推荐回答(4个)
设圆心为O,连接OC,OD,作OH垂直MN于H;
ΔOCD为等腰三角形,所以OH垂直平分CD,所以CH=DH;
AE平行OH平行BF,且O为AB的中点,所以EH=HF,CE=EH-CH=HF-HD=DF
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一直相等呀 过圆心o 做MN的垂线 交予H
那么可知H为 EF中点 由 AB为直径 O为中点 三条平行线可知
另外 H也是CD的中点 所以CE和DF一直相等
其次 当相交时情况没有发生根本改变 还是相等
过圆心o 做MN的垂线 交于点H 又以AB为直径 即O为中点 由三条平行线可知 因为H为 EF中点,也是CD的中点 所以CE和DF一直相等
平移时结果不变
过圆心o 做MN的垂线 交于H
又AB为直径 即O为中点 三条平行线可知 可知H为 EF中点
另外 H也是CD的中点 所以CE和DF一直相等
平移时不改变结果
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