Answer: (1) Proof: Because AB is the diameter of the circle O, BF⊥FH, …(2 points)

And DH⊥BD,

So the four points B, D, F and H are all on the circle with BH as the diameter.

So B, D, F and H are four * * * cycles ... (4 points)

(2) Solution: Because AH and circle B are tangent to point F,

AF2=AC from tangent theorem? AD, that is, (22)2=2? AD，

AD=4, …(6 points)

So BD= 12(AD? AC)= 1，BF=BD= 1，

And △AFB∽△ADH,

So DHBF=ADAF, DH=2, …(8 points)

Connect BH. From (1), BH is the diameter of the circumscribed circle of DBDF.

BH=BD2+DH2=3，

Therefore, the circumscribed circle radius of △BDF is 32...( 10 minute).