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\centerline{MATH 392 -- Geometry Through History}
\centerline{Saccheri Quadrilaterals and Consequences}
\centerline{Class Friday, February 12}
\vskip 10pt
\noindent
{\it Background}
\vskip 10pt
Recall that on Wednesday we were considering the proof of 

\begin{theorem}  The following are equivalent:
\begin{enumerate}
\item[(1)] Euclid's Postulates I-V (\emph{and} all the additional
facts such as Pasch's Axiom and the Axiom of Continuity 
that Euclid did not state explicitly, but that are needed for complete proofs of the 
Propositions in Book I of the {\it Elements}.)  
\item[(2)]  Euclid's Postulates I - IV (and all the facts as in (1)), plus:  The set of 
points equidistant from a given line is another line.
\end{enumerate}
\end{theorem}

We proved (1) $\Rightarrow$ (2) and saw the intuition behind (2) $\Rightarrow$ (1), but
we had some work left to establish the implication.  We had completed:

\begin{lemma}
\label{l1}
Let $CDKH$ be a quadrilateral with $CH \perp CD$, $DK\perp CD$ and $CH = DK$.  
Then $\angle CHK = \angle DKH$. 
\end{lemma}

Figures as in this Lemma 
are called ``Saccheri quadrilaterals'' because Girolamo Saccheri, S.J. considered
their properties extensively in his work on geometry.

\begin{lemma}[Saccheri-Legendre Theorem]
\label{l2}
Assume Euclid's Postulates I - IV (and all the facts as in (1) of the theorem) hold.
Then the angle sum in every triangle is $\le 180^\circ$.  
\end{lemma}

\noindent
To complete the proof of (2) $\Rightarrow$ (1), recall that we needed to show that 
non-parallel lines ``diverge'' on one side of their intersection point -- the distance 
between points on one line and the other line grows without bound.  This will follow from
our next statement:

\begin{lemma}
\label{l3}
Assume Euclid's Postulates I - IV (and all the facts as in (1) of the theorem) hold.
Let $\Delta ACM$ be a right triangle, with $\angle ACM = 90^\circ$.  Let $B$
be the midpoint of $AM$ and drop the perpendicular $BD \perp AC$ with $D$ on the 
segment $AC$.  Then $BD \le \frac{1}{2} CM$  (or equivalently $CM \ge 2 BD$).  
\end{lemma}

(The statement in parentheses is what guarantees the non-parallel 
lines $\overleftrightarrow{AM}$ and $\overleftrightarrow{AC}$ diverge as 
in the discussion above.)

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\noindent
{\it Questions}
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\noindent
I.  The proof of Lemma~\ref{l3} is based on this construction:
Extend $DB$ to $DH$ with $BH = DB$.  Arguing by contradiction, assume
$CM < 2 BD$.  Then we can extend the line segment $CM$ (Postulate II), lay off a
segment equal to $2 BD$ along that line, and join $HK$ to form a quadrilateral
$DCHK$. (Note that the order of the points $C,D$ has been reversed from the above -- I'm
listing the vertices moving counterclockwise around the quadrilateral.)
Follow this outline to provide a proof of Lemma~\ref{l3}:
\begin{enumerate}
\item[(A)]  Show that $\Delta ADB \cong \Delta MBH$.
\item[(B)]  Show that $\angle DHK > 90^\circ$.
\item[(C)]  Explain why $DCHK$ is a Saccheri quadrilateral, and deduce that 
the sum of the angles in that quadrilateral must be $> 360^\circ$.
\item[(D)]  However, show that (C) gives a contradiction to Lemma~\ref{l2}
(the Saccheri-Legendre Theorem).  Hint:  What does that result say about 
the angle sum in a quadrilateral?
\end{enumerate}

\vskip 10pt
\noindent
II.  Let $CDKH$ be a Saccheri quadrilateral and let $M$ be the midpoint of $CD$
and $N$ be the midpoint of $HK$.  Show that $MN \perp CD$ and $MN \perp HK$.

\vskip 10pt 
\noindent
Following Lemma~\ref{l1} Saccheri basically considered three alternatives in his work.
In a Saccheri quadrilateral $CDKH$ as above, recall $\angle CHK = \angle DKH$.
\begin{enumerate}
\item[] HAA: ``hypothesis of the acute angle'':  the equal angles $\angle CHK = \angle DKH$
are acute.
\item[] HRA: ``hypothesis of the right angle'':  the equal angles $\angle CHK = \angle DKH$
are right angles.  (This is the case in Euclidean geometry, where a Saccheri quadrilateral
is just an ordinary rectangle.)
\item[] HOA: ``hypothesis of the obtuse angle'':  the equal angles $\angle CHK = \angle DKH$
are obtuse.
\end{enumerate}

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\noindent
III.  Show that HOA contradicts the Saccheri-Legendre Theorem (Lemma~\ref{l2}).

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\noindent
IV.  Show that HAA implies $HK > CD$ in the Saccheri quadrilateral.  Hint:  Argue
by contradiction, starting from the assumption $HK \le CD$.  Use the result of II above.
If all this is true, you
can extend the line segment $HK$ to a point $T$ beyond $K$ with 
the distance from the midpoint $N$ satisfying $NT = \frac{1}{2} CD$.  What is 
now true about the quadrilateral $MDTN$?  

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Moreover, it follows that the sum of the angles in any triangle is $< 180^\circ$ under
HAA.  Saccheri then tried to eliminate HAA to show that Euclidean geometry was the only 
possibility (showing that Postulate V follows from I - IV).  However, even though
he eventually found a result that he claimed ``was repugnant to the nature of 
a straight line'' and he rejected HAA for that reason, he was actually proving
true theorems about \emph{hyperbolic geometry}.  We will begin to 
consider this next week(!)
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