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證明:(1))連接BD,DO,
∵AB是⊙O的直徑,
∴∠ADB=90°.
∴∠CDB=90°
又∵E為BC的中點,
∴DE=EB=EC,
∴∠EDB=∠EBD.
∵OD=OB,
∴∠ODB=∠OBD.
∵∠ABC=90°,
∴∠EDB+∠OBD=90°.
即OD⊥DE.
∴DE是⊙O的切線.
(2)設AF=x,則FD=
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=
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(切割線定理),
在RT△ABD中,BD=
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=2,
∵∠AFD=∠DFB,∠FDA=∠FBD,
∴△AFD∽△DFB,
∴
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=
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=
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,即
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=
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,
解得:x=
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,即線段AF的長度為
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;
(3)∵點D為EF中點,
∴BD=FD=DE(斜邊中線等于斜邊一半),
又∵ED=EB(切線的性質),
∴△EDB為等邊三角形,
∴∠DBE=60°,∠BCD=30°,
∴BC=
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AB;
分析:(1)連接BD,DO,則可得∠ODA=∠OAD,結合直徑所對的圓周角為90°,可得∠ADB=90°,從而可證明OD⊥DE,也可得出結論.
(2)設AF=x,則FD=
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=
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(切割線定理),在RT△ABD中,求出BD,然后判斷△AFD∽△DFB,利用相似三角形的性質可得出關于x的方程,解出即可得出答案;
(3)根據(jù)切線的性質及直角三角形中斜邊中線等于斜邊一半可判斷出△DEB為等邊三角形,然后可得出∠BCD=30°,繼而可得出線段AB與BC之間的數(shù)量關系.
點評:此題屬于圓的綜合題,涉及了解直角三角形、切割線定理切線的判定與性質、相似三角形的判定與性質,考察的知識點較多,解答第二問是本題的難點,關鍵是表示用AF表示出FD,難度較大.