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[几何] 反射三角形 九点圆圆心

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hbghlyj Posted 2023-3-15 04:55 |Read mode
Last edited by hbghlyj 2023-4-19 15:34$\triangle A'B'C'$是$\triangle ABC$关于平面上一点$P$的反射三角形.
过九点圆圆心$N$作$AA',BB',CC'$平行线,求证:这三条直线关于对应边反射后共点,这个点是$OP$中点,$O$为外心.
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Ly-lie Posted 2023-4-19 21:08
作$A$关于$BC$的对称点$A'$,有$AP'$与$A'P$关于$BC$对称,故$A'P$平行于$DQ$,延长$OD$交$A'P$于$E$,只需$OD=DE$,作$HE'$平行于$DN$交$A'F$于$E'$,$E'G\perp BC$于$G$,$OM\perp BC$于$M$,设$OE'$与$BC$交于$D'$,由$E'H$平行于$AF$得$$ \frac {E'G}{AK}=\frac {FE'}{A'F}=\frac {AH}{AA'}$$得$$ \frac {E'G}{AK}=\frac {2OM}{2AK} $$故$E'G=OM$,即$OD'=E'D'$,故$ ND' $平行于$ E'H $平行于$ AF $平行于$ ND $,得到$ D=D' $,进而$ E=E' $,于是$ OD=DE $,即得证.
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kuing Posted 2023-4-19 22:06
Ly-lie 发表于 2023-4-19 21:08
作$A$关于$BC$的对称点$A'$,有$AP'$与$A'P$关于$BC$对称,故$A'P$平行于$DQ$,延长$OD$交$A'P$于$E$,只需$OD ...
由E'H$平行于$AF$ 这里少了一个 $

另外,本论坛自定义了平行符号命令 \px

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