∵AC is the diameter⊙ O,

∴∠ADC=∠BDC=90。 ?

Point e is the midpoint of BC,

∴DE=BE=EC.

OA = OD，DE=BE，

∴∠ADO=∠A,∠DBE=∠BDE.

∵∠DBE+∠A=90，

∴∠BDE+∠ADO=90。

∴∠EDO=90，

∴OD⊥DE，

That is, DE is the tangent of ⊙O;

(2) Solution: Connect OE. Then OE∑AB, OE= 12AB,

∴△OEF∽△BDF，

∴EFFD=OEBD.

∵BC cuts⊙ O at point C,

∴∠ACB=90。

In Rt△ABC, AC=4 and BC=43,

According to Pythagorean theorem, AB=8,

∴OE=4，

∫∠A = 60，

∴△AOD is an equilateral triangle with a side length of 2.

∴AD=2,BD=AB-AD=6，

∴EFFD=OEBD=46=23.