Solution: (1) af = be. It is proved that in △AFC and △BEC, ∫△ABC and △CEF are equilateral triangles, ∴AC=BC, CF=CE, ∠ ACF = ∠ BCE = 60, which are in △AFC.
AC=BC
∠ACF=∠BCE
CF=CE
∴△AFC≌△BEC(SAS)，
∴AF=BE.
2) It holds. Reason: In △AFC and △BEC, ∫△ABC and △CEF are equilateral triangles, ∴AC=BC, CF=CE, ∠ACB=∠FCE=60 degrees, ∴AC b-∞.