Napoleon's Theorem,
A second proof with complex numbers

On each side of a triangle, erect an equilateral triangle, lying exterior to the original triangle. Then the segments connecting the centroids of the three equilateral triangles themselves form an equilateral triangle.

Let the original triangle be $ABC\;$ with equilateral triangle $ABC_{1},\;$ $CAB_{1},\;$ and $BCA_{1}\;$ built on its sides. Think of all the vertices involved as complex numbers. We shall apply a classical criterion to the three equilateral triangles. Let $j\;$ be a suitable rotation through $120°.\;$ Then the fact that triangles $ABC_{1},\;$ $CAB_{1},\;$ and $BCA_{1}\;$ are equilateral may be expressed as

$\begin{align} (1) & A + jB + j^{2}C_{1} = 0\\ (2) & C + jA + j^{2}B_{1} = 0\\ (3) & B + jC + j^{2}A_{1} = 0 \end{align}$

The center of $\Delta ABC_{1}\;$ is given by $P = (A + B + C_{1})/3,\;$ and similarly for centers $Q\;$ and $R\;$ of triangles $CAB_{1}\;$ and $BCA_{1}:\;$ $Q = (C + A + B_{1})/3\;$ and $R = (B + C + A_{1})/3.\;$ We want to show that $P + jQ + j^{2}R = 0.\;$ Indeed,

$\begin{align} 3(P + jQ + j^{2}R) &= A + B + C_{1} + j(C + A + B_{1}) + j^{2}(B + C + A_{1})\\ &= (B + jC + j^{2}A_{1}) + j(A + jB + j^{2}C_{1}) + j^{2}(C + jA + j^{2}B_{1})\\ &= 0. \end{align}$

Napoleon's Theorem

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