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\centerline{MATH 392 -- Geometry Through History}
\centerline{Solutions for Problem Set 1}
\centerline{Friday, February 5}
\vskip 10pt
\noindent
2.8. 
\begin{enumerate}
\item[(a)]  Proposition I.9:  With straightedge and compass, one can bisect a given angle.
Proof:  Say the angle is $\angle ABC$.  Pick any one point $P$ on the ray $BA$, and lay 
off a segment of length equal to $BA$ on the other ray $BC$ (Proposition I.3).  
Call the end point of that segment $Q$, and join $PQ$ (Postulate I).  Construct the equilateral
triangle $\Delta PQR$, with $R$ on the opposite side of $PQ$ from $B$ (Proposition I.1) 
and join $BR$.  We claim that $BR$ bisects
$\langle ABC$.  To prove this, note that in $\Delta PBQ$, we have $\angle BPQ = \angle BQP$
since $BP = BQ$ (Proposition I.5).  For a similar reason, in the equilateral triangle $\Delta PQR$,
$\angle QPR = \angle PQR$ (Proposition I.5).  Hence by Common Notion 2,
$$\angle BPR = \angle BPQ + \angle QPR = \angle BQP + \angle PQR = \angle BQR.$$
In addition, $BP = BQ$ by construction and $PR = QR$ since $\Delta PQR$ is equilateral.
Hence by SAS (Propositon I.4), $\Delta BPR \cong \Delta BQR$.  It follows that the corresponding
angles $\angle PBR$ and $\angle QBR$ are equal, so the segment $BR$ bisects the angle $\angle PBQ$. 
\item[(b)]  Proposition I.10:  One can bisect a given segment.  Proof:  Let $AB$ be the 
line segment.  Construct the equilateral triangle $\Delta ABC$ (Proposition I.1).  Construct
$CD$ bisecting $\angle ACB$ (Proposition I.9) and extend to $E$ lying on the segment $AB$.
(Note: That finding such a point $E$ is possible really doesn't follow from any of Euclid's axioms; for this you need something like \emph{Pasch's Axiom} on page 17 of McCleary.  We claim 
$AE = EB$, so we have bisected the segment $AB$.  To prove this, consider the triangles 
$\Delta ACE$ and $\Delta BCE$.  First, note that $CA = CB$ by 
the fact that $\Delta ABC$ is equilateral. Then $\angle ACE = \angle BCE$ since $CE$ is constructed
to bisect $\angle ACB$.  Finally $EC$ is common to the two triangles.  Hence
$\Delta ACE \cong \Delta BCE$ by SAS (Proposition I.4).  Therefore the corresponding sides
$AE$ and $BE$ are equal, so $E$ bisects $AB$.  
\item[(c)]  Proposition I.11:  One can ``erect'' a perpendicular to a given line from 
any point on that line.  Proof:  Call the point $E$ and take $A$ to be any point on
the line other than $E$.  Lay off a line segment $EB = EA$ on the other side of $E$
along the line (Proposition I.3).  Then construct the equilateral triangle $\Delta ABC$ (Proposition I.1) and join $CE$ (Postulate 1).  We claim $\angle AEC$ and $\angle BEC$ are right angles
so $EC$ is the required perpendicular.  This follows from the same proof given for Proposition
I.10 in part (b).  As there $\Delta ACE \cong \Delta BCE$ by Proposition I.8 (SSS) 
Hence the corresponding 
angles $\angle AEC$ and $\angle BEC$ are equal.  But also $\angle AEC + \angle BEC = $ 2 right angles
so each of $\angle AEC$ and $\angle BEC$ is a right angle (Common Notion 3).
\item[(d)]  Proposition I.18:  In any triangle the greater side is opposite the greater angle.
Proof:  That is, we have to show that if we have a triangle $\Delta ABC$ and (for instance)
 $BC > AC$, then $\angle BAC > \angle ABC$.  Since $AC < BC$ we can lay off a segment of length
$AC$ starting at $C$ and ending at a point $D$ between $B$ and $C$ (Proposition I.3).  Join
$AD$ (Postulate I).  The triangle $\Delta ADC$ is isosceles, so $\angle CDA = \angle CAD$.
The angle $\angle ADC$ is an exterior angle to the triangle $\Delta ABD$.  Hence by Proposition 
I.16, $\angle CDA > \angle ABD$.  But also by Common Notion 5, we have $\angle BAC > \angle CAD$.  
Putting these together we get $\angle BAC > \angle CAD = \angle CDA > \angle ABD$.  This
is what we wanted to show.
