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\begin{document}
\titlehead{Edited from \emph{The De Morgan Gazette} \textbf 8 no.\ 4 (2016) 27--78,
  \url{bit.ly/2hlyHzZ}, ISSN 2053--1451}
\title{Thales and the\\
Nine-point Conic}
\author{David Pierce}
\date{December 23, 2016\\
  July 25, 2017\\
March 14, 2018}
\publishers{Mathematics Department\\
Mimar Sinan Fine Arts University\\
%Istanbul\\
\url{http://mat.msgsu.edu.tr/~dpierce}}

\lowertitleback{
\chapter*{Abstract}
%\begin{abstract}
The nine-point circle is established by Euclidean means;
the nine-point conic, Cartesian.
Cartesian geometry is developed from Euclidean
by means of Thales\apo s Theorem.
A theory of proportion is given, and Thales\apo s Theorem proved,
on the basis of Book \textsc i of Euclid\apo s \emph{Elements,}
without the Archimedean assumption of Book \textsc v.
Euclid\apo s theory of areas is used,
although this is obviated by Hilbert\apo s theory of lengths.
It is observed how Apollonius relies on Euclid\apo s theory of areas.
The historical foundations of the name of Thales\apo s Theorem are considered.
Thales is thought to have identified water as a universal substrate;
his recognition of mathematical theorems as such
represents a similar unification of things.
%\end{abstract}
}
\maketitle

\chapter*{Preface}

I review here the theorem,
dating to the early nineteenth century,
that a certain collection of nine points,
associated with an arbitrary triangle,
lie on a circle.
Some of the nine points are defined
by means of the orthocenter of the triangle.
If the orthocenter is replaced by an arbitrary point,
not collinear with any two vertices of the triangle,
then the same nine points still lie on a conic section.
This was recognized in the late nineteenth century,
and I review this result too.
All of this is an occasion to consider
the historical and logical development of geometry,
as seen in work of Thales, Euclid, Apollonius,
Descartes, Hilbert, and Hartshorne.
All of the proofs here are based ultimately,
and usually explicitly,
on Book \textsc i of Euclid\apo s \emph{Elements.}
Even Descartes\apo s ``analytic'' methods are so justified,
although Descartes implicitly used the theory of proportion
found in Euclid\apo s Book \textsc v.
In particular,
Descartes relied on what is in some countries today called Thales\apo s Theorem.
Using only Book \textsc i of Euclid,
and in particular its theory of areas,
I give a proof of Thales\apo s Theorem in the following form:
\begin{quote}
If two triangles share a common angle,
the bases of the triangles are parallel
if and only if
the rectangles bounded by alternate sides of the common angle are equal.
\end{quote}
This allows an alternative development of a theory of proportion.

\begin{sloppypar}
An even greater logical simplification is possible.
As Hilbert shows, and as I review,
using the simpler arguments of Hartshorne,
one can actually \emph{prove}
Euclid\apo s assumption that a \emph{part}
of a bounded planar region is less than the whole region.
One still has to assume that a part of a bounded straight line
is less than the whole;
but one can now conceive of the theory of proportion
as based entirely on the notion of length.
This accords with Descartes\apo s geometry,
where length is the fundamental notion.
\end{sloppypar}

For Apollonius, by contrast,
as I show with an example,
areas are essential.
His proofs about conic sections are based on areas.
In Cartesian geometry,
we assign letters to lengths, known and unknown, 
so as to be able to do computations without any visualization.
This is how we obtain the nine-point conic.

We may however consider what is lost when we do not directly, 
\emph{visually,} consider areas.
Several theorems attributed to Thales
seem to be due to a visual appreciation of symmetry.
I end with this topic,
one that I consider further in \cite{Pierce-on-comm-symm}, 
an article not yet published when the present essay was published.

