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(1)證明:∵D是△ABC的邊BC的中點,
∴BD=CD,
∵BC∥EF,AD⊥EF,
∴AD⊥BC,
∴AB=AC;
(2)證明:連接BO,
∵BD=CD,AD⊥BC,
∴BO=CO,
∵AO=CO,
∴AO=BO=CO,
∴點O是△ABC的外接圓的圓心;
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(3)解:連接BE,
∵AB=5,BC=6,AD⊥BC,BD=CD,
∴BD=
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BC=3,
∴在Rt△ABD中,AD=4,
∵∠ABE=∠ADB=90°,∠BAE=∠DAB,
∴△ABE∽△ADB,
∴
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,
即
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,
∴AE=
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.
分析:(1)由BC∥EF,AD⊥EF,可證得AD⊥BC,又由D是△ABC的邊BC的中點,即可得AD是線段BC的垂直平分線,則可證得AB=AC;
(2)由AD是線段BC的垂直平分線,可證得OB=OC,又由AO=CO,則可得AO=BO=CO,則問題得證;
(3)首先求得AD的長,又由△ABE∽△ADB,根據(jù)相似三角形的對應(yīng)邊成比例,即可求得AE的長.
點評:此題考查了線段垂直平分線的性質(zhì),三角形內(nèi)接圓的性質(zhì)以及相似三角形的判定與性質(zhì)等知識.此題綜合較強,但難度不大,解題的關(guān)鍵是數(shù)形結(jié)合思想的應(yīng)用.