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%!TEX program = xelatex
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% 完整编译: xelatex -> biber/bibtex -> xelatex -> xelatex
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\documentclass[lang=cn,a4paper]{elegantpaper}
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\title{Pond}
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\author{作者 \\ Patrick}
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\date{\zhdate{2025/03/09}}
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% 本文档命令
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\usepackage{array}
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\usepackage{pgfplots}
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\usepackage{soul}
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\usepackage{physics}
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\usepackage{amssymb}
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\newcommand{\ccr}[1]{\makecell{{\color{#1}\rule{1cm}{1cm}}}}
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\addbibresource[location=local]{reference.bib} % 参考文献，不要删除
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\pgfplotsset{compat=1.15}
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\usetikzlibrary{arrows}
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\graphicspath{ {../images/} }
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\begin{document}
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\vspace*{\fill}
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\begin{figure}[h]
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\begin{center}
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\includegraphics[width=0.5\textwidth]{THE_LAST_PROOF.png}
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\end{center}
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\end{figure}
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\begin{center}
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\sffamily {
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\textbf{\LARGE{Silence TLP}}
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\Large{2025 / 03 / 09}
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}
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\end{center}
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\vspace*{\fill}
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\thispagestyle{empty}
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DOI: 11.4514/sil.tlp.2025.0000001
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\thispagestyle{empty}
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\newpage
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\setcounter{page}{1}
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\maketitle
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\begin{abstract}
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Pond 是一个试验性项目，使用了 ElegantPaper 模板。
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\keywords{Nonsense，Bullshit}
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\end{abstract}
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\section{题目}
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如图1，在$\triangle ABC$中，$BD$平分$\angle ABC$，$AD \bot BD$，$E$为$BC$中点，$EF$交射线$CA$于$F$，交$AB$于$G$，$GB = GE$，$DE = m$，$DF = n$.
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\begin{enumerate}
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\item 在图1中找到与$\angle BAC$相等的角，并证明；
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\item 求$BD$的长（用$m$，$n$表示）；
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\item 如图2，$AD$交$BC$的中垂线于$M$，连接$CM$，若$CB$平分$\angle ACM$，求$\dfrac{m}{n}$的值.
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\end{enumerate}
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% 图1
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\begin{figure}[h]
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\centering
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\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1cm,y=1cm,scale=4]
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\clip(-1.2,-0.1) rectangle (1.2,0.8);
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% 边框
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% \draw[red] (current bounding box.south west) rectangle (current bounding box.north east);
