∵AD⊥BC,EF⊥BC,OM⊥BC，

∴AD∥OM∥EF，

OA = OE，

∴DM=MF, so CM-DM=BM-MF, which means BF = CD.

(2) Solution: If BE is connected, then ∠ Abe = 90;

In Rt△ABD, AD=3 and BD=6, which is obtained by Pythagorean theorem:

AB = AD2+BD2 = 35；

Similarly, AC = 10.

∠∠C =∠AEB，∠ADC=∠ABE=90，

∴△ADC∽△ABE，

∴ ADAB = ACAE, that is, 335 = 10ae, and the solution is ae = 52.

That is, the diameter ⊙O is 52.