※ 引述《agusa (agusa)》之銘言：
: http://i.imgur.com/MV0gLMx.jpg
過D做BC垂直線交圓周於D'
CD = CD'
∠D'BC = ∠DBC
=> ∠COD' = ∠COA
=> CD' = AC
=> AC = CD
=> ∠ACD = 180 - 2∠CAB
= 2[90 - ∠CAB]
= 2∠ABC
= ∠AOC
=> △CAD ~ △OCA
=> AC : 4 = 5 : AC
=> CD = AC = 2√5
選(D)
: http://i.imgur.com/LuxYL3c.jpg
推文已有人解，略
: 謝謝大大
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