\documentclass[12pt]{amsart}
\usepackage{geometry}
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\oddsidemargin-.8in\textwidth8in
\thispagestyle{empty}
\parskip12pt\parindent0pt
\usepackage{tikz}
\usetikzlibrary{intersections,through,backgrounds}
\begin{document}
\begin{center}
\Huge\sf \textbf{A proof of the Pythagorean theorem}
\end{center}

\bigskip

\begin{center}\LARGE
\begin{tikzpicture}[thick,scale=1.8]
\draw[fill=red](0,0)--(3,0)--(3,4)--(0,0);
\draw[fill=red](7,4)--(10,4)--(10,8)--(7,4);
\draw[fill=blue](0,0)--(0,4)--(3,4)--(0,0);
\draw[fill=blue](3,7)--(3,11)--(6,11)--(3,7);
\draw[fill=green](10,8)--(10,11)--(6,11)--(10,8);
\draw[fill=green](3,7)--(7,7)--(7,4)--(3,7);
\draw[fill=yellow](3,7)--(3,4)--(7,4)--(3,7);
\draw(0,4)
--node[midway, above]{$a$}(3,4)
--node[midway, left]{$a$}(3,7)
--node[midway, below]{$a$}(0,7)
--node[midway, right]{$a$}(0,4);
\draw(3,0)
--node[midway, above]{$b$}(7,0)
--node[midway, left]{$b$}(7,4)
--node[midway, below]{$b$}(3,4)
--node[midway, right]{$b$}(3,0);
\draw(3,7)
--node[midway, above]{$c$}(7,4)
--node[midway, left]{$c$}(10,8)
--node[midway, below]{$c$}(6,11)
--node[midway, right]{$c$}(3,7);
\node at (1.5,5.5){$a^2$};
\node at (5,2){$b^2$};
\node at (6.5,7.5){$c^2$};
\end{tikzpicture}

\bigskip

\Large\bf Therefore \boldmath{$a^2+b^2=c^2$} in any right-angled triangle
\end{center}
\begin{center}\tiny 
Peter Jipsen -- math.chapman.edu/mathposters -- Get your own copy
\end{center}
\end{document}
