题目内容

如图2-2-8,在等腰三角形ABC中,AB =AC,DAC中点,DE平分∠ADB,交ABE,过ADE的圆交BDN.求证:BN =2AE.

图2-2-8

思路分析:要证BN =2AE,由已知有AB=AC =2AD,所以只需证=.而又因为AE =NE,所以只需证=,这可由△BNE∽△BAD证得.

证明:连结EN,∵四边形AEND是圆内接四边形,?

∴∠BNE =∠A.?

又∵∠ABD =∠EBN,∴△BNE∽△BAD.?

=.?

AB =AC,AC =2AD,∴AB =2AD.?

BN =2EN.?

又∵∠ADE =∠NDE,∴AE =EN,?

AE =EN,∴BN =2AE.

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