ILL 1984 Problem 4

Given a triangle $ABC$ , three equilateral triangles $AEB, BFC$ , and $CGA$ are constructed in the exterior of $ABC$ . Prove that: $(a) CE = AF = BG$ ; $(b) CE, AF$ , and $BG$ have a common point.

I could not find a separate topic for this question and I need one. http://en.wikipedia.org/wiki/Fermat_point of course.