如图等腰△ABC,AB=AC,AB为直径作⊙O交AC、BC于D、E,过B切线交OE延长线于F．第17

网友回答

选A∵OE=OB
所以∠OEB=∠OBE
∵AB=AC
所以∠ACB=∠OBE
所以∠OEB=∠ACB
所以OF‖CD
延长OF交⊙O于G
∵FG‖AC
所以⌒DE=⌒AG
⌒AG=⌒BE
所以⌒DE=⌒AG
======以下答案可供参考======
供参考答案1:
hgggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggg
供参考答案2:
选D 做辅助线，链接OD.BD，用三角形相似，对应边成比例证明，要注意圆的半径都是相等的，注意转化相等线段
供参考答案3:
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