题目

如图2-4-4,AD是⊙O的切线,AC是⊙O的弦,过C作AD的垂线,垂足为B,CB与⊙O相交于点E,AE平分∠CAB,且AE =2,求△ABC各边的长.图2-4-4 答案：思路解析:∠BAE为弦切角,于是∠BAE=∠C,再由AE平分∠CAB和△ABC是直角三角形可得∠C的度数,进而解直角三角形即可.解:∵AD为⊙O的切线,∴∠BAE=∠C.∵AE平分∠CAB,∴∠BAC=2∠BAE.又∵∠C+∠BAC =90°,∴∠BAE =∠C =30°.则有BE =1,AB =,BC =3,AC =2.

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