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\draw [line width=0.8pt] (-1.,0.)-- (1.,0.);
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\draw [line width=0.8pt] (-0.7071067811865475,0.7071067811865475)-- (0.,0.);
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||||
\draw [line width=0.8pt] (1.,0.)-- (-0.7071067811865475,0.7071067811865475);
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||||
\draw [line width=0.8pt] (-0.41421356237309503,0.5857864376269049)-- (-1.,0.);
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||||
\draw [line width=0.8pt] (-0.41421356237309503,0.5857864376269049)-- (-0.2928932188134525,0.2928932188134525);
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\draw [line width=0.8pt] (-0.2928932188134525,0.2928932188134525)-- (-1.,0.);
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\begin{scriptsize}
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\draw (-0.414213562373095, 0.585786437626905) node[xshift=0.1cm,yshift=0.2cm] (A) {$A$};
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\draw (-1,0) node[xshift=-0.2cm,yshift=-0.1cm] (B) {$B$};
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\draw (1,0) node[xshift=0.2cm,yshift=-0.1cm] (C) {$C$};
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\draw (-0.292893218813452, 0.292893218813452) node[xshift=0.15cm,yshift=0.1cm] (D) {$D$};
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\draw (0,0) node[xshift=0.1cm,yshift=0.2cm] (E) {$E$};
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\draw (-0.707106781186547, 0.707106781186547) node[xshift=-0.15cm,yshift=0.1cm] (F) {$F$};
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\draw (-0.5, 0.5) node[xshift=0cm,yshift=-0.25cm] (G) {$G$};
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\end{scriptsize}
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\end{tikzpicture}
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\caption{}
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\label{original_pic1}
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\end{figure}
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% 图2
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\begin{figure}[h]
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\centering
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\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1cm,y=1cm,scale=4]
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||||
\clip(-1.2,-0.5) rectangle (1.2,0.8);
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||||
% 边框
|
||||
% \draw[red] (current bounding box.south west) rectangle (current bounding box.north east);
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||||
\draw [line width=0.8pt] (-1.,0.)-- (1.,0.);
|
||||
\draw [line width=0.8pt] (-0.7071067811865476,0.7071067811865476)-- (0.,0.);
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||||
\draw [line width=0.8pt] (1.,0.)-- (-0.7071067811865476,0.7071067811865476);
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||||
\draw [line width=0.8pt] (-0.414213562373095,0.585786437626905)-- (-1.,0.);
|
||||
\draw [line width=0.8pt] (-0.414213562373095,0.585786437626905)-- (-0.2928932188134525,0.2928932188134525);
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||||
\draw [line width=0.8pt] (-0.2928932188134525,0.2928932188134525)-- (-1.,0.);
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||||
\draw [line width=0.8pt] (0.,-0.4142135623730955)-- (1.,0.);
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||||
\draw [line width=0.8pt] (-0.2928932188134525,0.2928932188134525)-- (0.,-0.4142135623730955);
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||||
\draw [line width=0.8pt] (0.,0.)-- (0.,-0.4142135623730955);
|
||||
\begin{scriptsize}
|
||||
\draw (-0.414213562373095, 0.585786437626905) node[xshift=0.1cm,yshift=0.2cm] (A) {$A$};
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\draw (-1,0) node[xshift=-0.2cm,yshift=-0.1cm] (B) {$B$};
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\draw (1,0) node[xshift=0.2cm,yshift=-0.1cm] (C) {$C$};
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\draw (-0.292893218813452, 0.292893218813452) node[xshift=0.15cm,yshift=0.1cm] (D) {$D$};
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\draw (0,0) node[xshift=0.1cm,yshift=0.2cm] (E) {$E$};
