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[几何] 等腰三角形求值

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力工 posted 2023-2-8 22:01 |Read mode
三角形$ABC$中,$AB=AC,CM$为角$ACB$的平分线,$AD,ME$均垂直于$BC$,垂足分别为$D,E$.若$CM$的垂线$MF$交$BC$于点$F$,且$CF=10$,求$DE$的长。 QQ图片20230208215940.png

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kuing posted 2023-2-8 23:51
我用麻烦的三角函数方法算出 DE:CF=1:4 😥 看来肯定有简单的几何解法,所以我的过程暂时就不写了😄

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Czhang271828 posted 2023-2-9 12:33 from mobile
简单写一下,暂时画不了图。延长 FM  CA 交于 P,将 ED 向上平移作 MN,则 MN 是三角形 PCF 中位线的一半

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果然简单👍  posted 2023-2-9 13:03

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original poster 力工 posted 2023-2-9 23:51
Czhang271828 发表于 2023-2-9 12:33
简单写一下,暂时画不了图。延长 FM  CA 交于 P,将 ED 向上平移作 MN,则 MN 是三角形 PCF 中位线的一半 ...
谢谢两位。

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乌贼 posted 2023-2-13 01:56
取$ CF $中点$ N $,在直角$ \triangle CMF $中有\[ MN=CN=\dfrac{1}{2}CF=5 \]\[ \angle MNF=2\angle MCF=\angle FBN \]即$ E $为$ BN $的中点,所以\[ DE=\dfrac{1}{2}(BC-BN)=\dfrac{1}{2}CN=\dfrac{5}{2} \]

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好久不见😃  posted 2023-2-13 12:14

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