>In Connes's proof of Morley's theorem is said that A + jB +
>j^2C = 0 is a classical characterization of equilateral
>triangles. I believe I saw this fact in Hahn's Complex Numbers & Geometry (MAA 1994).
On the page, I put an emphasis on the word classical mockingly, because, on one hand, it was the word used by Connes and, on the other, I never saw the assertion before.
>I want know where i can found one proof about this fact.
In fact, it's trivial.
Obviously, 1 + j + j2 = 0.
Therefore, we can move DABC so as to have A at the origin. The angle BAC = 60°. The orientation of DABC should be chosen in a manner that comforms with the behavior of j. Namely, j should take B "over" C, i.e. on the other side of AC, so that the angle CAB' = 60°. In this case, j2 will take C over B (in the opposite direction), so that the angle C'AB = 60°. Under those circumstances, DABC is equilateral iff B', A, C' form a straight line with A midway between B' and C'.