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\draw (-0.707106781186547, 0.707106781186547) node[xshift=-0.15cm,yshift=0.1cm] (F) {$F$};
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\draw (-0.5, 0.5) node[xshift=0cm,yshift=-0.25cm] (G) {$G$};
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\draw (0, -0.414213562373095) node[xshift=0cm,yshift=-0.15cm] (M) {$M$};
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\end{scriptsize}
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\end{tikzpicture}
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\caption{}
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\label{original_pic2}
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\end{figure}
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\section{解析}
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\subsection{第一小问}
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$\angle BAC = \angle BDE$.
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证明：如图3，以$AB$中点$H$为圆心，$HA$长度为半径作$\odot \mathrm{H}$，交$BC$于$K$，连接$HK$，$HE$，$HD$，
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所以$HA=HB=HK=\dfrac{1}{2}AB$，
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% 图3
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\begin{figure}[h]
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\centering
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\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1.0cm,y=1.0cm,scale=5]
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\clip(-1.2,-0.2) rectangle (1.2,0.8);
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% 边框
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% \draw[red] (current bounding box.south west) rectangle (current bounding box.north east);
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||||
\draw [line width=0.8pt] (-1.,0.)-- (1.,0.);
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||||
\draw [line width=0.8pt] (-0.7071067811865475,0.7071067811865475)-- (0.,0.);
|
||||
\draw [line width=0.8pt] (1.,0.)-- (-0.7071067811865475,0.7071067811865475);
|
||||
\draw [line width=0.8pt] (-0.41421356237309503,0.5857864376269049)-- (-1.,0.);
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||||
\draw [line width=0.8pt] (-0.41421356237309503,0.5857864376269049)-- (-0.2928932188134525,0.2928932188134525);
|
||||
\draw [line width=0.8pt] (-0.2928932188134525,0.2928932188134525)-- (-1.,0.);
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||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.2928932188134524) circle (0.414213562373095cm);
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||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.2928932188134524)-- (-0.2928932188134525,0.2928932188134525);
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||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.2928932188134524)-- (-0.41421356237309503,0.);
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||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (0.,0.)-- (-0.7071067811865475,0.2928932188134524);
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||||
\begin{scriptsize}
|
||||
\draw (-0.414213562373095, 0.585786437626905) node[xshift=0.1cm,yshift=0.2cm] (A) {$A$};
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||||
\draw (-1,0) node[xshift=-0.2cm,yshift=-0.1cm] (B) {$B$};
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||||
\draw (1,0) node[xshift=0.2cm,yshift=-0.1cm] (C) {$C$};
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||||
\draw (-0.292893218813452, 0.292893218813452) node[xshift=0.15cm,yshift=0.1cm] (D) {$D$};
|
||||
\draw (0,0) node[xshift=0.1cm,yshift=0.2cm] (E) {$E$};
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||||
\draw (-0.707106781186547, 0.707106781186547) node[xshift=-0.1cm,yshift=0.15cm] (F) {$F$};
|
||||
\draw (-0.5, 0.5) node[xshift=0cm,yshift=-0.25cm] (G) {$G$};
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||||
\draw (-0.707106781186547, 0.292893218813452) node[xshift=-0.1cm,yshift=0.15cm] (H) {$H$};
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||||
\draw (-0.414213562373095, 0) node[xshift=0.15cm,yshift=-0.15cm] (K) {$K$};
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||||
\end{scriptsize}
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||||
