(1)證明:∵連接CD,在⊙O中,
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∵∠ABC=∠ADC,∠1=∠3,
∴△ABE∽△CDE,
∴
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∵AE•DE=BE•CE;
解:(2)BD=CD,
理由:連接OD、BD,
∵MN切⊙O于點D,
∴OD⊥MN,
∵MN∥BC,
∴OD⊥BC,
∴在⊙O中,
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=
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,
∴BD=CD;
(3)∵在⊙O中,
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=
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,
∴∠1=∠2,
在⊙O中,
∵∠ADB=∠4,
∵MN∥BC,
∴∠C=∠4,
∴∠ADB=∠C,
∴△ABD∽△ADN,
∴
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,
∴AD
2=AB•AN=6×15=90,
∴AD=
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.
分析:(1)連接CD,利用已知條件證明△ABE∽△CDE,由相似三角形的性質(zhì):對應(yīng)邊的比值相等即可證明:AE•DE=BE•CE;
(2)BD和CD的數(shù)量關(guān)系是BD=CD,根據(jù)切線的性質(zhì)和平行線的性質(zhì)證明,
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=
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即可;
(3)首先證明△ABD∽△ADN,所以可得:
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,即AD
2=AB•AN問題得解.
點評:本題考查的是切線的判定,要證某線是圓的切線,已知此線過圓上某點,連接圓心和這點(即為半徑),再證垂直即可.乘積的形式通�?梢赞D(zhuǎn)化為比例的形式,通過相似三角形的性質(zhì)得出.