Proving this theorem was a class assignment. My proof may or may not be original (probably not).

Theorem (BONAPARTE)
Let ABC be a triangle. Erect equilateral triangles A'BC, AB'C, ABC' so either all overlap with Δ ABC or none overlap, with centroids a, b, c respectively. Then Δ abc is an equilateral triangle (known as Napoleon's triangle).
```                                          _.B'
_ ./  |
_ ./      /
_ ./           |
_ ./               /
C'________________A /                    |
\                /\             .b     /
\              /   \                  |
\     .c     /      \               /
\          /         \             |
\        /            \          /
\      /               \        |
\    /                  \     /
\  /                     \   |
\/________________________\/
B \                        /C
\                      /
\                    /
\                  /
\       .a       /
\              /
\            /
\          /
\        /
\      /
\    /
\  /
\/
A'
```
Proof
Make two comparisons between Δ cAb and Δ BAB' to see that they are similar triangles:
cAb = ∠ cAB + ∠ BAC + ∠ CAb = π/6 + ∠ BAC + π/6 = ∠ BAC + π/3
BAB' = ∠ BAC + ∠ CAB' = ∠ BAC + π/3
(same angles)
By the fixed ratios of lengths in an equilateral triangle:
|c - A| = |B - A| / sqrt(3)
|b - A| = |B' - A| / sqrt(3)
∴ |c - A| / |b - A| = |B - A| / |B' - A|
(preserved ratio of side lengths)
They are similar triangles. Hence the third length must have the same ratio:
|c - b| = |B' - B| / sqrt(3)
By reasons of symmetry,
|a - b| = |B' - B| / sqrt(3)
Hence |a - b| = |b - c| and by reasons of symmetry, all sides of Δ abc are of equal length.

This proof works for the case where the constructed equilateral triangles point inward, too.

QED

Napoleon Points
Lines laA , lbB , and lcC meet at the first Napoleon point when the three equilateral triangles point away from Δ ABC. When they point toward Δ ABC, the lines meet at a different point known as the second Napoleon point.

Fermat Point
The point P of a triangle ABC that minimizes the sum of distances |A - P| + |B - P| + |C - P| is known as the Fermat point. When each of the three angles of Δ ABC are less than 120°, the Fermat point is the concurrent point of lines lA'A, lB'B, and lC'C. It is also where the circumcircles of the erected equilateral triangles meet. When one of the vertices has an angle greater than 120°, then that vertex is the Fermat point. The Fermat point was not discovered by Fermat himself. Instead, he was the one who asked where such a point would exist, to which Toricelli was the first to answer this question.