\end{tikzpicture}
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\caption{}
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\label{step_pic1}
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\end{figure}
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$\because AD \bot BD$，$\therefore \angle ADB=\dfrac{\pi}{2}$，
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$\because H$是$AB$中点，$\therefore HD=\dfrac{1}{2}AB$，
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$\therefore HD=HA=HB=HK=\dfrac{1}{2}AB$，
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$\therefore$点$D$在$\odot \mathrm{H}$上；
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设$\angle EBD = \alpha$，
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$\because BD$平分$\angle ABC$，
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$\therefore \angle ABD = \angle EBD = \alpha$，
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$\therefore \angle ABC = 2\alpha$，
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$\because HB=HD$，
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$\therefore \angle ABD = \angle HDB = \alpha$，
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$\therefore \angle EBD = \angle HDB$，
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$\therefore HD \mathop{//} BC$，
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$\because GB=GE$，
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$\therefore \angle ABC = \angle GEB = 2\alpha$，
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$\because HB=HK$，
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$\therefore \angle ABC = \angle HKB = 2\alpha$，
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$\therefore \angle GEB = \angle HKB$，
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$\therefore HK \mathop{//} EF$，
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$\because HD \mathop{//} BC$，
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$\therefore$四边形$HDEK$是平行四边形，
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$\therefore HD=HK=EK$，
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$\therefore \angle KHE = \angle KEH$，
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$\because H$是$AB$的中点，$E$是$BC$的中点，
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$\therefore HE \mathop{//} AC$，
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$\therefore \angle KEH = \angle C$，
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$\therefore \angle KHE = \angle KEH = \angle C$，
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$\therefore \angle HKB = \angle KHE + \angle KEH = 2\angle C = 2\alpha$，
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$\therefore \angle C = \alpha$，
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$\because \angle GEB = \angle C + \angle F = \alpha + \angle F = 2\alpha$，
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$\therefore \angle F = \alpha$，
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$\therefore \angle C = \angle F = \alpha$；
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$
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\begin{aligned}
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\because & \angle BDE = \pi - \angle EBD - \angle GEB = \pi - \alpha - 2\alpha = \pi - 3\alpha \\
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& \angle BAC = \pi - \angle ABC - \angle C = \pi - 2\alpha - \alpha = \pi - 3\alpha
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\end{aligned}
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$，
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$\therefore \angle BAC = \angle BDE$. $\square$
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\subsection{第二小问}
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如图4，连接$AK$，
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% 图4
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\begin{figure}[h]
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\centering
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||||
|
||||
\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1.0cm,y=1.0cm,scale=5]
|
||||
\clip(-1.2,-0.2) rectangle (1.2,0.8);
|
||||
% 边框
|
||||
% \draw[red] (current bounding box.south west) rectangle (current bounding box.north east);
|
||||
\draw [line width=0.8pt] (-1.,0.)-- (1.,0.);
|
||||
