\documentclass[12pt]{article}
\usepackage{graphicx,amssymb,amsmath}
\newtheorem{prop}{Proposition}
\setlength{\topmargin}{-1in}
\setlength{\textheight}{9in}
\begin{document}
\begin{figure}
\label{Prop3}
\includegraphics[width=4truein,height=3truein]{ParallelTrans.png}
\caption{Figure for Propositition~\ref{trans}.}
\end{figure}

\noindent
\setcounter{prop}{2}
\begin{prop} 
\label{trans}
If $\overleftrightarrow{PQ}$ is parallel to $\overleftrightarrow{AB}$ in the direction 
of $B$ and $\overleftrightarrow{UV}$ is parallel to $\overleftrightarrow{PQ}$ in the direction of 
$Q$ (on the same side as $B$), then $\overleftrightarrow{UV}$ is parallel to $\overleftrightarrow{AB}$.
\end{prop}

(In other words, parallelism is transitive.)  There are two cases, depending on whether
$\overleftrightarrow{PQ}$ lies between the two other lines or outside the region between
them.  We will do the first of these; for the other, refer to McCleary's text.   

\begin{itemize}
\item $PA \perp \overleftrightarrow{AB}$ and $UP \perp \overleftrightarrow{PQ}$ as in 
the Figure above.  
\item Say $S$ is in $PQVU$.  The goal is to show that $\overleftrightarrow{US}$
meets $\overleftrightarrow{AB}$. 
\item Since $\overleftrightarrow{UV}$ is parallel to $\overleftrightarrow{PQ}$ in the 
direction of $Q$, line $\overleftrightarrow{US}$ meets $\overleftrightarrow{PQ}$
at some point $T$.  If we continue that line across $\overleftrightarrow{PQ}$, then
by Proposition 1 from last time, the line must continue and meet $\overleftrightarrow{AB}$.
\item On the other hand, suppose $S$ is in $ABQP$.  
\item Postulate I says we can join $US$ with a line segment crossing 
$\overleftrightarrow{PQ}$ at some $T$ and Proposition 1 again shows
that line must continue and cross $\overleftrightarrow{AB}$.
\end{itemize}

\vfill\eject
\begin{figure}
\label{Prop4}
\includegraphics[width=5.5truein,height=3.5truein]{AngleOfPar.png}
\caption{Figure for Proposition~\ref{AofPsymm}.}
\end{figure}


\begin{prop} 
\label{AofPsymm}
The angles of parallelism $\Pi(AP)$ in the direction of $B$ and $\Pi(AP)$
in the direction of $B'$ are equal if $B,B'$ lie along the same line through $A$, but on 
opposite sides of $A$. 
\end{prop}

\begin{itemize}
\item The proof consists of showing that the angles $\angle APS$ with 
$S$ on the right side of $PA$ yielding lines that meet 
$\overleftrightarrow{AB} = \overleftrightarrow{AB'}$ are exactly the same
as the angles $\angle APS'$ with $S'$ on the other side yielding lines that meet
$\overleftrightarrow{AB}$. 
\item Suppose $\overrightarrow{PS}$ meets $\overrightarrow{AB}$ at $C$.  
\item Lay off a segment $AC' = AC$ on the other side of $PA$ and join $PC'$.
\item Then $\Delta PAC \cong \Delta PAC'$ (why?)
\item This shows that $\angle S'PA = \angle SPA$ is the same angle on the other side
and the line through $S'$ meets $\overleftrightarrow{AB'}$ at $C'$.  
\item Hence the angles that ``work'' on the left contain all the angles that ``work'' on 
the right.    
\item Now, reverse the roles of the two sides and repeat the same argument. 
\end{itemize}
\vfill\eject

\begin{figure}
\label{Prop6}
\includegraphics[width=5truein,height=3.5truein]{Mon.png}
\caption{Figure for Proposition~\ref{monotone}.}
\end{figure}

\begin{prop} 
\label{PiLength}
The angle of parallelism $\Pi(AP)$ depends only on the length of the segment $AP$
and not on the exact locations of the points $A$ or $P$.  
\end{prop}

\begin{prop}
\label{monotone}
If Saccheri's HAA holds and $AP > AQ$, then $\Pi(AP) < \Pi(AQ)$.  (That is, the 
angle of parallelism is monotone decreasing as a function of the length $AP$.  
\end{prop}

\begin{itemize}
\item The proof of Proposition~\ref{monotone} consists of showing that both 
$\Pi(AP) = \Pi(AQ)$ and $\Pi(AP) > \Pi(AQ)$ lead to contradictions.
\item If $\Pi(AP) = \Pi(AQ)$, then the result from Euclid I.27 shows
that the parallels $\overrightarrow{QQ'}$ and $\overrightarrow{PP'}$ are
themselves parallel.  This means they have a common perpendicular line and 
that contradicts Theorem 3.14 in McCleary (one of Saccheri's results assuming HAA). 
\item If $\Pi(AP) > \Pi(AQ)$, then refer to the Figure above.
\item There exists $R$ in $ABP'P$ such that $\angle APR = \angle AQQ'$ (angles marked in blue
in the Figure)
\item But then $\overrightarrow{PR}$ must meet $\overleftrightarrow{AB}$ at
some point because that line lies below the parallel $\overleftrightarrow{PP'}$, and 
it must also meet $\overleftrightarrow{QQ'}$.
\item However the result of Euclid I.27 says that $\overleftrightarrow{PR}$ and
$\overleftrightarrow{QQ'}$ cannot meet since they are parallel in Euclid's sense.
\item This is a contradiction and it finishes the proof.
\end{itemize}

\end{document}
