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[几何] 几何证明题2

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zhiwen posted 2023-11-9 22:03 |Read mode

A,B为对径点, CD∈圆O, CC',BB'关于AB,C'P对称, AB'F,ABE,DCE, DC'F共线。
求证EF⊥C'D

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战巡 posted 2023-11-9 23:14
p0193.png

令$AB, DC'$交于$G$,$BB', DC'$交于$H$,其他连线如图

1、
由梅涅劳斯定理,可以得到
\[\frac{BG}{GA}\cdot\frac{AF}{FB'}\cdot\frac{B'H}{BH}=1\]
$B'H=BH$是个显然的是,因此即有
\[\frac{FB'}{AF}=\frac{BG}{AG}\]

2、
易证$DB$平分$\angle EDF$,故此由角平分线定理有
\[\frac{BG}{BE}=\frac{DG}{DE}\]
又易证$\Delta BED\sim\Delta CEA$,有
\[\frac{BE}{CE}=\frac{DE}{AE}\]
这两个式子相乘会得到
\[\frac{BG}{CE}=\frac{DG}{AE}\]
\[BG\cdot AE=CE\cdot DG\]

3、
易证$\Delta AGD\sim \Delta C'GB$,以及$CG=C'G$,因此有
\[\frac{DG}{AG}=\frac{BG}{GC'}=\frac{BG}{CG}\]
又显然$\angle GCB=\angle GC'B=\angle DAB=\angle ECB$,即$CB$平分$\angle ECG$,有
\[\frac{BG}{CG}=\frac{BE}{CE}\]

\[\frac{DG}{AG}=\frac{BE}{CE}\]
结合2的结果,就有
\[DG\cdot CE=AG\cdot BE=BG\cdot AE\]
再加上1的结果
\[\frac{FB'}{AF}=\frac{BG}{AG}=\frac{BE}{AE}\]
\[BB'\parallel EF\]
\[EF\perp DF\]

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kuing posted 2023-11-10 01:39
Last edited by kuing 2023-11-10 13:55题目好像可以简化一点,像这样:
如下图,`PB` 平分 `\angle MPN`,`PB'` 与 `PB` 关于 `PM` 对称,`PA\perp PB`,直线 `AB`, `AB'` 分别交 `PN`, `PM` 于 `E`, `F`,则 `EF\perp PF`。
QQ截图20231110013425.png

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kuing posted 2023-11-10 15:49

续楼上

用三角很容易证出来,由张角定理,有
\begin{align*}
\frac{\sin\angle APB}{PE}+\frac{\sin\angle EPB}{PA}&=\frac{\sin\angle APE}{PB},\\
\frac{\sin\angle APB'}{PF}+\frac{\sin\angle FPB'}{PA}&=\frac{\sin\angle APF}{PB'},
\end{align*}
记 `\angle BPF=x`,则以上两式变成
\begin{align*}
\frac1{PE}+\frac{\sin x}{PA}&=\frac{\cos x}{PB},\\
\frac{\cos2x}{PF}+\frac{\sin x}{PA}&=\frac{\cos x}{PB'},
\end{align*}
而 `PB=PB'`,故由以上两式立得
\[\frac{PF}{PE}=\cos2x=\cos\angle FPE,\]
这就证明了 `EF\perp PF`。

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乌贼 posted 2023-11-11 12:02
如图: 36.png
令$ BB_1 $中点为$ N $,$ C_1D $与$ AB $交点为$ M $,圆心为$ O $,有\[ \dfrac{MN}{NF}=\dfrac{MO}{OA} \]由熟知的结论(可能不熟,见链接证明forum.php?mod=viewthread&tid=4714&pid … &extra=#pid5704626楼)即有\[ \dfrac{MN}{NF}=\dfrac{MO}{OA}=\dfrac{MB}{BE}\riff EF\px BN\riff DF\perp EF \]

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