\draw [line width=0.8pt] (-0.7071067811865475,0.7071067811865475)-- (0.,0.);
|
||||
\draw [line width=0.8pt] (1.,0.)-- (-0.7071067811865475,0.7071067811865475);
|
||||
\draw [line width=0.8pt] (-0.41421356237309503,0.5857864376269049)-- (-1.,0.);
|
||||
\draw [line width=0.8pt] (-0.41421356237309503,0.5857864376269049)-- (-0.2928932188134525,0.2928932188134525);
|
||||
\draw [line width=0.8pt] (-0.2928932188134525,0.2928932188134525)-- (-1.,0.);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.2928932188134524) circle (0.414213562373095cm);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.2928932188134524)-- (-0.2928932188134525,0.2928932188134525);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.2928932188134524)-- (-0.41421356237309503,0.);
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||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (0.,0.)-- (-0.7071067811865475,0.2928932188134524);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.41421356237309503,0.5857864376269049)-- (-0.41421356237309503,0.);
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||||
% \draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.7071067811865475)-- (-1.,0.);
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||||
% \draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.7071067811865475)-- (-0.7071067811865475,0.2928932188134524);
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||||
\begin{scriptsize}
|
||||
\draw (-0.414213562373095, 0.585786437626905) node[xshift=0.1cm,yshift=0.2cm] (A) {$A$};
|
||||
\draw (-1,0) node[xshift=-0.2cm,yshift=-0.1cm] (B) {$B$};
|
||||
\draw (1,0) node[xshift=0.2cm,yshift=-0.1cm] (C) {$C$};
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||||
\draw (-0.292893218813452, 0.292893218813452) node[xshift=0.15cm,yshift=0.1cm] (D) {$D$};
|
||||
\draw (0,0) node[xshift=0.1cm,yshift=0.2cm] (E) {$E$};
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||||
\draw (-0.707106781186547, 0.707106781186547) node[xshift=-0.1cm,yshift=0.15cm] (F) {$F$};
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||||
\draw (-0.5, 0.5) node[xshift=0cm,yshift=-0.25cm] (G) {$G$};
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||||
\draw (-0.707106781186547, 0.292893218813452) node[xshift=-0.1cm,yshift=0.15cm] (H) {$H$};
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||||
\draw (-0.414213562373095, 0) node[xshift=0.15cm,yshift=-0.15cm] (K) {$K$};
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||||
\end{scriptsize}
|
||||
\end{tikzpicture}
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||||
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\caption{}
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||||
\label{step_pic2}
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||||
\end{figure}
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||||
$\because AB$是$\odot \mathrm{H}$直径，
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||||
$\therefore \angle AKB = \dfrac{\pi}{2}$，
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||||
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||||
$\because HD = HK$，
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||||
$\therefore$平行四边形$HDEK$是菱形，
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||||
$\because DE=m$，
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||||
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||||
$\therefore HD = EK = DE = m$，
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||||
$\therefore HF = HA = HB = HD = HK = m$，
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||||
$\therefore AB = 2m$，
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||||
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$\because DF = n$，
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$\therefore EF = DE + DF = m + n$，
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$\because EB = EK + BK$，$EK = m$，
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$\therefore EB = m + BK$，
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||||
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||||
$\therefore BK = n$，
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||||
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$\therefore \cos \angle ABC = \cos 2\alpha = \dfrac{BK}{AB} = \dfrac{n}{2m}$，
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||||
