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\begin{document}


\title{Right Triangles}

\maketitle

\section{An Ancient discovery}

\begin{figure}
\includegraphics{r-triang1.gif} 


\caption{\label{RT}}
\end{figure}

\textbf{Phil} Today we talk about right triangles. Here we have a
right triangle \( ABC \) where the angle \( \angle BAC \) is the
right angle. 
Thousands of years before the time of Christ, someone drew, from
the vertex $A$ of the right triangle, a streight line
perpendicular to the opposite side and discovered that the two
triangle \( DBA \) and \( DAC \) are similar to the original
triangle \( ABC \). This dicovery was not so much difficult, but
implies a lot of important things. 

First we notice that the smaller two trinagles divide the
original triangle, so that the following relation is clear about
the areas of the three triangles. 


\[
|\triangle DBA|+|\triangle DAC|=|\triangle ABC|.\]
Now the areas of similar triangles in general are proportional to
the squares of corresponding edges. Therefore, for the
corresponing edges. \( BA \), \( AC \), \( BC \),  


\[
|BA|^{2}+|AC|^{2}=|BC|^{2}.\]


This is the famous and ubiquitous Pythagorian theorem.

Now let's turn the triangle and place it on a coordinate system
with two perpendicular $X$-axis and $Y$-axis as in the following
figure, where $AB$ is parallel to the $X$-axis and $AC$ is
parallel to the $Y$-axis. 

\vspace{0.3cm}
{\begin{figure}
{\centering \includegraphics{r-triang3.gif} \par}


\caption{\label{Perpend}}
\end{figure}\par}
\vspace{0.3cm}

Now we draw, from the vertex \( D \) of \( \triangle DAE \),
another line
\( DE \) perpendicular to the edge \( CA \). Therefore the small
\( \triangle EAD \)
is similar to the original \( \triangle ABC \). The slope of the
line passing
through points \( B \) and \( C \) is by definition \( |AC|/|BA|
\). Similarly,
the slope of the line passing through \( A \) and \( D \), which
is perpendicular
to the first line, is \( -|EA|/|DE| \). Therfore, by the
similarity of the
two triangles, the slope of the line passing through \( A \)and
\( D \) is
\( -|BA|/|AC| \). we have thus obtained.

\textbf{Proposition.} If the slope of a line is \( \alpha \ne 0
\), then the
slope of the line perpendicular to the first line is \( -1/\alpha 
\).


\section{Translate!}

\textbf{Phil} The great Descartes taught us how important it is
to translate
a geometric problem into an algebraic problem, and thus founded
the method of
modern mathematics. Here we will work on translation from lengths
into angles
and vice versa.

\textbf{Problem 2.1.} In the triangle \( ABC \), let \( M \) be
the middle
point of \( BC \). Prove that the triangle is a right triangle if
\( AM \)
is equal to \textbf{\( BM \)}. 

\vspace{0.3cm}
{\begin{figure}
{\centering \includegraphics{r-trian4.gif} \par} 


\caption{\label{3ET}}
\end{figure}\par}
\vspace{0.3cm}

\textbf{Phil} Formulate the conclusion. 

\textbf{Jessica} \( \angle A \) or \( \angle B \) or \( \angle C
\) is the
right angle. 

\textbf{Phil} Formulate the given conditions.

\textbf{Jessica} \( |BM|=|CM|=|AM| \) .

\textbf{Phil} The conditions are given in terms of line segments.
The conclusion
to be proved is given in terms of angles. Therefore, try to
translate given
relations among lengths into relations among angles.

\textbf{Hint} Two isosceles triangles \( MAB \) and \( MAC \)
provide such
a translation.

\textbf{Jessica} \( \angle MAB=\angle B \), and \( \angle
MAC=\angle C \).

\textbf{Phil} Do these relations seem to be enough? 

\textbf{Jessica} I don't know.

\textbf{Phil} Are there any other conditions about these
angles?

\textbf{Jessica} \( \angle A+\angle B+\angle C=180^{\circ } \),
because the
sum of the angles in any triangle is \( 180^{\circ } \). Also \(
\angle A=\angle MAB+\angle MAC \).

