# Three equilateral triangles

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.

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?

• 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. – Orange Mushroom Jan 20 at 6:49

## 3 Answers

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.

• A somewhat similar issue was posted here shortly after this one and I had, like you, the idea to use complex numbers... – Jean Marie Jan 21 at 23:06

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.

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.