Aside from the formatting (I prefer A5 paper,
so that two pages can easily be read side by side on a computer screen,
or for that matter on an A4 printout)---aside from the formatting,
the present version of my essay differs from the version published in the \emph{De Morgan Gazette} as follows.
\begin{compactenum}
\item
  Reference \cite{Pierce-on-comm-symm}
  is now to a published paper,
  not to the preliminary version on my own webpage.
\item
    In the Introduction,
to the last clause of the 23rd paragraph,
I have supplied the missing \enquote{not,}
so that the clause now reads,
\enquote{doing this does not represent
  the accomplishment of a preconceived goal.}
\item
  In the last paragraph of \S\ref{sect:ortho},
  to what was a bare reference to \emph{Elements} \textsc i.15,
  I have added notice of Proclus's attribution to Thales
  and a reference to Chapter \ref{ch:unity}.
\item
  In Figure \ref{fig:Thales-full} in \S\ref{sect:TT},
  labelled points of intersection are no longer emphasized with dots.
\item
  In \S\ref{sect:Th-D},
  \begin{compactenum}
  \item
    Towards the end of the first paragraph,
    where the relation defined by \eqref{eqn:adbc} is observed to be transitive
    if we allow passage to a third dimension,
    the original passage was to a fourth dimension,
    in which $a\cdot d\cdot e\cdot f=b\cdot c\cdot d\cdot e$;
    \item
in the paragraph before Theorem \ref{thm:rect-prop},
after
\enquote{The definition also ensures that \eqref{eqn:abcd}
  \emph{implies} \eqref{eqn:adbc},}
I have added the clause,
\enquote{let $(x,y)=(c,d)$ in \eqref{eqn:ef}.}
  \end{compactenum}
\item
  In \S\ref{sect:L-p},
  \begin{compactenum}
    \item
the displayed proportion $(a:b)\comp(b:c)\as a:c$ is no longer numbered;
\item
I have deleted the second \enquote{this}
in the garbled sentence,
\enquote{Descartes shows this implicitly, 
  this in order to solve ancient problems.}
  \end{compactenum}
\item
  In \S\ref{sect:9-p-c},
  \begin{compactenum}
    \item
      at the end of the first paragraph,
      the reference to \enquote{p.\ 284}
      now has the proper spacing
      (the \TeX\ code is now
\verb*+p.\ 284+,
      but I had forgotten the backslash before);
      \item
  in the second paragraph,
in the clause that was
\enquote{the point now designed by $(x,y)$
  is the one that was called $(x-hy/a,y)$ before,}
I have corrected \enquote{designed} to \enquote{designated}
and made the minus sign a plus sign.
  \end{compactenum}
\item
  In \S\ref{sect:A},
  \begin{compactenum}
    \item
I have enlarged and expanded Figure \ref{fig:ay=x2};
\item
  in the fourth paragraph,
the clause beginning,
\enquote{If $d$ is the distance between the new origin,}
continues now with \enquote{and} (not \enquote{to});
\item
  in the fifth paragraph, the letter $\grm B$ at the end
  is now thus, a Beta, rather than $B$;
\item
  in the sixth paragraph, after the beginning,
  \enquote{A length $\grm H$,}
  the parenthesis \enquote{not depicted} is added;
\item
  the proportion \eqref{eqn:KL:}
  formerly had the same internal label as \eqref{eqn:KL},
  which meant that the ensuing \emph{proof} of \eqref{eqn:KL}
  was wrongly justified by a reference to itself,
  rather than to \eqref{eqn:KL:}.
  \end{compactenum}
\item
  In the third paragraph of Chapter \ref{ch:unity},
  \enquote{direction} has thus been made singular.
\end{compactenum}


\tableofcontents
\listoffigures
\chapter{Introduction}

According to Herodotus of Halicarnassus,
a war was ended by a solar eclipse,
and Thales of Miletus had predicted the year \cite[\textsc i.74]{Herodotus-Loeb}.
The war was between the Lydians and the Medes in Asia Minor;
the year was the sixth of the war.
The eclipse is thought to be the one that must have occurred 
on May 28 of the Julian calendar, 
in the year 585 before the Common Era
\cite[p.\ 15, n.\ 3]{Heath-Aristarchus}.

The birth of Thales is sometimes assigned to the year 624 \textsc{b.c.e.}
This is done in the \enquote{Thales} article on \emph{Wikipedia}
\cite{Thales-Wikipedia};
but it was also done in ancient times
(according to the reckoning of years by Olympiads).
The sole reason for this assignment
seems to be the assumption 
that Thales must have been forty when he predicted the eclipse
\cite[p.\ 76, n.\ 1]{Kirk}.

