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連接BD,∵AB為⊙O的直徑,∴∠BCA=∠BDA=90°,∵CD平分∠ACB,∴∠ACD=45°,∴∠ABD=45°,∴△ABD為等腰直角三角形,∴AD2+BD2=AB2,∵AB=2cm,∴AD=2cm.故答案為2.
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