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[几何] 定三角形三边上作已知角的三角形

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hejoseph posted 2023-7-21 15:46 |Read mode
Last edited by hejoseph 2023-7-21 16:16已知 $\triangle ABC$ 以及三个角 $\alpha$、$\beta$、$\gamma$,其中 $\alpha+\beta+\gamma=180^\circ$,求作一点 $P$,使其到直线 $BC$、$CA$、$AB$ 的垂足分别为 $D$、$E$、$F$,且满足 $\angle FDE=\alpha$,$\angle DEF=\beta$,$\angle EFD=\gamma$。

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original poster hejoseph posted 2023-7-24 14:05
Last edited by hejoseph 2023-7-24 17:31作图问题是这样得到的:若点 $D$、$E$、$F$ 分别在直线 $BC$、$CA$、$AB$ 上,且 $\triangle DEF$ 的三个内角是固定的已知值,$\odot AEF$、$\odot BFD$、$\odot CDE$ 共点于 $P$,那么点 $P$ 是一个定点,与点 $D$、$E$、$F$ 的位置无关。若点 $D$、$E$、$F$ 共线,那么点 $P$ 必定在 $\triangle ABC$ 的外接圆上,这也是 Simson 定理的逆定理。

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