The nine-point circle was found early in the nineteenth century of the Common Era.
The 9-point conic,
a generalization,
was found later in the century.
The existence of the curves
is proved with mathematics that can be traced to Thales
and that is learned today in high school.

In the Euclidean plane,
every triangle determines a few points individually:
the incenter, circumcenter, orthocenter, and centroid.
In addition to the vertices themselves,
the triangle also determines various triples of points,
such as those in Figure \ref{fig:9}:
\begin{figure}
\centering
%\psset{unit=15mm}
\begin{pspicture}(0,-0.3)(7,5.7)%\psgrid
\psset{PointName=none,dotscale=1.6} 
\pstTriangle[PointSymbol=none,PosAngleA=60,PosAngleB=-90,PosAngleC=-90](4.5,1.9){A}(0,0){B}(7,0){C} 
%\psset{PointSymbol=*} 
\pstProjection BCA[D]
\pstProjection[PosAngle=135] CAB[E]
\pstProjection[PosAngle=0] ABC[F]
\psset{PointSymbol=diamond*} 
\pstMiddleAB[PosAngle=-90] BCG
\pstMiddleAB[PosAngle=-90] CAH
\pstMiddleAB[PosAngle=-90] ABK
\pstInterLL[PointSymbol=none,PosAngle=90] BECFP
\psset{RightAngleSize=0.21} % default is 0.28 unit
%\pstRightAngle ADC
%\pstRightAngle AEP
%\pstRightAngle CFA
\psset{linestyle=dotted}
\pstLineAB PA
\pstLineAB PB
\pstLineAB PC
\psset{PointSymbol=triangle*} 
\pstMiddleAB[PosAngle=45] PAL
\pstMiddleAB[PosAngle=180] PBM
\pstMiddleAB[PosAngle=45] PCN
\pstLineAB AD
\pstLineAB AE
\pstLineAB AF
\end{pspicture}
\caption{The nine points}\label{fig:9}
\end{figure}
the feet $\bullet$ of the altitudes,
the midpoints $\blacklozenge$ of the sides,
and the midpoints $\blacktriangle$ 
of the straight lines running from the orthocenter to the three vertices.
It turns out that these nine points lie on a circle,
called the \textbf{nine-point circle} 
or \textbf{Euler circle} of the triangle.
The discovery of this circle
seems to have been published first by Brianchon and Poncelet in 1820--1
\cite{BP-9},
then by Feuerbach in 1822 \cite{Feuerbach,F-9}.

Precise references for the discovery of the nine-point circle 
are given in Boyer\apo s
\emph{History of Mathematics}
\cite[pp.\ 573--4]{Boyer} and Kline\apo s
\emph{Mathematical Thought from Ancient to Modern Times}
\cite[p.\ 837]{MR0472307}.
However, 
as with the birth of Thales,
precision is different from accuracy.
Kline attributes the first publication 
on the nine-point circle to \enquote{Gergonne and Poncelet.}
In consulting his notes, 
Kline may have confused an author with the publisher,
who was himself a mathematician.
Boyer mentions that the joint paper of Brianchon and Poncelet
was published in Gergonne\apo s \emph{Annales.}
The confusion here is a reminder 
that even seemingly authoritative sources may be in error.

The mathematician is supposed to be a skeptic,
accepting nothing before knowing its proof.
In practice this does not always happen, even in mathematics.
But the ideal should be maintained,
even in subjects other than mathematics, like history.

