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\centerline{MATH 392 -- Geometry Through History}
\centerline{Class Monday, February 8}
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\noindent
{\it Background}
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Recall that on Friday we had started into a rather complicated proof of a result
showing that the usual fact about the sum of the angles in a triangle cannot
be assumed if we want to try to prove Postulate V from the other postulates.
In fact, Postulates I-IV plus that statement is equivalent to Postulates I-V as
Euclid stated them.  This proof is attributed
to the medieval Islamic mathematician Nasir al-Din al-Tusi (although 
this has sometimes been questioned).  

\begin{theorem}  Assume that Euclid's Postulates I-IV (\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 Propositions
1 - 28) hold.  Assume in addition that the angle sum in every triangle is two
right angles ($180^\circ$).  Then the statement of Postulate V also holds (as a theorem).  
\end{theorem}

What we will do today is work out the remainder of this proof in our discussion groups.
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\noindent
{\it Questions}
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\noindent
We said to start that, given a transversal line falling on two lines making angles
on one side summing to less than $180^\circ$, (if necessary) we could replace that transversal with
a different transversal for which one of the angles was a right angle and for which the angle sum
on that one side did not change.  (Note that showing the two lines meet using this other transversal
is sufficient for what we are trying to show!)  
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\noindent 
\emph{I.  Prove that we can always construct this other transversal.  (Hint:  Drop a perpendicular.  
You will need to use the assumption that the angle sum in all triangles is $180^\circ$.)}
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\noindent
Now, assuming the transversal line $AC$ makes a right angle with one line $CD$ and an 
acute angle with the other line $AB$, we began the argument as follows:  Let $G_1$
be an arbitrary point on the line $AB$ on the side of the transversal with the angle sum
less than $180^\circ$.  Drop a perpendicular to the transversal $AC$ from $G_1$ and call the 
foot $H_1$.  If $AH_1 > AC$, then the line $CD$ enters the triangle $\Delta AH_1G_1$ along
the side $AH_1$.  The line $CD$ is parallel to $H_1G_1$ since both make right angles
with $AH_1$ (which is $AC$, extended).  Hence Pasch's Axiom implies $CD$ must exit the
triangle $\Delta AH_1G_1$ through the other side and we are done in this case.  
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\noindent
\emph{II.  Why can't we just stop there?  Why do we need to do the next part of the 
argument?}
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\noindent
If $AH_1 \le AC$, then we argued as follows.  (By the Axiom of Continuity, also called the 
 \emph{Archimedean Axiom} -- see McCleary, p. 17), using Euclid's Proposition I.3, 
 we can lay off enough equal segments
 $$AH_1 = H_1H_2 = \cdots = H_{n-1}H_n$$
 to make $H_n$ lie ``strictly past'' $C$ along the line $AC$ (extended using 
 Postulate II).  We can also lay off equal segments 
 $$AG_1 = G_1G_2 = \cdots = G_{n-1}G_n$$
 (the same number $n$) along the line $AB$ (extending it as needed using Postulate II).  
 The theorem will be proved if we can show that for all $i \ge 2$, the point $H_i$ is the foot of the perpendicular
 dropped from $G_i$ to the line $AC$, extended to $AH_n$.  (Reason:  We will have $CD$ entering
 one side of the triangle $\Delta AH_nG_n$ and we can argue as before, using Pasch's Axiom, that
 the line $CD$, extended using Postulate II, must exit that triangle along the side $AG_n$, which
 is the extension of the line $AB$.)
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 So we need to show that $G_iH_i$ is perpendicular to $AH_n$ for all $i = 2, \ldots, n$.  
 To start, suppose $K$ is the foot of the perpendicular from $G_2$ to $AH_n$.  We must show $K = H_2$.
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 \noindent
 \emph{III.  Construct $AL$ perpendicular to $AH_n$ with $AL = H_1G_1$.  Show that 
 $\angle G_1AL = \angle AG_1H_1$ (using the assumption about the angle sum in triangles).
 Conclude that $\Delta G_1H_1A \cong \Delta G_1LA$, hence $\angle G_1LA$ is a right angle.}
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 \noindent
 Now construct a point $M$ on the line segment $KG_2$ so that $KM = H_1G_1$.   
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 \noindent
 \emph{IV.  Show that $\angle H_1G_1K = \angle MKG_1$ (again use the assumption about
 angle sums in a triangle).  Deduce that $\Delta MG_1K\cong H_1KG_1$ and $\angle KMG_1$
 is a right angle.}
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 \noindent
 \emph{V.  Explain why $M,G_1$, and $L$ must be collinear.  (Note:  that was not assumed, but
 it follows from what we have done to this point.)}
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 \noindent
 \emph{VI.  Next, show that $\Delta MG_2G_1 \cong \Delta LG_1A$.  Which congruence criterion 
 are you using?  Be sure it depends only on Postulates I - IV and Propositions 1 - 28.}
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 \noindent
 \emph{VII.  Now deduce that $H_1K = AH_1$ which implies that $K = H_2$ from the construction
 of the points $H_i$}.  
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 \noindent
 \emph{VIII.  What technique of proof would be most efficient to continue and show
 $H_i$ is the foot of the perpendicular from $G_i$ to $AH_n$ for all $i$?  Can you 
 see how that would go, without writing out all of the details?}
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