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13问答网 > 圆心不在同一直线、且直径不相同的三个圆；分别两两做外公切线，相交于三个点。试证此三点共线。

圆心不在同一直线、且直径不相同的三个圆；分别两两做外公切线，相交于三个点。试证此三点共线。

2025-03-15 22:21:37

推荐回答（1个）

回答1：

设圆O1、圆O2、圆O3的半径分别为a、b、c。
由圆切线性质可知AO1/AO2=a/b，BO1/BO3=a/c，CO2/CO3=b/c
于是AO1/AO2·BO1/BO3·CO2/CO3=a/b·a/c·b/c=1
由梅涅劳定理各A、B、C三点共线。

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