There is however another side to this ideal.
We shall refer often in this essay to Euclid\apo s \emph{Elements}
\cite{Euclid-Heiberg,MR17:814b,Euclid-homepage,bones,MR1932864},
and especially to the first of its thirteen books.
Euclid is accused of accepting things without proof.
Two sections of Kline\apo s history 
are called \enquote{The Merits and Defects of the \emph{Elements}} \cite[ch.~4, \S10, p.~86]{MR0472307}
and \enquote{The Defects in Euclid} (ch.~42, \S1, p.~1005).
One of the supposed \enquote{defects} is, 
\begin{quote}
he uses dozens of assumptions that he never states and undoubtedly did not recognize.
\end{quote}
We all do this, all the time;
and it is not a defect.
We cannot state everything that we assume;
even the possibility of stating things is based on assumptions about language itself.
We try to state some of our assumptions,
in order to question them,
as when we encounter a problem with the ordinary course of life.
By the account of R. G. Collingwood \cite{Collingwood-EM},
the attempt to work out our fundamental assumptions is metaphysics.

\begin{sloppypar}
Herodotus says the Greeks learned geometry from the Egyptians,
who needed it in order to measure how much land they lost
to the flooding of the Nile.
Herodotus\apo s word \gr{gewmetr'ih} means also surveying,
or \enquote{the art of measuring land} \cite[\textsc{ii}.109]{Herodotus-Loeb}.
According to David Fowler in
\emph{The Mathematics of Plato\apo s Academy}
\cite[\S7.1(d), pp.\ 231--4; \S8.1, pp.\ 279--81]{MR2000c:01004},
the Egyptians defined the area of a quadrilateral field
as the product of the averages of the pairs of opposite sides.
\end{sloppypar}

The Egyptian rule is not strictly accurate.
Book \textsc i of the \emph{Elements} corrects the error.
The climax of the book is the demonstration,
in Proposition 45,
that every straight-sided field
is \emph{exactly} equal to a certain parallelogram with a given side.

Euclid\apo s demonstrations take place in a world where,
as Archimedes postulates \cite[p.~36]{MR2093668},
\begin{quote}
among unequal [magnitudes], 
the greater exceeds the smaller by such [a difference] 
that is capable, added itself to itself, of exceeding everything set forth\lips
%(of those which are in a ratio to one another).
\end{quote}
This is the world in which the theory of proportion 
in Book \textsc v of the \emph{Elements} is valid.
A theory of proportion is needed for the Cartesian geometry
whereby the nine-point conic is established.

One can develop a theory of proportion 
that does not require the Archimedean assumption.
Is it a defect that Euclid does not do this?
We shall do it,
trying to put into the theory
only enough to make \textbf{Thales\apo s Theorem} true.
This is the theorem that,
if a straight line cuts two sides of a triangle,
it cuts them proportionally
if and only if it is parallel to the third side.
I shall call this Thales\apo s Theorem for convenience,
and because it is so called in some countries today \cite{Thales-Theorem}.
There is also some historical basis for the name:
we shall investigate how much.

The present essay is based in part on notes
prepared originally for one of several twenty-minute talks 
at the \tur{Thales Bulu\c smas\i} 
(Thales Meeting),
held in the Roman theater 
in the ruins of Thales\apo s home town,
September 24, 2016.
The event was arranged by 
the Tourism Research Society
(\tur{Turizm Ara\c st{\i}rmalar{\i} Derne\u gi, TURAD})
and the office of the mayor of Didim.
Part of the Ayd\i n province of Turkey,
the district of Didim encompasses the ancient Ionian cities 
of Priene and Miletus,
along with the temple of Didyma.
This temple was linked to Miletus,
and Herodotus refers to the temple 
under the name of the family of priests,
the Branchidae.

My essay is based also on notes from a course
on Pappus\apo s Theorem and projective geometry given at my home university,
Mimar Sinan, in Istanbul,
and at the Nesin Mathematics Village,
near the Ionian city of Ephesus.

