\documentclass{article}
\newtheorem{theorem}{Theorem}
\title{Proof of Desargues's Theorem}
\author{Po-Shen Loh}
\date{7 March 2001}
\begin{document}
\maketitle

We embed the problem in three space.  Consider the following
generalization:

\begin{theorem}
   Let $\mathcal{P}, \mathcal{Q}$ be planes, and let $ABC \in
   \mathcal{P}$, $DEF \in \mathcal{Q}$ such that $X = AD \cap BE \cap
   CF$ is defined.  Let $\ell = \mathcal{P} \cap \mathcal{Q}$.  Then
   $AB \cap DE \in \ell$, $BC \cap EF \in \ell$, and $CA \cap FD \in
   \ell$.  Complete the space by letting parallel lines meet at the
   point at infinity.
\end{theorem}

\noindent \textbf{Proof of Theorem 1:}

Since $AD$ and $BE$ intersect, the four points $\{A, D, B, E\}$ are
coplanar.  In particular, that means that $AB$ and $DE$ are not skew;
they therefore intersect somewhere.  Yet $AB \in \mathcal{P}$ and $DE
\in \mathcal{Q}$, so $AB \cap DE \in \mathcal{P} \cap \mathcal{Q} =
\ell$.  The other claims follow similarly.  


\vspace{2cm}

Desargues's Theorem follows immediately from this result; simply
project our configuration onto a plane, and we will have Desargue's
Theorem!  The line of collinearity will be the image of $\ell$, and
the point of concurrence will be $X$'s image.

\end{document}
