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\begin{figure}
\includegraphics[width=5truein,height=4.1truein]{ParallelSymm.png}
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\noindent
\setcounter{prop}{1}
\begin{prop} If $\overleftrightarrow{PQ}$ is parallel to $\overleftrightarrow{AB}$ in the direction 
of $B$ (according to Gauss's definition of parallelism), then
$\overleftrightarrow{AB}$ is also parallel to $\overleftrightarrow{PQ}$ in the direction 
of $Q$, provided $B,Q$ are on the same side of the segment $AP$.  
\end{prop}

The construction:

\begin{itemize}
\item Drop $AR \perp PQ$
\item Say $S$ is in $ABQR \subset ABQP$.  The goal is to show that $\overleftrightarrow{AS}$
meets $\overleftrightarrow{PQ}$.  Call $\angle SAB = \alpha$.
\item Choose $C$ so that $\angle RAC = \frac{1}{2} \alpha$
\item Assume $\overrightarrow{AC}$ meets $\overleftrightarrow{PQ}$ at $D$.
\item Let $RE = RD$, with $E$ on the other side of $D$.  Hence $\angle SAB = \angle EAD = \alpha$
and $AE = AD$. (Why?)
\item Choose $F$ so $\angle ADF = \angle AED$.  Since $\overleftrightarrow{PQ}$ 
is the parallel to $\overleftrightarrow{AB}$ in 
the direction of $B$, the line $\overleftrightarrow{DF}$ meets the line 
$\overleftrightarrow{AB}$ at some point $G$ (see Proposition 1 from Monday).
\item Let $H$ be on $\overleftrightarrow{PQ}$ with $EH = DG$ and join $AH$.   
\end{itemize}

\noindent
{\bf Claim.}  $\Delta AEH$ (outlined in blue) $\cong \Delta ADG$ (outlined in red).  (Why?)  

\vskip 10pt 
\noindent
{\bf Conclusion.}  $\angle GAH = \angle GAS$, so $S$ is on the line $\overleftrightarrow{AH}$, 
which meets $\overleftrightarrow{PQ}$, which is what we wanted to show.  (Why?)

\noindent
What if $\overrightarrow{AC}$ does not meet $\overleftrightarrow{PQ}$?
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