Equilateral triangle circumscribed to ABC

1) Let Fe be a Fermat point of ABC.
2) Take A' in the circle {FeBC}, in the external arc with respect to the triangle.
3) The line CA' cuts the circle {FeCA} at C and B'.
4) Let C' be the second intersection between BA' and the circle {FeAB}. A'B'C' is the requested triangle.