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||||
$\therefore \cos \alpha = \cos \dfrac{2\alpha}{2} = \sqrt{\dfrac{1+\dfrac{n}{2m}}{2}} = \sqrt{\dfrac{2m+n}{4m}}$，
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||||
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||||
$\therefore BD = AB \times \cos \angle ABD = AB \times \cos\alpha = 2m \times \sqrt{\dfrac{2m+n}{4m}} = \sqrt{2m^2 + mn}$. $\square$
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||||
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||||
\subsection{第三乐章}
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||||
% 图5
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||||
如图5，记$AM$交$BC$于R，
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\begin{figure}[h]
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||||
\centering
|
||||
|
||||
\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1.0cm,y=1.0cm,scale=5]
|
||||
\clip(-1.2,-0.5) rectangle (1.2,0.8);
|
||||
% 边框
|
||||
% \draw[red] (current bounding box.south west) rectangle (current bounding box.north east);
|
||||
\draw [line width=0.8pt] (-1.,0.)-- (1.,0.);
|
||||
\draw [line width=0.8pt] (-0.7071067811865475,0.7071067811865475)-- (0.,0.);
|
||||
\draw [line width=0.8pt] (1.,0.)-- (-0.7071067811865475,0.7071067811865475);
|
||||
\draw [line width=0.8pt] (-0.41421356237309503,0.5857864376269049)-- (-1.,0.);
|
||||
\draw [line width=0.8pt] (-0.41421356237309503,0.5857864376269049)-- (-0.2928932188134525,0.2928932188134525);
|
||||
\draw [line width=0.8pt] (-0.2928932188134525,0.2928932188134525)-- (-1.,0.);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.2928932188134524) circle (0.414213562373095cm);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.2928932188134524)-- (-0.2928932188134525,0.2928932188134525);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.2928932188134524)-- (-0.41421356237309503,0.);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (0.,0.)-- (-0.7071067811865475,0.2928932188134524);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.41421356237309503,0.5857864376269049)-- (-0.41421356237309503,0.);
|
||||
\draw [line width=0.8 pt] (-0.2928932188134525,0.2928932188134525)-- (0.,-0.414213562373095);
|
||||
\draw [line width=0.8 pt] (0.,-0.414213562373095)-- (1.,0.);
|
||||
\draw [line width=0.8 pt] (0.,0.)-- (0.,-0.414213562373095);
|
||||
\begin{scriptsize}
|
||||
\draw (-0.414213562373095, 0.585786437626905) node[xshift=0.1cm,yshift=0.2cm] (A) {$A$};
|
||||
\draw (-1,0) node[xshift=-0.2cm,yshift=-0.1cm] (B) {$B$};
|
||||
\draw (1,0) node[xshift=0.2cm,yshift=-0.1cm] (C) {$C$};
|
||||
\draw (-0.292893218813452, 0.292893218813452) node[xshift=0.15cm,yshift=0.1cm] (D) {$D$};
|
||||
\draw (0,0) node[xshift=0.1cm,yshift=0.2cm] (E) {$E$};
|
||||
\draw (-0.707106781186547, 0.707106781186547) node[xshift=-0.1cm,yshift=0.15cm] (F) {$F$};
|
||||
\draw (-0.5, 0.5) node[xshift=0cm,yshift=-0.25cm] (G) {$G$};
|
||||
\draw (-0.707106781186547, 0.292893218813452) node[xshift=-0.1cm,yshift=0.15cm] (H) {$H$};
|
||||
\draw (-0.414213562373095, 0) node[xshift=0cm,yshift=-0.2cm] (K) {$K$};
|
||||
\draw (0, -0.414213562373095) node[xshift=0cm,yshift=-0.15cm] (M) {$M$};
|
||||
\draw (-0.17157287525381, 0) node[xshift=-0.2cm,yshift=-0.2cm] (R) {$R$};
|
||||
\end{scriptsize}
|
||||
\end{tikzpicture}
|
||||
|
||||
\caption{}
|
||||
\label{step_pic3}
|
||||
\end{figure}
|
||||
|
||||
$\because BD \bot AR$，
|
||||
|
||||
$\therefore \angle ADB = \angle RDB = \dfrac{\pi}{2}$，
|
||||
|
||||
在$\mathrm{Rt}\triangle ADB$与$\mathrm{Rt}\triangle RDB$中
|
||||
|
||||
$$
|
||||
\begin{cases}
|
||||
\angle ADB = \angle RDB \\
|
||||
BD = BD \\
|
||||
\angle ABD = \angle EBD
|
||||
\end{cases}
|
||||
$$
|
||||
|
||||
$\therefore \triangle ADB \cong \triangle RDB (ASA)$
|
||||
|
||||
$\therefore AB = RB = 2m$，
|
||||
|
||||
$\because EB = BK + EK = m + n$，
|
||||
|
||||
$\because E$是$BC$的中点，
|
||||
|
||||
$\therefore BC = 2EB = 2m + 2n$，
|
||||
|
||||
$\therefore RC = BC - RB = 2m + 2n - 2m = 2n$，$EC = BC - EB = 2m + 2n - (m + n) = m + n$，
|
||||
|
||||
$\because EM$是$BC$的中垂线，
|
||||
|
||||
$\therefore EM \bot BC$，
|
||||
|
||||
$\therefore \angle MEC = \dfrac{\pi}{2}$，
|
||||
|
||||
$\because CB$平分$\angle ACM$，
|
||||
|
||||
$\therefore \angle ACB = \angle ECM = \alpha$，
|
||||
|
||||
$\therefore CM = EC \times \sec\alpha = RC \times \cos\alpha$，
