三角形内接与圆O,AB是圆O直径,点D在圆O上,过点C的切线交AD的延长线与点E,且AE垂直于CE,连接CD
求证(1)AC平分角BAE(2)DC等于BC(3)若AB=5,AC=4,求sin角CDE的值
题目补充“三角形ABC内接与圆O”
人气:287 ℃ 时间:2019-09-02 09:41:14
解答
证明:(1)、连接OC∵ CE是圆O切线∴OC⊥CE ∵AE⊥CE∴OC‖AE∴∠OCA=∠EAC∵OA=OB∴∠OCA=∠OAC∴∠EAC=∠OAC即AC平分角BAE(2)、∵∠EAC=∠OAC∴弧CD=弧BC∴DC=BC(3)、作CF⊥AB于F∴∠CFB=∠CED=90°∵∠EAC=∠B...
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