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[几何] 完全四边形的等角线

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hbghlyj Post time 2023-3-24 21:31 |Read mode
知乎
设射线 $OA,OB,OP,OQ$ 满足 $\angle AOP=\angle BOQ$ 且 $OP,OQ$ 同时在 $\angle AOB$ 的内侧或外侧(也即关于角平分线对称),则称 $OP,OQ$ 为 $\angle AOB$ 的一对等角线.

下面介绍等角线的一个重要性质.
v2-a5868a63570cfeae509a3bd0c33c5658_1440w[1].jpg
定理 如图, $OC,OD$ 是 $\angle AOB$ 的一对等角线, $AC$ 与 $BD$ 交于点 $E,$ $AD$ 与 $BC$ 交于点 $F,$ 则 $OE,OF$ 也是 $\angle AOB$ 的一对等角线.

证明:用同一法.

设 $F'$ 是 $AD$ 上的一点,使得 $OE,OF'$ 是 $\angle AOB$ 的一对等角线,则

$\angle AOC=\angle BOD,$ $\angle DOF'=\angle COE,$ $\angle BOE=\angle AOF'.$

所以 $\frac{\sin\angle AOC}{\sin\angle COE}\cdot\frac{\sin\angle EOB}{\sin\angle BOD}\cdot\frac{\sin\angle DOF'}{\sin\angle F'OA}=1.$

故由 Menelaus 定理的第二角元形式, $B,F',C$ 三点共线.

所以 $F'$ 与 $F$ 重合,即 $OE,OF$ 为 $\angle AOB$ 的一对等角线.

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 Author| hbghlyj Post time 2023-3-24 21:32
artofproblemsolving.com/community/c2771h1181729_isogonal_line_lemma_updated

Lemma
Let $AP,AS$ and $AQ,AR$ be two pairs of isogonal lines with respect to $\angle BAC$.
Let $PR \cap QS = X$ and $PQ \cap RS = Y$.
Then $AX,AY$ are isogonal line with respect to $\angle BAC$

Proof
Without loss of generality let $AY$ intersect $PR,QS$ in $K,L$ respectively.
Considering the perspectivity that sends line $PR \to QS$ ($Y$ is the centre of perspectivity).
$\implies \frac{PK \cdot XR}{XK \cdot PR} = \frac { QL \cdot SX}{XL \cdot SQ}$

Now using the property of cross ratio w.r.t point $A$.
Let $\angle PAQ = \angle RAS = x, \angle QAL = y, \angle XAR = z$
Using the cross ratios,
$\sin(x+y) \cdot \sin z = \sin y \cdot \sin (x+z)$

$\implies \sin y = \sin z$
$\implies y=z$ (configuration matters here)
$\implies \angle QAY = \angle RAX$
$\implies$ our lemma.


X Y C A Q P B S R

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 Author| hbghlyj Post time 2023-3-24 21:32
Advanced Lemmas in Geometry,Fedir Yudin第7页的证明应该差不多:
Proof. We use trigonometric Ceva's theorem in $\triangle O A B$.$$\frac{\sin A O E}{\sin E O B} \frac{\sin A O F}{\sin F O B}=\frac{\sin O A E \sin A B E}{\sin E A B \sin E B O} \frac{\sin O A F \sin A B F}{\sin F A B \sin F B O}=\frac{\sin O A C \sin A B C}{\sin O B C \sin B A C} \frac{\sin O A D \sin A B D}{\sin O B D \sin B A D}=\frac{\sin A O C}{\sin C O B} \frac{\sin A O D}{\sin D O A}=1$$and the lemma follows.

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 Author| hbghlyj Post time 2023-3-24 21:33
又见 教师作品专栏第六期:等角线-萧振纲.pdf 第9页
nsmath.cn下载到 $type 教师作品专栏第六期:等角线-萧振纲.pdf (956.06 KB, Downloads: 7)

三角形中的等角线
$type 三角形中等角线的性质及应用_华接春.pdf (951.46 KB, Downloads: 2)

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kuing Post time 2023-3-24 21:53


看到你这个图,我总觉得你以前发过🤔

=====
哦,原来是这帖又发了一次:kuing.orzweb.net/viewthread.php?tid=8938 (新论坛上这帖删了,只能贴旧论坛链接)

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 Author| hbghlyj Post time 2023-3-24 22:02

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 Author| hbghlyj Post time 2024-3-9 01:32
kuing 发表于 2023-3-24 13:53
原来是这帖又发了一次

其实,是发了新帖,又想起旧帖,所以把新、旧帖手动合并了。

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