From AB as the diameter, it can be known that ∠ AGB = 90.

And CD⊥AB, so ∠ BEF = ∠ AGB = 90,

So e, f, g and b are four * * * cycles.

(2) solution: connecting BC, which consists of four * * * circles of E, F, G and B,

So AF? AG=AE？ Ba,

In Rt△ABC, AC2=AE? Ba,

Since GF=2FA=4, AF=2 and FG=4, that is, AG=6,

So AC2=2×6,

So AC = 23.