AB = AC，

∴∠B=∠ACB.

OC = OE，

∴∠OEC=∠ACB.

∴∠B=∠OEC，

∴OE∥AB.

∵AB⊥GF，

∴OE⊥GF.

∵ point e is on ⊙O,

∴ The straight line FG is tangent to⊙ O. 。

(2) Let the radius of ⊙O be r, then OE=r and AB = AC = 2r.

∫BF = 1，CG=2，

∴AF=2r- 1,OG=r+2,AG=2r+2.

∫OE∨AB，

∴△GOE∽△GAF，

∴OEAF=OGAG，

∴r2r？ 1=r+22r+2，

The solution is r=2,

That is, the radius of ⊙O is 2.