(1)证明:连接OC······································································· 1分
∵OA=OC ∴∠OAC=∠OCA ∵CE是⊙O的切线 ∴∠OCE=90° ·············································· 2分 ∵AE⊥CE ∴∠AEC=∠OCE=90° ∴OC∥AE ·················································· 3分 ∴∠OCA=∠CAD ∴∠CAD=∠BAC ∴ ∴DC=BC ··························································································· 4分 (2)∵AB是⊙O的直径 ∴∠ACB=90° ∴·························································· 5分 ∵∠CAE=∠BAC ∠AEC=∠ACB=90° ∴△ACE∽△ABC······················································································ 6分 ∴ ∴ ······················································ 7分 ∵DC=BC=3 ∴····················································· 8分 ∴-----------9分 (其它解法参考得分) |