\end{enumerate} 
\vskip 10pt
\noindent
2.9.  Say $AB$ and $CD$ are two lines with transversal line $AC$ and 
$\angle BAC + \angle DCA < 180^\circ$
(i.e. less than two right angles).  We need to show that $AB$ and $CD$, if extended indefinitely
will meet on the side containing $B$ and $D$.  If not, then those two lines would be parallel
(they cannot meet on the other side of the transversal -- the sum of the angles on
that side is greater than two right angles so the lines must diverge on that side by 
Proposition I.17).  However,
then Proposition I.29 implies that $\angle DAC = \angle ECA$ (for any point $E$ on the 
other side of $C$ on the line $CD$.  But then Proposition I.13 implies 
$\angle DCA = 180^\circ - \angle ECA = 180^\circ - \angle BAC$, so 
$\angle BAC + \angle DCA = 180^\circ$, which contradicts the hypothesis.  
\vskip 10pt
\noindent
2.13.  Let $\Delta ABC$ be inscribed in a circle with $AB$ a diameter and $C$ on
a semicircle cut out by the diameter.  We claim $\angle ACB$ is a right angle, \emph{assuming
Postulate V}.  Let $O$ be the center of the circle which is a point on $AB$ and join
$OC$ (Postulate 1).  Then $\Delta AOC$ and $\Delta BOC$ are isosceles triangles since 
$OA = OC = OB$ (definition of a circle).  By Proposition I.5, we have 
$\angle OAC = \angle OCA$ and $\angle OCB = \angle OBC$.  By Proposition I.32 (which depends on
Postulate V), the sum of the angles in $\Delta ABC$ is equal to two right angles.  But that sum 
equals $2 (\angle OCA + \angle OCB) = 2\angle ACB$.  Hence $\angle ACB$ is a right angle 
(Common Notion 3).
\vskip 10pt
\noindent
2.14.  There is a solution in the back of McCleary's book that uses several different
sorts of constructions based on Euclid's Proposition I.43 that we did not discuss. 
 It has the merit of producing what are probably the
\emph{most efficient} constructions.  But there is also another way that is based entirely on 
the fact proved in 2.13.  Here's an outline of this alternative method:  
\vskip 10pt
Step 0: First, note that given 
segments of lengths $x > y$, it is easy to construct segments of lengths $x + y$ and
$x - y$ by laying off segments along lines using Proposition I.3.  Then 
note that if we have any triangle
$\Delta ABC$ with $AB$ a diameter of the circumscribed circle, then $\angle ACB$ is
a right angle.  Hence if we drop the perpendicular $CD$ from $C$ to $AB$ (using Proposition I.12)    
then $\Delta ABC$, $\Delta ACD$ and $\Delta BCD$ are all \emph{similar triangles}. 
\vskip 10pt
Step 1:  given a segment $AD$ of length $x$ and a segment $DB$ of length $1$, make the segment
$AB$ of length $1 + x$, construct the circle with that diameter, and erect the perpendicular
to $AB$ from the point $D$, call (one of) the intersection points with the circle $C$ and
consider the triangle $\Delta ABC$ and the two smaller triangles $\Delta ACD$ and $\Delta BCD$.
These are similar by the observation above.
If $z$ is the length of $CD$, then by the fact that corresponding parts of similar triangles
are in the same proportion:
$$\frac{x}{z} = \frac{z}{1},\ {\rm or}\ x = z^2.$$
This can be read two ways:
\begin{itemize}
\item If we know $x$, it gives a construction of $z = \sqrt{x}$.
\item On the other hand if we know $z = DC$, we can start from 
$BD = 1$, do the construction of the perpendicular to the diameter, lay off $DC$
on that line, construct the perpendicular line to $BC$, find the point $A$ where 
that intersects the $BD$ extended, and call $AD = x$.  This gives a construction of $z^2$
if we know $z$.   
\end{itemize}
\vskip 10pt
Step 2:  Now suppose segments of lengths $x > y$ are known.  Consider this construction:
\begin{itemize}
\item Construct segments of lengths $x + y$ and $x - y$ using Step 0
\item Construct segments of lengths $(x + y)^2$ and $(x - y)^2$ using Step 1,
\item Construct a segment of length 
$$(x + y)^2 - (x - y)^2 = 4xy$$
by Step 0 again
\item Construction a segment of length $xy$ by bisecting the previous segment twice
(Proposition I.10).  Putting all of this together gives a construction of a segment of 
length $xy$.
\end{itemize}
\vskip 10pt
The only remaining part is to construct a segment of length $\frac{1}{x}$ given a 
segment of length $x$.  To do this proceed as in the second part of Step 1, except 
make $z = 1$ and $x$ the known segment along the diameter of the circle.  By similar
triangles, if $y$ is the remaining part of the diameter, we will have  
$$\frac{y}{1} = \frac{1}{x}$$
so $y$ is the reciprocal of $x$.

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