# $\triangle ABC$ and equilateral triangles $\triangle ABD$, $\triangle BCE$, $\triangle ACF$, prove that

On the sides of a certain $$\triangle ABC$$, the equilateral $$\triangle ABD$$, $$\triangle BCE$$, $$\triangle ACF$$ are drawn outside the $$\triangle ABC$$. Show that the triangles ABC and DEF have the same center of gravity.

I started to solve it in the following way: Let be $$G$$ the center of gravity of $$\triangle ABC$$. Then, I get $$\overline{GD} + \overline{GE} + \overline{GF} = \overline{GA} + \overline{AD} + \overline{GB} + \overline{BE} + \overline{GC} + \overline{CF} = \overline{AD} + \overline{BE} + \overline{CF}$$ And I have to prove that $$\overline{AD} + \overline{BE} + \overline{CF}=0$$ I constructed M the symmetrical point of E with respect to BC. I suppose that from figure, $$MC=BE$$ (that is trivial) and $$FM=AD$$. If $$FM=AD$$, then $$\overline{AD} + \overline{BE} +\overline{CF} = \overline{FM}+ \overline{MC} +\overline{CF} =0$$ and the problem is solved.

How to prove that $$FM=AD$$?

• You can prove $\Delta MCF \cong \Delta BCA \; (SAS)$ by showing $FC=AC,MC=BC,\angle FCM=\angle FCA + \angle ACM=\dfrac{\pi}{3} + \angle ACM = \angle MCB + \angle ACM = \angle ACB$. Commented Feb 12, 2022 at 17:57
• Working in the complex plane, let $\,\omega\,$ be a complex cube root of unity so that $\,\omega^3=1\,$ and $\,1+\omega+\omega^2=0\,$. Then $\,d = -\omega^2 a - \omega b\,$, and similar for $\,e,f\,$. Adding them together gives $\,d+e+f=a+b+c\,$.
– dxiv
Commented Feb 12, 2022 at 18:44
• It is a consequence of Napoleon's theorem Commented Feb 13, 2022 at 22:12

After noting that you need $$\vec{AD}+\vec{BE}+\vec{CF}=0$$ you can immediately end the proof by noting that these are the original triangle's sides rotated by $$60^\circ$$ in the same direction, so they must sum to the zero vector.