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 $FM=AD$?enter image description here

  • 1
    $\begingroup$ You can prove $\Delta MCF \cong \Delta BCA \; (SAS)$ by showing $FC=AC,MC=BC,\angle FCM=\angle FCA + \angle ACM=\dfrac{\pi}{3} + \angle ACM = \angle MCB + \angle ACM = \angle ACB$. $\endgroup$
    – Zerox
    Commented Feb 12, 2022 at 17:57
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    $\begingroup$ 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 $\,d+e+f=a+b+c\,$. $\endgroup$
    – dxiv
    Commented Feb 12, 2022 at 18:44
  • $\begingroup$ It is a consequence of Napoleon's theorem $\endgroup$
    – Jean Marie
    Commented Feb 13, 2022 at 22:12

1 Answer 1


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.


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