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(1)證明:∵B、D關于AC對稱,∴AC⊥BD,
又∵CF∥BD,
∴AC⊥CF,
∵點C在⊙O上,
∴CF是⊙O的切線;
(2)解:∵AC是⊙O的直徑,
∴∠ADC=90°(直徑所對的圓周角是直角),
∴∠CDF=90°;
又∵CF∥BD(已知),
∴∠BDC=∠DCF(兩直線平行,內錯角相等);
∵∠BAC=∠BDC(同弧所對的圓周角相等),
∴∠BAC=∠DCF(等量代換),
∴sin∠BAC=sin∠DCF=
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=
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,
∴CF=5;
∴CD=4;
∵B、D關于AC對稱,
∴BC=CD=4,
∴sin∠BAC=
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=
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,
∴AC=
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,
∴⊙O的半徑長=
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AC=
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.
分析:(1)欲證CF是⊙O的切線,只需證明⊥CF即可;
(2)由圓周角定理、平行線的性質以及等量代換推知∠BAC=∠DCF;然后根據三角函數的定義、軸對稱圖形的性質求得DC=BC=4;最后在直角三角形ABC中利用正弦三角函數的定義求得該圓的直徑AC的長度.
點評:本題考查了切線的判定.要證某線是圓的切線,已知此線過圓上某點,連接圓心與這點(即為半徑),再證垂直即可.