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\centerline{MATH 392 -- Geometry Through History}
\centerline{Saccheri Quadrilaterals and Consequences}
\centerline{Lecture Notes/Solutions From 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[]  {\it Solution:}  We have $AB = BM$ and $DB = BH$ by construction.
Moreover $\angle ABD = \angle MBH$ by equality of vertical angles at intersection
of two lines (Proposition I.15).  Therefore this congruence
is true by Proposition I.4 (SAS).  
\item[(B)]  Show that $\angle DHK > 90^\circ$.
\item[] {\it Solution:}  $\angle DHM = \angle ABD = 90^\circ$ 
by the congruence in part (A).
Then $\angle DHK > 90^\circ$ by Common Notion 5 (``the whole is greater than the part.'')
\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[]  The first part is true by the definition of a Saccheri quadrilateral:  $DH \perp DC$, 
$KC \perp DC$, and $DH = CK$.  Then Lemma 1 implies that the angle sum is equal to 
two right angles, plus two angles greater than a right angle.     
\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?
\item[]  The contradiction comes from the Saccheri-Legendre theorem.  If you 
join two opposite vertices of a quadrilateral to form two triangles, then 
the sum of the angles in the two quadrilaterals is $\le 2\cdot 180^\circ = 360^\circ$. 
But (C) says we have a Saccheri quadrilateral where the two summit angles are 
equal by Lemma 1, and also obtuse.  This would give an angle sum $> 360^\circ$. 
Therefore it must be true that $CM \ge 2 BD$, which is what we wanted to show.  (The 
contradiction came from assuming $CM < 2 BD$.) 
\end{enumerate}

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\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
{\it Solution:}  Join $CN$ and $DN$ first to form triangles $\Delta CHN$ and
$\Delta DKN$.  We have $CH = DK$ by definition of a Saccheri quadrilateral, 
$\angle CHN = \angle DKN$ by Lemma 1, and $HN = KN$ since $N$ is the midpoint of $HK$.  Therefore 
$\Delta CHN \cong \Delta DKN$ by SAS (Proposition I.4).  It follows that $CN = DN$ since
those are corresponding parts of the congruent triangles.  Now, we also have
$\Delta CMN \cong \Delta DMN$ by SSS (Proposition I.8).  It follows that 
$\angle CMN = \angle DMN$, so both of them are right angles.  The proof that 
$\angle HNM = \angle KNM$ are right angles
is similar, but it proceeds by starting from the diagonals $HM$ and $KM$, showing
$\Delta HMC \cong \Delta KMD$, then deducing $\Delta HMN \cong \Delta KMN$.

\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
{\it Solution:}  As in I above, HOA would imply that the angle sum in the Saccheri
quadrilateral is $> 360^\circ$.  But that is not possible because the Saccheri-Legendre
theorem implies that the sum is $\le 360^\circ$ (by splitting the quadrilateral
into two triangles with a diagonal).  
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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 $MDTN$? 
\vskip 10pt
\noindent
{\it Solution:}  Following the Hint:  Assume HAA, but $HK \le CD$.  Then 
$\frac{1}{2} HK \le \frac{1}{2} CD$ as well and we can extend $HK$ to a point $T$ beyond $K$
with $NT = \frac{1}{2} CD$.  Join $TD$ and consider the quadrilateral $MDTN$.  By II, the angles 
$\angle TNM$ and $\angle DMN$ are both right angles.  Moreover $NT = MD$ by construction.
So $MDTN$ is a Saccheri quadrilateral.  But the summit angles are now $\angle MDT > 90^\circ$
(Common Notion 5 again)
and $\angle NTD$ which must be equal.  This is a contradiction since we were assuming HAA.   

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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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