To seal the Peace of the Eclipse,
the Lydian King Alyattes gave his daughter Aryenis
to Astyages, son of the Median King Cyaxares
\cite[\textsc i.74]{Herodotus-Loeb}.
It is not clear whether Aryenis was the mother of Astyages\apo s daughter Mandane,
whom Astyages married to the Persian Cambyses,
and whose son Astyages tried to murder,
because of the Magi\apo s unfavorable interpretation of certain dreams
\cite[\textsc i.107--8]{Herodotus-Loeb}.
That son was Cyrus,
who survived and grew up to conquer his grandfather.
Again Herodotus is not clear that this was the reason
for the quarrel with Cyrus by Croesus
\cite[\textsc i.75]{Herodotus-Loeb},
who was son and successor of Alyattes and thus
brother of Astyages\apo s consort Aryenis.
But Croesus was advised by the oracles at Delphi and Amphiaraus
that, if he attacked Persia,
a great empire would be destroyed,
and that he should make friends with the mightiest of the Greeks
\cite[\textsc i.52--3]{Herodotus-Loeb}.
Perhaps it was in obedience to this oracle
that Croesus sought the alliance with Miletus
mentioned by Diogenes Laertius,
who reports that Thales frustrated the plan, and
\enquote{this proved the salvation of the city when Cyrus obtained the victory}
\cite[\textsc i.25]{Diogenes-I}.
Nonetheless,
Herodotus reports a general Greek belief---%
which he does not accept---%
that Thales helped Croesus\apo s army march to Persia
by diverting the River Halys (today\apo s K\i z\i l\i rmak) around them
\cite[\textsc i.75]{Herodotus-Loeb}.
But Croesus was defeated,
and thus his own great empire was destroyed.

When the victorious Cyrus returned east from the Lydian capital of Sardis,
he left behind a Persian called Tabulus to rule,
but a Lydian called Pactyes to be treasurer
\cite[\textsc i.153]{Herodotus-Loeb}.
Pactyes mounted a rebellion,
but it failed,
and he sought asylum in the Aeolian city of Cyme.
The Cymaeans were told by the oracle at Didyma to give him up
\cite[\textsc i.157--9]{Herodotus-Loeb}.
In disbelief, a Cymaean called Aristodicus 
began driving away the birds that nested around the temple.
\begin{quote}
But while he so did, a voice (they say) came out of the inner shrine
calling to Aristodicus, and saying,
\enquote{Thou wickedest of men, wherefore darest thou do this?
wilt thou rob my temple of those that take refuge with me?}
Then Aristodicus had his answer ready:
\enquote{O King,} said he,
\enquote{wilt thou thus save thine own suppliants,
yet bid the men of Cyme deliver up theirs?}
But the god made answer,
\enquote{Yea, I do bid them, that ye may the sooner perish for your impiety,
and never again come to inquire of my oracle 
concerning the giving up of them that seek refuge with you.}
\end{quote}
As the temple survives today,
so does the sense of the injunction of the oracle, 
in a Turkish saying
\cite[p.\ 108]{Ozdemir}:
\begin{quote}\centering
\tur{\.Isteyenin bir y\"uz\"u kara, vermeyenin iki y\"uz\"u.}\\
Who asks has a black face, but who does not give has two.
\end{quote}

From his studies of art, history, and philosophy,
Collingwood concluded that \enquote{all history is the history of thought}
\cite[p.\ 110]{Collingwood-Auto}.
As a form of thought, mathematics has a history.
Unfortunately this is forgotten in some \emph{Wikipedia} articles,
where definitions and results may be laid out
as if they have been understood since the beginning of time.
We can all rectify this situation, if we will,
by contributing to the encyclopedia.
On May 13, 2013,
to the article \enquote{Pappus\apo s Hexagon Theorem}
\cite{Pappus-Theorem-Wikipedia}, 
I added a section called \enquote{Origins,} giving Pappus\apo s own proof.
The theorem can be seen as lying behind Cartesian geometry.

In his \emph{Geometry} of 1637,
Descartes takes inspiration from Pappus, whom he quotes in Latin,
presumably from Commandinus\apo s edition of 1588
\cite[p.\ 6, n.\ 9]{Descartes-Geometry}; 
the 1886 French edition of the \emph{Geometry} has a footnote
\cite[p.\ 7]{Descartes-Geometrie}, 
seemingly in Descartes\apo s voice, although other footnotes are obviously from an editor:
\enquote{I cite rather the Latin version than the Greek text,
so that everybody will understand me more easily.}
%Few understand Latin today, and I am not one of them;
%in fact my Greek is better.
%This is an historical change.

