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\documentclass[lang=cn,a4paper]{elegantpaper}

\title{Pond}
\author{作者 \\ Patrick}
\date{\zhdate{2025/03/09}}

% 本文档命令
\usepackage{array}
\usepackage{pgfplots}
\usepackage{soul}
\usepackage{physics}
\usepackage{amssymb}
\newcommand{\ccr}[1]{\makecell{{\color{#1}\rule{1cm}{1cm}}}}
\addbibresource[location=local]{reference.bib} % 参考文献，不要删除
\pgfplotsset{compat=1.15}
\usetikzlibrary{arrows}
\graphicspath{ {../images/} }

\begin{document}
\vspace*{\fill}
\begin{figure}[h]
  \begin{center}
    \includegraphics[width=0.5\textwidth]{THE_LAST_PROOF.png}
  \end{center}
\end{figure}
\begin{center}
  \sffamily {
    \textbf{\LARGE{Silence TLP}}

    \Large{2025 / 03 / 09}
  }
\end{center}
\vspace*{\fill}
\thispagestyle{empty}

DOI: 11.4514/sil.tlp.2025.0000001
\thispagestyle{empty}

\newpage

\setcounter{page}{1}
\maketitle

\begin{abstract}
  Pond 是一个试验性项目，使用了 ElegantPaper 模板。
  \keywords{Nonsense，Bullshit}
\end{abstract}

\section{题目}
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$G$，$GB = GE$，$DE = m$，$DF = n$.
\begin{enumerate}
  \item 在图1中找到与$\angle BAC$相等的角，并证明；

  \item 求$BD$的长（用$m$，$n$表示）；

  \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.
\end{enumerate}

% 图1 
\begin{figure}[h]
  \centering

  \begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1cm,y=1cm,scale=4]
    \clip(-1.2,-0.1) 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.);
    \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=-0.15cm,yshift=0.1cm] (F) {$F$};
      \draw (-0.5, 0.5) node[xshift=0cm,yshift=-0.25cm] (G) {$G$};
    \end{scriptsize}
  \end{tikzpicture}

  \caption{}
  \label{original_pic1}
\end{figure}

% 图2
\begin{figure}[h]
  \centering

  \begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1cm,y=1cm,scale=4]
    \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] (-0.7071067811865476,0.7071067811865476)-- (0.,0.);
    \draw [line width=0.8pt] (1.,0.)-- (-0.7071067811865476,0.7071067811865476);
    \draw [line width=0.8pt] (-0.414213562373095,0.585786437626905)-- (-1.,0.);
    \draw [line width=0.8pt] (-0.414213562373095,0.585786437626905)-- (-0.2928932188134525,0.2928932188134525);
    \draw [line width=0.8pt] (-0.2928932188134525,0.2928932188134525)-- (-1.,0.);
    \draw [line width=0.8pt] (0.,-0.4142135623730955)-- (1.,0.);
    \draw [line width=0.8pt] (-0.2928932188134525,0.2928932188134525)-- (0.,-0.4142135623730955);
    \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) {$A$};
      \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=-0.15cm,yshift=0.1cm] (F) {$F$};
      \draw (-0.5, 0.5) node[xshift=0cm,yshift=-0.25cm] (G) {$G$};
      \draw (0, -0.414213562373095) node[xshift=0cm,yshift=-0.15cm] (M) {$M$};
    \end{scriptsize}
  \end{tikzpicture}

  \caption{}
  \label{original_pic2}
\end{figure}

\section{解析}
\subsection{第一小问}

$\angle BAC = \angle BDE$.

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$HK$，$HE$，$HD$，

所以$HA=HB=HK=\dfrac{1}{2}AB$，

% 图3
\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);
    \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_pic1}
\end{figure}

$\because AD \bot BD$，$\therefore \angle ADB=\dfrac{\pi}{2}$，

$\because H$是$AB$中点，$\therefore HD=\dfrac{1}{2}AB$，

$\therefore HD=HA=HB=HK=\dfrac{1}{2}AB$，

$\therefore$点$D$在$\odot \mathrm{H}$上；

设$\angle EBD = \alpha$，

$\because BD$平分$\angle ABC$，

$\therefore \angle ABD = \angle EBD = \alpha$，

$\therefore \angle ABC = 2\alpha$，

$\because HB=HD$，

$\therefore \angle ABD = \angle HDB = \alpha$，

$\therefore \angle EBD = \angle HDB$，

$\therefore HD \mathop{//} BC$，

$\because GB=GE$，

$\therefore \angle ABC = \angle GEB = 2\alpha$，

$\because HB=HK$，

$\therefore \angle ABC = \angle HKB = 2\alpha$，

$\therefore \angle GEB = \angle HKB$，

$\therefore HK \mathop{//} EF$，

$\because HD \mathop{//} BC$，

$\therefore$四边形$HDEK$是平行四边形，

$\therefore HD=HK=EK$，

$\therefore \angle KHE = \angle KEH$，

$\because H$是$AB$的中点，$E$是$BC$的中点，

$\therefore HE \mathop{//} AC$，

$\therefore \angle KEH = \angle C$，

$\therefore \angle KHE = \angle KEH = \angle C$，

$\therefore \angle HKB = \angle KHE + \angle KEH = 2\angle C = 2\alpha$，

$\therefore \angle C = \alpha$，

$\because \angle GEB = \angle C + \angle F = \alpha + \angle F = 2\alpha$，

$\therefore \angle F = \alpha$，

$\therefore \angle C = \angle F = \alpha$；

$
  \begin{aligned}
    \because & \angle BDE = \pi - \angle EBD - \angle GEB = \pi - \alpha - 2\alpha = \pi - 3\alpha \\
             & \angle BAC = \pi - \angle ABC - \angle C = \pi - 2\alpha - \alpha = \pi - 3\alpha
  \end{aligned}
$，

$\therefore \angle BAC = \angle BDE$. $\square$

\subsection{第二小问}
如图4，连接$AK$，
% 图4
\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_pic2}
\end{figure}

$\because AB$是$\odot \mathrm{H}$直径，

$\therefore \angle AKB = \dfrac{\pi}{2}$，

$\because HD = HK$，

$\therefore$平行四边形$HDEK$是菱形，

$\because DE=m$，

$\therefore HD = EK = DE = m$，

$\therefore HF = HA = HB = HD = HK = m$，

$\therefore AB = 2m$，

$\because DF = n$，

$\therefore EF = DE + DF = m + n$，

$\because EB = EK + BK$，$EK = m$，

$\therefore EB = m + BK$，

$\therefore BK = n$，

$\therefore \cos \angle ABC = \cos 2\alpha = \dfrac{BK}{AB} = \dfrac{n}{2m}$，

$\therefore \cos \alpha = \cos \dfrac{2\alpha}{2} = \sqrt{\dfrac{1+\dfrac{n}{2m}}{2}} = \sqrt{\dfrac{2m+n}{4m}}$，

$\therefore BD = AB \times \cos \angle ABD = AB \times \cos\alpha = 2m \times \sqrt{\dfrac{2m+n}{4m}} = \sqrt{2m^2 + mn}$. $\square$

\subsection{第三乐章}
% 图5
如图5，记$AM$交$BC$于R，
\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,-0.5) 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);
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    \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);
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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.);
    \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] (-0.7071067811865475,0.7071067811865475)-- (-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}