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\centerline{MONT 104Q -- Mathematical Journeys:  From Known to Unknown}
\centerline{Proofs from Class on Friday, October 2}
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Here are writeups of the various proofs the groups presented in class
today on the problems from Wednesday's handout so that everyone has a 
``clean record'' of what we did.   The figures are on the back of this page.
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\noindent
B) Let $\Delta ABC$ be an equilateral triangle and let $D$ be a point on the side 
$BC$ different from $B,C$.  Prove that the line segment $AD$ is shorter than the sides of the triangle. 
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\noindent
\emph{Proof 1:}  (by contradiction).  Assume $AD$ is greater than the sides $AB,AC,BC$ of
the equilateral triangle $\Delta ABC$.  By Proposition 18, $\angle ACD > \angle ADC$ and $\angle ABD > \angle ADB$.
By ``Common Notion 6,'' this implies $\angle ACD + \angle ABD > \angle ADC + \angle ADB$.  By Proposition
13, $\angle ADC + \angle ADB = $ two right angles.  So  $\angle ACD + \angle ABD > $ two right angles.  But
that contradicts Proposition 17.  Therefore $AD$ is shorter than the sides of the triangle.
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\noindent
\emph{Proof 2:}  (also by contradiction).   Assume $AD$ is greater than the sides $AB,AC,BC$ of
the equilateral triangle $\Delta ABC$.  Then by Proposition 16, the exterior angle $\angle ADB$
to triangle $\Delta ADC$ is greater than the interior angle $\angle ACD$.  By Proposition 18, 
in the triangle $\Delta ABD$, we have $\angle ADB < \angle ABD$ since the side $AD > AB$
by assumption.  But Proposition 5 implies $\angle ABD = \angle ACD$ since $AB = AC$.  This is 
a contradiction because we have $\angle ADB > \angle ACD = \angle ABD > \angle ADB$.  (The angle
$\angle ADB$ cannot be greater than itself.)

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C)  Prove the following important fact:  Given a line and a point $P$ not on the line, 
the perpendicular from $P$ to the line (the result of the construction from Proposition 12) is the 
{\it shortest} line segment joining $P$ to any point on the line.  ({\it Note}: You will probably be
tempted to use the Pythagorean Theorem here.  We have not proved that!)

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\emph{Proof 1:}  (This is the one I saw first and I think this is where Isabel was headed -- another proof by contradiction):  Assume $PC \ge PQ$ (or equivalently $PQ \le PC$).  Then in the triangle $\Delta PQR$, 
by Proposition 18, we have $\angle PQC \ge \angle PCQ$.  But then $\angle PQC + \angle PCQ \ge $ 2 right angles.
This contradicts Proposition 17.  

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\emph{Proof 2:}  (This is what Katie was saying at the end.)  A direct proof (not by contradiction):  Since $\angle PCQ$
is a right angle, by Proposition 17, $\angle PQC$ is less than a right angle.  (If it weren't less, then it, plus
$\angle PCQ$ would be more than two right angles and that contradicts Proposition 17).  But since
$\angle PCQ > \angle PQC$, Proposition 19 implies that $PQ > PC$.

\begin{figure}
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\includegraphics[width = .5\linewidth]{diagB.png}
\caption{Figure for Question B}
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\begin{figure}
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\includegraphics[width = .5\linewidth]{diagC.png}
\caption{Figure for Question C}
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\end{figure}



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