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20 13 mathematics for senior high school entrance examination in inner Mongolia
Analysis:

(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.