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\begin{document}
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Math 202 \hfill Jerry L. Kazdan
\bigskip

{\hfill \large\bf Pythagorean Theorem: Using the Area of Similar Triangles \hfill}
\bigskip

\comments{
The standard figure associated with the Pythagorean Theorem is a right
triangle with squares drawn on each side and one compares the areas of
theses squares
\medskip
}

Euclid's Second Proof of the Pythagorean Theorem uses the
following figure:
\smallskip

\parbox{4.3in}{ Let $T$ be a right triangle whose sides have length
 $a,\, b,$ and $c$ ($c$ is the hypotenuse).  Partition it into two
 triangles, $T_1$ and $T_2$ by drawing a line (the altitude) from the
 right angle perpendicular to the hypotenuse.  We claim that
 the obvious formula
 \begin{equation} \label{area}
\text{Area}(T) =\text{Area}(T_1) + \text{Area}(T_2)
\end{equation}
is exactly the Pythagorean Theorem.
}
\parbox{2.3in}{\vskip -15pt
\quad \includegraphics[scale=0.60]{pyth-sim1a.eps}
}
\medskip

To see this, first, following Euclid, observe that the right
triangles $T_1$ and $T_2$ are both similar to $T$ (they have the same
angles), Thus we need to relate the areas of similar right triangles.
\bigskip

\parbox{4.3in}{
Let $T'$ be a right triangle with sides $a',\, b'$, and $c'$ that is
similar to $T$.  Then the length of the corresponding sides of $T$ and
$T'$ are proportional, that is, there is a \emph{scaling factor} $t>0$
so that
\[
a'=ta, \quad b'=tb,\quad c'=tc.
\]
}
\qquad \parbox{2.3in}{\vskip -25pt
\includegraphics[scale=0.50]{pyth-sim4R.eps} 
}

Every right triangle is half of a rectangle: \hskip-0pt
 \includegraphics[scale=0.21]{rectangle2.eps}  , 
so Area$(T)=\frac{1}{2}ab$ and 
\begin{equation*}%\label{area1}
\text{Area}(T') = \tfrac{1}{2}a'b'=
\tfrac{1}{2}(ta)(tb)  =t^2\text{Area}(T).
\end{equation*}
%
But $t =\frac{c'}{c}$, so we can restate this as
\begin{equation}\label{area2}
\text{Area}(T')=
\left[\frac{\text{hypotenuse}(T')} {\text{hypotenuse}(T)} \right] ^2
\text{Area}(T).
\end{equation}
Since for us
 $\text{hypotenuse}(T)=c,\ \text{hypotenuse}(T_1)=a, \text{ and }
\text{ hypotenuse}(T_2)=b$,
 equation \eqref{area2} implies
\[
\text{Area}(T_1)=(a/c)^2\text{Area}(T)\quad \text{and}\quad
\text{ Area}(T_2)=(b/c)^2\text{Area}(T).
\]
Thus, equation \eqref{area} states that
\[ 
\text{Area}(T) =(a/c)^2 \,\text{Area}(T) + (b/c)^2]\,\text{Area}(T).
\]
Dividing by $\text{Area}(T)$ reveals the \emph{Pythagorean formula}
for the right triangle $T$
\begin{equation}\label{Pyth}
c^2 = a^2 + b^2
\end{equation}
Note that the importance of the Pythagorean formula is not for
comparing areas, but, taking the square root of \eqref{Pyth}, for
finding the distance between two points in an orthogonal coordinate
system.

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\vfill

-----

[POLYA] \ Polya, G. {\it Induction and Analogy in Mathematics}, 
Princeton University Press 1954.

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