(1) connects OC, where C is the midpoint of arc AB and AB is the diameter ⊙O, then CO⊥AB and BD are the tangents of ⊙O, and BD⊥AB is obtained, so oc∨BD can prove that ac = cd.

(2) According to the point E is the midpoint of OB, OE=BE can prove that △ Coe △ FBE (ASA), BF=CO, BF=2, AF = √ (ab 2+BF 2) is based on Pythagoras theorem, and BH⊥AF is based on AB, which can be proved.

Answer:

(1) proof: connect oc,

∵C is the midpoint of arc AB, and AB is the diameter⊙ O,

∴CO⊥AB，

∵BD is the tangent of⊙ O,

∴BD⊥AB，

∴OC∥BD，

OA = OB，

∴AC=CD.

(2) Solution:

E is the midpoint of of ob,

In delta division and delta △FBE,

{∠CEO=∠FEB

OE=BE

{∠COE=∠FBE，

∴△COE≌△FBE(ASA)，

∴BF=CO，

OB = 2，

∴BF=2，

∴AF=√(AB^2+BF^2)=2√5，

∫AB is the diameter,

∴BH⊥AF，

∴△ABF∽△BHF，

∴AB/BH=AF/BF，

∴AB？ BF=AF？ BH，

∴BH=(AB？ BF)/AF=(4×2)/(2√5)=(4√5)/5。