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[几何] 正方形$EFGH$内接于$ABCD$,若$EB=FC=GD=HA$,求证:$ABCD$是正方形

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isee posted 2014-8-9 11:14 |Read mode
如图,$EFGH$是(平面)四边形$ABCD$的内接正方形,若$EB=FC=GD=HA$,求证:$ABCD$亦是正方形。
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kuing posted 2014-8-9 11:18
又是这种……之前那个正三角形的……还是闪了哎……

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original poster isee posted 2014-8-9 11:25
回复 2# kuing


嗯,思索之后的升级版,比原三角形时,相对易入手,个人觉得,然后,看看能不能顺便把一般情况都解决掉

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