圆01与圆O2交与A,B,O1在圆O2上,AC不是圆O1直径,CB的延长线交与圆O2于D,求证:O1D垂直AC
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![圆01与圆O2交与A,B,O1在圆O2上,AC不是圆O1直径,CB的延长线交与圆O2于D,求证:O1D垂直AC](/uploads/image/z/5026971-3-1.jpg?t=%E5%9C%8601%E4%B8%8E%E5%9C%86O2%E4%BA%A4%E4%B8%8EA%2CB%2CO1%E5%9C%A8%E5%9C%86O2%E4%B8%8A%2CAC%E4%B8%8D%E6%98%AF%E5%9C%86O1%E7%9B%B4%E5%BE%84%2CCB%E7%9A%84%E5%BB%B6%E9%95%BF%E7%BA%BF%E4%BA%A4%E4%B8%8E%E5%9C%86O2%E4%BA%8ED%2C%E6%B1%82%E8%AF%81%EF%BC%9AO1D%E5%9E%82%E7%9B%B4AC)
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圆01与圆O2交与A,B,O1在圆O2上,AC不是圆O1直径,CB的延长线交与圆O2于D,求证:O1D垂直AC
圆01与圆O2交与A,B,O1在圆O2上,AC不是圆O1直径,CB的延长线交与圆O2于D,求证:O1D垂直AC
圆01与圆O2交与A,B,O1在圆O2上,AC不是圆O1直径,CB的延长线交与圆O2于D,求证:O1D垂直AC
设O1D与AC交于Q.
则角OAC=(180-角AOC)/2
=(180-2角ABC)/2 (圆周角是圆心角一半)
=90-角ABC
=90-(180-角ABD)=角ABD-90
=角AO1D-90 (对同弧的圆周角相等)
所以 角AQO1=角AO1D-角OAC=90
证毕