The admirable 
\emph{Princeton Companion to Mathematics} \cite[pp.\ 47--76]{MR2467561}
says a lot about where mathematics is now in its history.
In one chapter, editor Timothy Gowers discusses 
\enquote{The General Goals of Mathematical Research.} 
He divides these goals among nine sections:
  \begin{inparaenum}[(1)]
    \item
Solving Equations,
\item
Classifying,
\item
Generalizing,
\item
Discovering Patterns,
\item
Explaining Apparent Coincidences,
\item
Counting and Measuring,
\item
Determining Whether Different
Mathematical Properties Are Compatible,
\item
Working with Arguments That
Are Not Fully Rigorous,
\item
Finding Explicit Proofs and Algorithms.
  \end{inparaenum}
These are some goals of research \emph{today.}
There is a tenth section of the chapter, but its title is general:
\enquote{What Do You Find in a Mathematical Paper?}
As Gowers says,
the kind of paper that he means
is one written on a pattern 
established in the twentieth century.

The \emph{Princeton Companion} is expressly not an encyclopedia.
One must not expect every species of mathematics 
to meet one or more of Gowers\apo s enumerated goals.
Geometrical theorems like that of the nine-point circle 
do not really seem to meet the goals.
They are old-fashioned.
The nine-point circle itself is not the \emph{explanation} of a coincidence;
it \emph{is} the coincidence that a certain set of nine points 
all happen to lie at the same distance from a tenth point.
A proof of this coincidence may be all the explanation there is.
The proof might be described as \emph{explicit,}
in the sense of showing how that tenth point can be found;
but in this case, there can be no other kind of proof.
From any \emph{three} points of a circle,
the center can be found.

One can \emph{generalize} the nine-point circle,
obtaining the \textbf{nine-point conic,}
which is determined by any four points in the Euclidean plane,
provided no three are collinear.
As in Figures \ref{fig:9-conic} and \ref{fig:9-ellipse},
\begin{figure}
\centering
\psset{xunit=8mm,yunit=6.4mm}
\begin{pspicture}(-6.5,-4)(7,7)%\psgrid
\pstGeonode[PointSymbol=none,PosAngle={180,0,-90,90}](-3,0)A(4,0)B(0,-1)C(0,6)D
\psset{PointName=none}
\psset{dotscale=1.6}
\pstInterLL[PosAngle=135] ABCDE
\pstInterLL[PosAngle=45] ACBDF
\pstInterLL[PosAngle=135] ADBCG
\ncline AB\ncline CD
\ncline AF\ncline BG
\ncline DF\ncline DG
\psset{PointSymbol=triangle*} 
\pstMiddleAB[PosAngle=45] ABH
\pstMiddleAB[PosAngle=-90] ACK
\pstMiddleAB[PosAngle=-165] ADL
\pstMiddleAB[PosAngle=-90] BCM
\pstMiddleAB[PosAngle=-15] BDN
\pstMiddleAB[PosAngle=-135] CDP
\psset{linestyle=dashed}
\psplot[plotpoints=200]{-6.5}7{x 0.25 sub dup mul 2 div 49 32 div add sqrt 1.25 add}
\psplot[plotpoints=200]{-6.5}7{x 0.25 sub dup mul 2 div 49 32 div add sqrt neg 1.25 add}
\end{pspicture}
\caption{A nine-point hyperbola}\label{fig:9-conic}
\end{figure}
\begin{figure}
\centering
\psset{xunit=12mm,yunit=11mm}
\begin{pspicture}(-1,-1)(7,5.2)%\psgrid
\pstGeonode[PointSymbol=none,PosAngle={90,-90,180,180}](5,0)A(7,0)B(0,-1)C(0,5)D
\psset{PointName=none}
\psset{dotscale=1.6}
\pstInterLL[PosAngle=135] ABCDE
\pstInterLL[PosAngle=45] ACBDF
\pstInterLL[PosAngle=135] ADBCG
\ncline EB\ncline CD
\ncline FC\ncline BD
\ncline GD\ncline BC
\psset{PointSymbol=triangle*} 
\pstMiddleAB[PosAngle=45] ABH
\pstMiddleAB[PosAngle=-90] ACK
\pstMiddleAB[PosAngle=-165] ADL
\pstMiddleAB[PosAngle=-90] BCM
\pstMiddleAB[PosAngle=-15] BDN
\pstMiddleAB[PosAngle=-135] CDP
\psset{linestyle=dashed}
\psplot[plotpoints=200]{-1}7{x dup 6 mul 7 add exch dup mul sub 7 div sqrt 1 add}
\psplot[plotpoints=200]{-1}7{x dup 6 mul 7 add exch dup mul sub 7 div sqrt 1 sub neg}
\end{pspicture}
\caption{A nine-point ellipse}\label{fig:9-ellipse}
\end{figure}
where the four points are $A$, $B$, $C$, and $D$,
the conic passes through the midpoints $\blacktriangle$
of the straight lines bounded by the six pairs formed out of the four points,
and it also passes through the intersection points $\bullet$
of the pairs of straight lines that together pass through all four points.

