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\lhead{\sc{Math 4310}} 
\chead{8/22/2012} 
\rhead{\sc{Another direct proof}}
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\begin{document}

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The Greek mathematician Pythagoras is often credited with proving the following theorem.  We give a proof
that is essentially a single figure.  We include the full details explaining why this figure establishes
the Pythagorean Theorem.

\begin{pyth-theorem}
Consider a right triangle with leg lengths $a$ and $b$, and hypotenuse length $c$.  These lengths satisfy
$$
a^2+b^2 = c^2.
$$
\end{pyth-theorem}


\begin{proof}
We may arrange four copies of the right triangle into a square, as shown in the figure below.
\begin{figure}[h]
  \begin{center}
 \includegraphics[width=2.5in]{pythagorean-theorem.pdf}
 \caption{Four copies the right triangle, arranged in a square.}
 \end{center}
\end{figure}
We notice that the angle marked $\gamma$ is a right angle, since  $\alpha+\beta+\gamma=180^\circ$, and the angles $\alpha$ and $\beta$  are the two acute
angles in a right triangle, so their sum is $\alpha+\beta = 90^\circ$.  Therefore,
$\gamma=90^\circ$, and 
the quadrilateral in the center of the figure must be a square.

The entire square has side lengths $a+b$, and the central square has side length $c$.  We may compute the area $A$ of 
the entire square as $A=(a+b)^2$, or by adding up the areas of the four triangles plus the area of the central square,
$$
A = 4\cdot \left(\frac{1}{2} ab\right) + c^2.
$$
Thus we have
$$
(a+b)^2 = 2ab+c^2.
$$
We now multiply out the left-hand side to get
$$
a^2 + 2ab +b^2 = 2ab + c^2.
$$
Subtracting $2ab$ from each side, we get the desired equality, $a^2+b^2=c^2$.
\end{proof}

\end{document}