saltos de línea al usar lema o prueba

2024-6-17 • tag-icon

Perdón por el largo MWE, pero solo la última parte sale mal, pero no estaba seguro si es mi preámbulo lo que hace que salga mal, el texto en mi lema o las pruebas se salen de la página y no puedo ver el final de la oración. o mi árbitro.

\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}

Gracias por cualquier ayuda.

Respuesta1

\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}

No align*hace nada y \textfuerza todo el texto en un hbox de una sola línea.

Solo usa

\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}

Respuesta1

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