(2) Even for OA, we get ∠ bad = 90 according to the inference of the theorem of the circle angle, and then calculate BD with Pythagorean theorem, and get ∠ d = 30, and it is easy to get that △OAB is an equilateral triangle, so we have AB=BF=BO. According to the inference of the theorem of circle angle, we get that △OAF is a right triangle, that is △ OAF.
Solution: solution: (1)∵AB=AC,
∴∠ABC=∠ACB，
∠∠C and∠ D are the angles of the circle opposite to the same arc,
∴∠C=∠D，
∴∠ABE=∠D，
And ∠BAE=∠DAB,
∴△BAE∽△DAB，
∴ AB: AD = AE: AB, that is, AB2=AD? AE，
AE = 2，ED = 4。
∴AD=6，
∴AB2=2×6= 12，
∴AB=2
three
(2) The straight line FA is tangent to ≧O for the following reasons:
Even OA, as shown in the figure,
∫BD is the diameter,
∴∠BAD=90，
∴BD=
AB2+AD2
=
(2
three
)2+62
=4
three
∴∠D=30，
∴∠AOB=60，
△ OAB is an equilateral triangle,
∴AB=BO，
BF = BO，
∴AB=BF=BO，
∴∠ABO=∠AOB=60，∠F=∠FAB，
∴∠F=∠FAB=
1
2 ∠ABO=30，
∴∠OAF=∠FAB+∠BAO=90，
∴ The straight line AF is the tangent of⊙ O. 。