(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。