Prove:

( 1)

AB = AC

∴∠Abe =∞∠ADB (in the same circle, the chords are equal and the circumferential angles are equal)

∵∠BAE =∠ It's broken again.

∴△ABE∽△ADB

∴AE:AB=AB:AD

Namely AB? =AE×AD

AB？ =2×(2+4)

AB？ = 12

The solution is AB=2√3 or AB=-2√3 (negative numbers are irrelevant and omitted).

∴AB=2√3

(2) The straight line FA is tangent to ≧O for the following reasons:

Connecting OA

∵BD is the diameter of⊙ O.

∴∠ Difference = 90 (the circumferential angle of the diameter is 90).

∴△Abd in Rt, AB=2√3, AD=AE+ED=2+4=6.

BD=√AB？ +AD？ =√48=4√3 (Pythagorean theorem)

∴AB=2√3= 1/2BD=BO

∴AB=BO=BF= 1/2EO (if the midline of one side of a triangle is equal to half of this side, then this triangle is a right triangle).

△ FAO is a right triangle.

That is ∠ FAO = 90.

And ∵OA is the radius of ⊙ o.

∴ The straight line FA is tangent to⊙ O.