Napoleon's theorem

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 l_aA , l_bB , and l_cC 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 l_A'A, l_B'B, and l_C'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.