|
||||
|
||||
$\because \cos\alpha = \sqrt{\dfrac{2m+n}{4m}}$
|
||||
|
||||
$\therefore (m + n)\sec\alpha = 2n\sqrt{\dfrac{2m+n}{4m}}$，
|
||||
|
||||
整理得
|
||||
\[
|
||||
\qty(\dfrac{4m^2}{2n^2})=1
|
||||
\]
|
||||
\[
|
||||
\dfrac{m^2}{n^2}=\dfrac{1}{2}
|
||||
\]
|
||||
\[
|
||||
\dfrac{m}{n}=\pm\dfrac{\sqrt{2}}{2}
|
||||
\]
|
||||
|
||||
因为$m>0$，$n>0$，所以
|
||||
$$
|
||||
\dfrac{m}{n}=\dfrac{\sqrt{2}}{2}
|
||||
$$
|
||||
|
||||
$\square$
|
||||
|
||||
\subsection{点\texorpdfstring{$F$}{}在\texorpdfstring{$\odot \mathrm{H}$}{}上的证明}
|
||||
|
||||
如图6，连接$HF$，$FB$，
|
||||
% 图6
|
||||
\begin{figure}[h]
|
||||
\centering
|
||||
|
||||
\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1.0cm,y=1.0cm,scale=5]
|
||||
\clip(-1.2,-0.2) rectangle (1.2,0.8);
|
||||
% 边框
|
||||
% \draw[red] (current bounding box.south west) rectangle (current bounding box.north east);
|
||||
\draw [line width=0.8pt] (-1.,0.)-- (1.,0.);
|
||||
\draw [line width=0.8pt] (-0.7071067811865475,0.7071067811865475)-- (0.,0.);
|
||||
\draw [line width=0.8pt] (1.,0.)-- (-0.7071067811865475,0.7071067811865475);
|
||||
\draw [line width=0.8pt] (-0.41421356237309503,0.5857864376269049)-- (-1.,0.);
|
||||
\draw [line width=0.8pt] (-0.41421356237309503,0.5857864376269049)-- (-0.2928932188134525,0.2928932188134525);
|
||||
\draw [line width=0.8pt] (-0.2928932188134525,0.2928932188134525)-- (-1.,0.);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.2928932188134524) circle (0.414213562373095cm);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.2928932188134524)-- (-0.2928932188134525,0.2928932188134525);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.2928932188134524)-- (-0.41421356237309503,0.);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (0.,0.)-- (-0.7071067811865475,0.2928932188134524);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.41421356237309503,0.5857864376269049)-- (-0.41421356237309503,0.);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.7071067811865475)-- (-1.,0.);
|
||||
\draw [line width=0.8 pt,dash pattern=on 3pt off 3pt] (-0.7071067811865475,0.7071067811865475)-- (-0.7071067811865475,0.2928932188134524);
|
||||
\begin{scriptsize}
|
||||
\draw (-0.414213562373095, 0.585786437626905) node[xshift=0.1cm,yshift=0.2cm] (A) {$A$};
|
||||
\draw (-1,0) node[xshift=-0.2cm,yshift=-0.1cm] (B) {$B$};
|
||||
\draw (1,0) node[xshift=0.2cm,yshift=-0.1cm] (C) {$C$};
|
||||
\draw (-0.292893218813452, 0.292893218813452) node[xshift=0.15cm,yshift=0.1cm] (D) {$D$};
|
||||
\draw (0,0) node[xshift=0.1cm,yshift=0.2cm] (E) {$E$};
|
||||
\draw (-0.707106781186547, 0.707106781186547) node[xshift=-0.1cm,yshift=0.15cm] (F) {$F$};
|
||||
\draw (-0.5, 0.5) node[xshift=0cm,yshift=-0.25cm] (G) {$G$};
|
||||
\draw (-0.707106781186547, 0.292893218813452) node[xshift=-0.1cm,yshift=0.15cm] (H) {$H$};
|
||||
\draw (-0.414213562373095, 0) node[xshift=0.15cm,yshift=-0.15cm] (K) {$K$};
|
||||
\end{scriptsize}
|
||||
\end{tikzpicture}
|
||||
|
||||
\caption{}
|
||||
\label{step_pic4}
|
||||
\end{figure}
|
||||
|
||||
设$\angle ABF = \beta$，
|
||||
|
||||
$\because E$是$BC$的中点，
|
||||
|
||||
$\therefore EB = EC$，
|
||||
|
||||
$\because \angle C = \angle EFC = \alpha$，
|
||||
|
||||
$\therefore EF = EC$，
|
||||
|
||||
$\therefore EF = EB$，
|
||||
|
||||
$\therefore \angle BFE = \angle EBF$，
|
||||
|
||||
$\because \angle EBF = \angle ABC + \angle ABF = 2\alpha + \beta$，
|
||||
|
||||
$\therefore \angle BFE = 2\alpha + \beta$，
|
||||
|
||||
$\therefore \angle BFC = \angle BFE + \angle EFC = 2\alpha + \beta + \alpha = 3\alpha + \beta$，
|
||||
|
||||
$\because \angle EBF + \angle C = \angle ABF + \angle ABC + \angle C = \beta + 2\alpha + \alpha = 3\alpha + \beta$，
|
||||
|
||||
$\therefore \angle BFC = \angle EBF + \angle C$，
|
||||
|
||||
$\because \angle BFC + \angle EBF + \angle C = \pi$，
|
||||
|
||||
$\therefore 2\angle BFC = \pi$，
|
||||
|
||||
$\therefore \angle BFC = \dfrac{\pi}{2}$，
|
||||
|
||||
$\because H$是$AB$中点，$\therefore HF=\dfrac{1}{2}AB$，
|
||||
|
||||
$\therefore HF = HA = HB = HD = HK = \dfrac{1}{2}AB$，
|
||||
|
||||
$\therefore$点$F$在$\odot \mathrm{H}$上. $\square$
|
||||
|
||||
\nocite{*}
|
||||
\printbibliography[heading=bibintoc, title=\ebibname]
|
||||
|
||||
\appendix
|
||||
%\appendixpage
|
||||
\addappheadtotoc
|
||||
|
||||
\end{document}
|
||||
Reference in New Issue
Block a user