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

问题补充：

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题.

答案：

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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