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使用引理或证明时换行

tag-icon • tag-icon text-manipulation proof-package lemma
使用引理或证明时换行

抱歉，MWE 太长了，但只有最后一部分出了问题，但我不确定是不是我的序言出了问题，我的引理或证明中的文本超出了页面范围，我看不到句子的结尾或我的参考文献。

\documentclass{book}

\usepackage{etoolbox}
\usepackage{parskip} %% <-- added
\usepackage{blindtext}
\usepackage{tkz-euclide}
\usepackage{latexsym}
\usepackage{amsmath,amsthm,amssymb}

% Standard packages
\usepackage{
float, 
graphicx
}
% Set page margins
\usepackage[top=1.0in, bottom=1.0in, left=1.0in, right=1.0in]{geometry}
\geometry{a4paper}
% Set nice page headers
\usepackage{fancyhdr}

\newcommand{\Tri}[8]{%
\tkzDefPoint(0,0){#1}
\tkzDefPoint({#2,#3}){#4}
\tkzDefPoint({#5,#6}){#7}
\tkzDrawPolygon[fill=#8](#1,#4,#7)
\tkzLabelPoint[left](#1){$#1$} 
\tkzLabelPoint[below](#4){$#4$}
\tkzLabelPoint[above](#7){$#7$}
\tkzDrawPoints(#1,#4,#7)
}

\setlength{\marginparwidth}{0pt}
% Paragraph style}
\setlength{\parindent}{0em}
\setlength{\parskip}{1em}

\pagestyle{fancy}

% Proposition environment
\newenvironment{proposition}
{\begin{center}\em}
{\end{center}}
%Diagram environment
\newenvironment{diagram}
{\stepcounter{CountDiag}\vspace*{10pt}

\begin{center}
\begin{tikzpicture}}
    {\end{tikzpicture}\vspace*{-5pt}\par Diagram~\theCountDiag
\end{center}}
%Step environment
\newenvironment{step}
{\stepcounter{CountStep}\vspace*{10pt} Step~\theCountStep\par
{\raggedright}}

\newcounter{CountStep} 
\newcounter{CountDiag}

%Reset counters   
\AtBeginEnvironment{proposition}{\setcounter{CountStep}{0}}
\AtBeginEnvironment{proposition}{\setcounter{CountDiag}{0}} 

\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}{Corollary}[theorem]
\newtheorem{lemma}[theorem]{Lemma}

\begin{document}

\section{Proposition 18}

\begin{proposition} 
In any triangle the angle opposite the greater side is greater.
\end{proposition}

\begin{lemma}
In any triangle $ABC$,if side $\overline{AC}$ greater than $\overline{AB}$,  then     $\angle{ABC}$ is also greater than $\angle{BCA}$.
\end{lemma}

\begin{diagram}
\Tri{B}{-2}{3}{A}{3}{0}{C}{blue,  opacity=0.1}
\end{diagram}

\begin{lemma}
\begin{align*}
&\text{If $\overline{AC}$ is greater than $\overline{AB}$,  then a point $D$ can be     constructed such that $\overline{AD}$ is equal to $\overline{AB}$} &&   I.3,     I.Post.1\\
\end{align*}
\end{lemma}

\begin{diagram}
\Tri{B}{-2}{3}{A}{3}{0}{C}{blue,  opacity=0.1}
%\tkzDrawCircle(A,B)
\tkzInterLC(A,C)(A,B)   \tkzGetPoints{E}{D}
\tkzDrawPoints(D)
\tkzLabelPoints[above](D)
\tkzDrawSegment(B,D)
\end{diagram}
\begin{proof}
\begin{lemma}
\begin{align*}
&\text{In $\triangle{ABC}$, if $\overline{AD}$ is constructed to be equal to     $\overline{AB}$, and $\overline{BD}$ is joined, then the exterior $\angle{ADB}$ is     greater than the interior and opposite $\angle{DCB}$.} &&  I.16\\
&\text{In $\triangle{BCD}$, the exterior angle $\angle{ADB}$ is equal to     $\angle{ABD}$ since $\overline{AD}$ is equal to $\overline{AB}$} && I.5
\end{align*}
\end{lemma}
Therefore in any triangle the angle opposite the greater side is greater.
\end{proof}

\end{document}

谢谢你的帮助。

在此处输入图片描述

答案1

在

\begin{lemma}
\begin{align*}
&\text{If $\overline{AC}$ is greater than $\overline{AB}$,
then a point $D$ can be
     constructed such that $\overline{AD}$ 
is equal to $\overline{AB}$} &&   I.3,     I.Post.1\\
\end{align*}
\end{lemma}

不align*执行任何操作并\text强制将所有文本放入单行水平盒子中。

只需使用

\begin{lemma}
If $\overline{AC}$ is greater than $\overline{AB}$,
then a point $D$ can be
     constructed such that $\overline{AD}$ 
is equal to $\overline{AB}$   I.3,     I.Post.1
\end{lemma}

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