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## Core latex/pdflatex auxiliary files:
*.aux
*.lof
*.log
*.lot
*.fls
*.out
*.toc
*.fmt
*.fot
*.cb
*.cb2
.*.lb
.DS_Store
clean.bat
elegant*-cn.pdf
elegant*-en.pdf
*.dates
## Intermediate documents:
*.dvi
*.xdv
*-converted-to.*
# these rules might exclude image files for figures etc.
# *.ps
# *.eps
# *.pdf
## Generated if empty string is given at "Please type another file name for output:"
.pdf
## Bibliography auxiliary files (bibtex/biblatex/biber):
*.bbl
*.bcf
*.blg
*-blx.aux
*-blx.bib
*.run.xml
## Build tool auxiliary files:
*.fdb_latexmk
*.synctex
*.synctex(busy)
*.synctex.gz
*.synctex.gz(busy)
*.pdfsync
## Auxiliary and intermediate files from other packages:
# algorithms
*.alg
*.loa
# achemso
acs-*.bib
# amsthm
*.thm
# beamer
*.nav
*.pre
*.snm
*.vrb
# changes
*.soc
# cprotect
*.cpt
# elsarticle (documentclass of Elsevier journals)
*.spl
# endnotes
*.ent
# fixme
*.lox
# feynmf/feynmp
*.mf
*.mp
*.t[1-9]
*.t[1-9][0-9]
*.tfm
#(r)(e)ledmac/(r)(e)ledpar
*.end
*.?end
*.[1-9]
*.[1-9][0-9]
*.[1-9][0-9][0-9]
*.[1-9]R
*.[1-9][0-9]R
*.[1-9][0-9][0-9]R
*.eledsec[1-9]
*.eledsec[1-9]R
*.eledsec[1-9][0-9]
*.eledsec[1-9][0-9]R
*.eledsec[1-9][0-9][0-9]
*.eledsec[1-9][0-9][0-9]R
# glossaries
*.acn
*.acr
*.glg
*.glo
*.gls
*.glsdefs
# gnuplottex
*-gnuplottex-*
# gregoriotex
*.gaux
*.gtex
# htlatex
*.4ct
*.4tc
*.idv
*.lg
*.trc
*.xref
# hyperref
*.brf
# knitr
*-concordance.tex
# TODO Comment the next line if you want to keep your tikz graphics files
*.tikz
*-tikzDictionary
# listings
*.lol
# makeidx
*.idx
*.ilg
*.ind
*.ist
# minitoc
*.maf
*.mlf
*.mlt
*.mtc[0-9]*
*.slf[0-9]*
*.slt[0-9]*
*.stc[0-9]*
# minted
_minted*
*.pyg
# morewrites
*.mw
# nomencl
*.nlg
*.nlo
*.nls
# pax
*.pax
# pdfpcnotes
*.pdfpc
# sagetex
*.sagetex.sage
*.sagetex.py
*.sagetex.scmd
# scrwfile
*.wrt
# sympy
*.sout
*.sympy
sympy-plots-for-*.tex/
# pdfcomment
*.upa
*.upb
# pythontex
*.pytxcode
pythontex-files-*/
# thmtools
*.loe
# TikZ & PGF
*.dpth
*.md5
*.auxlock
# todonotes
*.tdo
# easy-todo
*.lod
# xmpincl
*.xmpi
# xindy
*.xdy
# xypic precompiled matrices
*.xyc
# endfloat
*.ttt
*.fff
# Latexian
TSWLatexianTemp*
## Editors:
# WinEdt
*.bak
*.sav
# Texpad
.texpadtmp
# Kile
*.backup
# KBibTeX
*~[0-9]*
# auto folder when using emacs and auctex
./auto/*
*.el
# expex forward references with \gathertags
*-tags.tex
# standalone packages
*.sta
# generated if using elsarticle.cls
*.spl
# build
./build
./output

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% This work may be distributed and/or modified under the
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# Pond
Pond is an experimental project that uses the ElegantPaper template.
Pond 是一个试验性项目，使用了 ElegantPaper 模板。
## License
This project is released under the LaTeX Project Public License, v1.3c or later.
本项目发布遵循 LaTeX 项目公共许可证 1.3c 或更高版本。

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% Author: Dongsheng Deng
% Homepage: https://ddswhu.me/
% Email: elegantlatex2e@gmail.com
% Lastest Version: https://github.com/ElegantLaTeX/ElegantPaper
% --- Class structure: identification part
\ProvidesClass{elegantpaper}[2022/12/31 v0.11 ElegantLaTeX Paper class]
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%%%
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\DeclareVoidOption{founder}{\ekv{chinesefont=founder}}
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\DeclareDefaultOption{\PassOptionsToClass{\CurrentOption}{article}}
\ProcessKeyvalOptions*\relax
\LoadClass{article}
\RequirePackage{hyperref}
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pdfborder={0 0 0},
colorlinks=true,
linkcolor={winered},
urlcolor={winered},
filecolor={winered},
citecolor={winered},
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% \RequirePackage{linegoal}
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% caption settings
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% add the email cmd
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% font settings
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\AfterEndPreamble{
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%%% use yhmath pkg, uncomment following code
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% \let\widering\oldwidering
%%% use esvect pkg, uncomment following code
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%!TEX program = xelatex
% 完整编译: xelatex -> biber/bibtex -> xelatex -> xelatex
\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);
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\draw (0,0) node[xshift=0.1cm,yshift=0.2cm] (E) {$E$};
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\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);
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\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 (0,0) node[xshift=0.1cm,yshift=0.2cm] (E) {$E$};
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\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);
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\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) {$K$};
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\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
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\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
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\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}

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