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[几何] 垂心四面体的一个等量关系

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力工 Posted 2024-7-19 17:46 |Read mode
求指导下综合证法,不知怎么处理,向量还可动动。
已知四面体$ABCD$各面上的垂线交于一点$O$,外接球半径为$R$,求证:$OA^2+OB^2+OC^2+OD^2=4R^2$

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1+1=? Posted 2024-7-19 18:13 From mobile phone
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那实际就是向量法了。想知道几何法如何证.  Posted 2024-7-20 13:43

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lihpb Posted 2024-7-20 15:52

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 Author| 力工 Posted 2024-7-20 17:07
lihpb 发表于 2024-7-20 15:52
https://kuing.cjhb.site/forum.php?mod=viewthread&tid=12411
谢谢,向量法知道的。我想学习点纯正的几何法。

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