\textbf{Phil} Can you combine and simplify all these conditions?

\textbf{Jessica} Yes, by substituting the first conditions to the
last condition,
we can eliminate \( \angle MAB \) and \( \angle MAC \) , which
are not explicit
in the conclusion. The result is \( \angle A=\angle B+\angle C
\). Oh, we can
further simplify the conditions. The result is \( 2\angle
A=180^{\circ } \)
or \( \angle A=90^{\circ } \). That's it. That's the desired
conclusion.

\textbf{Stephan} It seems to me in the beginning through drawing
that $\angle A$ should
be the right angle. Therefore I tried to prove \( \angle
MAB+\angle MAC=90^{\circ } \).
I tried it to move these into a constant place and relate them.

Phil That's a good idea.

\textbf{Stephan} Since an external angle of a triangle is the sum
of opposing internal
angles, \( \angle AMB=\angle MAC+\angle C=2\angle MAC \) and \(
\angle AMC=\angle MAB+\angle B=2\angle MAB \).
Therefore \( 180^{\circ }=\angle AMB+\angle AMC=2(\angle
MAC+\angle MAB) \).
Hence \( \angle A=90^{\circ } \) .

\textbf{Phil} That's a good intuitive proof. 

\textbf{Problem 2.2.} In the triangle \( ABC \), let \( M \) be
the middle
point of \( BC \). Prove \( |AM|=|BM| \) if and only if \(
|BC|^{2}=|AB|^{2}+|AC|^{2} \).

\textbf{Phil} The conditions and the conclusions of this problem
are about length.
However, it is far easier to solve the problem indirectly through
transforming
these conditions and conclusions into relations about angles. 

\textbf{Problem 2.3.} Formulate and prove the inverse of the
Pythagorean theorem.

Phil The inverse of the Pythagorean theorem is usually proved by
means of the
Cosine Rule. However, we will try to prove it without using the
Cosine Rule. 


\section{Generalize! and Specialize!}

\begin{figure}
{\centering \includegraphics{r-trian5.gif} \par}


\caption{\label{CircumT}}
\end{figure}

\textbf{Phil} You may have noticed that the condition of the
Problem 2.1 means that the vertices \( A \), \( B \), \( C \) are
on a circle with center \( M. \) That observation may not help
you to prove the theorem, but helps you to draw a right triangle,
when its hypotenese \( BC \) is given. For example,
in the xfig program, you can draw a circle by clicking two
diameter points, in this case \textbf{\( B \)} and \( C \). then
the conclusion of the problem claims that any triangle with a
third vertex \( A \) on the circle is a right
triangle with a right angle \( \angle A \). 

\vspace{0.3cm}
{\begin{figure}
{\centering \includegraphics{cir-tri1.gif} \par}


\caption{\label{CircA}}
\end{figure}\par}

A next question is, ``What happens if \( BMC \) is not
necessarily a streight-line
segment as in Figure 5? '' This is a generalization of Fig. 4.
The problem seems not easy. Therefore, we again specialize the
general case, not exactly
the same way back to Fig. 4 but in a different way. Let \( BMC \)
stay as in Fig. 4, but move the vertex \( A \) so that \( AMB \)
is a streight-line segment.
We have obtained Fig. 6. In this figure, \( \triangle CAB \)
satisfies the corresponding condition of \( \triangle ABC \) in
Fig. 4.
However, this time,
we focus on \( \angle A \) and not on \( \angle C \), which is a
right angle.\begin{figure}
{\centering \includegraphics{cir-tri2.gif} \par}


\caption{\label{SCircA}}
\end{figure} 

\vspace{0.3cm}
{\begin{figure}
{\centering \includegraphics{cir-tri3.gif} \par}


\caption{\label{GCircA1}}
\end{figure}\begin{figure}
{\centering \includegraphics{cir-tri4.gif} \par}


\caption{\label{GCircT2}}
\end{figure}\par}

\end{document}
