(1) connects OC, and the tangent of ∠ ABC+∠ BAC = 90 and CM⊙O gives ∠ ACM+∠ ACO = 90, and then draws a conclusion with ∠BAC=∠AOC.
(2) Connect OC to obtain that △AEC is a right triangle and the diameter of the circumscribed circle of △AEC is AC. Get AC with △ABC∽△CDE.
Answer: (1) Proof: As shown in the figure, connect OC,
∵AB is the diameter⊙ O,
∴∠ACB=90,
∴∠ABC+∠BAC=90,
And ∵CM is the tangent of ⊙O,
∴OC⊥CM,
∴∠ACM+∠ACO=90,
CO = AO,
∴∠BAC=∠ACO,
∴∠acm=∠abc;
(2) solution: BC = CD,
∴OC∥AD,
∵OC⊥CE,
∴AD⊥CE,
∴△AEC is a right triangle,
The diameter of the circumscribed circle of AEC is AC,
∠∠ABC+∠BAC = 90,∠ ACM+∠ ECD = 90,
∴△ABC∽△CDE,
∴AB/CD=BC/ED,
⊙O has a radius of 3,
∴AB=6,
∴6/CD=BC/2,
∴BC^2= 12,
∴BC=2√3,
∴AC=√(36? 12)=2√6,
∴△△△ The radius of the circumscribed circle of AEC is ∴ 6.