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[几何] 外接圆 中点 求证: NM=MD

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hbghlyj Posted at 2020-3-16 00:02:14 |Read mode
Last edited by hbghlyj at 2024-11-3 13:59:00$\odot O$是△ABC外接圆,D是AB上一点,G是AD中点,H是DB中点,EG⊥AB交AC于E,FH⊥ AB交BC于F,DN⊥EF交EF于M,交$\odot O$于N,求证: NM=MD

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乌贼 Posted at 2020-3-17 02:32:56
回复 8# hbghlyj
没证出来,但发现一些有趣……
如图: 211.png
    $ \triangle ABC $,$ D $为$ AB $上一点,$ AD $的垂直平分线交$ AC $于$ E $,$ DB $的垂直平分线交$ BC $于$ F $,$ G、H、P、Q $分别为$ AD、DB、AC、BC $的中点,$ O $为$ \triangle ABC $外接圆圆心,$ K $为$ EF $与$ PQ $的交点。则有:(1)$ OK\perp EF $且$ EOFC $四点共园(2)$ DE\px HK $

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乌贼 Posted at 2020-3-17 13:26:29
如图: 212.png
延长$ ND $交外接圆于$ Q $,取$ DQ $中点$ P $。连接$ AN 、BN、ED、GM、MH、DF、GP、PH $,有\[ \angle GMH=\angle GED+\angle DFH=\angle AEG+\angle BFH=\angle ACB=\angle ANB \]又\[ \angle GPH=\angle AQB\riff\angle GMH+\angle GPH=\angle ANB+\angle AQB=180\du  \]即$ GMHP $四点共园,故\[ \angle GMP=\angle GHP=\angle ABQ=\angle ANQ\\\riff AN\px GM\\\riff DM=MN \]复 8# hbghlyj [/b]

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