_1H = \lambda YE$ and $G_1I = \lambda XD$ for a scaling factor $\lambda = 2/3$. This, all together, shows

  1. $\angle(\overline{G_1I}, \overline{G_1H}) = 60^{\circ}$.
  2. $G_1H = G_1I$.

We're done.

enter image description here

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Three equilateral triangles

Let $L$ and $M$ be the midpoints of side $AB$ and $BC$ respectively. Draw medians $AM$ and $CL$.The point at which they intersect will be the centroid of $\triangle ABC$, point $G$. Draw $GH$, $LE$, $GI$ and $MD$. Let the points at which $GH$ and $LE$ intersect $MD$ be $P$ and $Q$ respectively.Let the point at which $GI$ intersect $LE$ be $R$. Draw $BQ$.

Observe that, $\triangle MDB\cong \triangle LBE$. $\Rightarrow LE=MD$

Also, observe that, since $\frac {CG}{GL}=\frac {CH}{HE}=\frac {2}{1}$, $GH\parallel LE$ and $GH=\frac 23LE$. Similarly, $GI\parallel MD$ and $GI=\frac 23MD$.

$LE=MD$ $\Rightarrow GH=GI$

Quadrilateral $GPQR$ is a parallelogram since its opposite sides are parallel.

$\Rightarrow \angle HGI=\angle PGR=\angle PQR=\angle EQD$

Notice that, in quadrilateral $QEDB$, $\angle QEB=\angle LEB=\angle MDB=\angle QDB$; Hence, $\angle EQD=\angle EBD=60^{\circ}$.

$\Rightarrow \angle HGI=\angle EQD=60^{\circ}$

$HG=GI$ and $\angle HGI=60^{\circ}$ ; Therefore, $\triangle HGI$ is equilateral.

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