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\centerline{MATH 392 -- Geometry Through History}
\centerline{Solutions/Lecture Notes for 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 (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}

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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
{\it Solution:}  Say the lines are $AB$ and $CD$ where $AC$ is the transversal line and
$\angle BAC + \angle DCA < 180^\circ$.  Drop a perpendicular $AK$ to $CD$ (where $K$ is the foot, 
a point lying on the line $CD$).  In the triangle $\Delta AKC$, note that 
$\angle KAC + \angle ACK = 90^\circ$ under the assumption that all triangles have angle sum 
equal to $180^\circ$.  Using the line $AK$ as the new transversal, we have angle $\angle BAK$ and 
the right angle $\angle AKD$.  But 
$$\angle BAK + 90^\circ = \angle BAK + \angle KAC + \angle ACK = \angle BAC + \angle ACD$$
so the sum of the angles for the new transversal is the same the angle sum for the old transversal
on that side.   (This leads to a diagram like the one we were considering in class last Friday.)
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\noindent
{\it It will help to refer to the diagram we drew on the board, or to page 32 in McCleary's book.}
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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 -- $AG_1 = $ the extension of $AB$ -- 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
{\it Solution:}  We cannot just stop there since there is nothing in Postulates I - IV (or Pasch's 
Axiom for that matter) that implies directly that the point $G_1$ can be chosen so that $AH_1 > AC$.  In other words,
the next steps are necessary.
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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 
 $$AG_1 = G_1G_2 = \cdots = G_{n-1}G_n$$
 (with 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 $AH_1$ 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
 {\it Solution:}  Extending $AC$ beyond the line $AL$ to a point $N$, we have $\angle NAL = 90^\circ$ and then
 $$180^\circ = 90^\circ + \angle LAG_1 + \angle G_1AH_1.$$
 On the other hand, by the assumption about the angle sum in the triangle $\Delta G_1H_1A$, we also have
 $$180^\circ = \angle G_1H_1A + \angle G_1AH_1 + \angle H_1G_1A = 90^\circ + \angle G_1AH_1 + \angle H_1G_1A.$$
 Comparing the last two equations yields $\angle LAG_1 = \angle H_1G_1A$.  Now we also have $AL = G_1H_1$  by construction and the side $AG_1$ is shared by the two triangles $\Delta G_1H_1A$ and $\Delta G_1LA$.
 By SAS (Proposition I.4), the triangles are congruent.  It follows that the corresponding angles
 $\angle G_1H_1A$ and $\angle G_1LA$ are congruent, so they are both right angles.
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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 $\Delta MG_1K\cong \Delta H_1KG_1$ and $\angle KMG_1$
 is a right angle.}
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 \noindent
 {\it Solution:}  This is similar to step III above:  At $K$ we have a straight angle along
 the line $AC$, so 
 $$180^\circ = 90^\circ + \angle MKG_1 + \angle G_1KH_1$$
 But in the triangle $\Delta H_1KG_1$ we have
 $$180^\circ = 90^\circ + \angle H_1G_1K + \angle G_1KH_1$$
 Combining these two equations, we get $\angle MKG_1 = \angle H_1G_1K$.  It follows by SAS 
 that $\Delta MG_1K\cong \Delta H_1KG_1$ since $KM = H_1G_1$ by construction and the side $G_1K$ is 
 common to the triangles.  Then, the corresponding angles $\angle KH_1G_1$ and $\angle KMG_1$
 are equal, so $\angle KMG_1$ 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
 {\it Solution:}  This follows because the sum 
 $$\angle MG_1K + \angle KG_1H_1 + \angle H_1G_1A + \angle AG_1L$$
 is $180^\circ$.  The first two add to $90^\circ$ because they equal the other two angles
 in one of the congruent right triangles $\Delta KMG_1$ or $\Delta KH_1G_1$; and the other two also 
 add to $90^\circ$ for the same sort of reason using the congruent right 
 triangles $\Delta G_1H_1A$ and $\Delta G_1LA$.
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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
 {\it Solution:}  Since $\angle KMG_1$ is a right angle, so is $\angle G_1MG_2$.  This is 
 the same as $\angle G_1LA$ by step III above.  Moreover, $\angle G_2G_1M = \angle AG_1L$
 since those are two vertical angles at the intersection of two lines as in Proposition I.15 
 ($A,G_1,G_2$ lie along one line, and $L,G_1,M$ lie along another by step V).  Finally $AG_1 = G_1G_2$
 by construction.  Hence $\Delta MG_2G_1 \cong \Delta LG_1A$ by ASA (Proposition I.26 -- note that 
 that proposition comes before I.29, so it depends only on Postulates I - IV(!))
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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
 {\it Solution:}  Using the triangle congruences from Steps IV, VI, III (in that order) we have
 $$H_1K = MG_1 = LG_1 = AH_1.$$
 Since point $H_2$ was constructed to make $H_1H_2 = AH_1$, and $K,H_2$ both lie
 along the line $AH_n$, we get $K = H_2$.   
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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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 \noindent
 {\it Solution:} This would be a great candidate for a proof by mathematical induction.
 What we did above is essentially the base case for the induction.  The induction step
 would be to prove that $H_k$ is the foot of the perpendicular to $AH_n$ through $G_k$, 
 under the assumption that $H_{k-1}$ is the foot of the perpendicular to $AH_n$ through
 $G_{k-1}$.  The proof would be similar, but it would require considering more triangles. 
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