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).