\begin{sloppypar}
The discovery of the nine-point circle itself
would seem not to be the accomplishment of any particular goal,
beyond simple enjoyment.
Indeed, 
one might make an alternative list of goals of mathematical research:
\begin{inparaenum}[(1)]
  \item
personal satisfaction,
\item
satisfying collaboration with friends and colleagues,
\item
\emph{impressing} those friends and colleagues,
\item
serving science and industry,
\item
winning a grant,
\item
earning a promotion,
\item
finding a job in the first place.
\end{inparaenum}
These may be goals in any academic pursuit.
But none of them can come to be recognized as goals
unless the first one or two have actually been achieved.
First you have to find out by chance that something is worth doing for its own sake,
before you can put it to some other use.
\end{sloppypar}

The present essay is an illustration of the general point.
A fellow alumnus of St John\apo s College \cite{Pierce-SJC}
expressed to me, along with other alumni and alumnae,
the pleasure of learning the nine-point circle.
Not having done so before,
I learned it too, and also the nine-point conic.
I wrote out proofs for my own satisfaction.
The proof of the circle uses the theorem
that a straight line bisecting two sides of a triangle
is parallel to the third.
This is a special case of Thales\apo s Theorem.
The attribution to Thales actually obscures some interesting mathematics;
so I started writing about this,
using the notes I mentioned.
Thales\apo s Theorem allows
a theoretical justification of the multiplication
that Descartes defines in order to introduce algebra to geometry.
The nine-point conic is an excellent illustration of the power of Descartes\apo s geometry;
Thales\apo s Theorem can make the geometry rigorous.
However, David Hilbert takes another approach.

The first-year students in my department in Istanbul
read Euclid for themselves in their first semester.
They learn implicitly about Descartes in their second semester,
in lectures on analytic geometry.
I have read Pappus with older students too, as I mentioned.
All of the courses have been an opportunity and an impetus 
to clarify the transition from Euclid to Descartes.
The nine-point circle and conic provide an occasion to bring the ideas together;
but doing this does not represent the accomplishment of a preconceived goal.

The high-school geometry course that I took in 1980--1 in Washington
could have included the nine-point circle.
The course was based on the text by Weeks and Adkins,
who taught proofs in the two-column, \enquote{statement--reason} format
\cite[pp.\ 47--8]{Weeks-Adkins}.
A 1982 edition of the text is apparently still available,
albeit from a small publisher.
The persistence of the text is satisfying,
though I was not satisfied by the book as a student.
The tedious style
had me wondering why we did not just read Euclid\apo s \emph{Elements.}
I did this on my own, 
and I did it three years later as a student at St John\apo s College.
%Book \textsc i of the \emph{Elements} is going to be the basis on which we establish the nine-point circle.

\begin{sloppypar}
After experiencing Euclid,
both as student and teacher,
I have gone back to detect
foundational weaknesses
in the Weeks--Adkins text.
One of them is the confusion of equality with sameness.
I discuss this in detail elsewhere \cite{Pierce-on-comm-symm}.
The distinction between equality and sameness is important,
in geometry if not in algebra.
In geometry, equal line segments have the same \emph{length};
but the line segments are still not in any sense the same segment.
An isosceles triangle has \emph{two} equal sides.
But in Euclid, 
two ratios are never equal, although they may be the \emph{same.}
This helps clarify what can be meant by a ratio in the first place.
In modern terms, a ratio is an equivalence class;
so any definition of ratio must respect this.
\end{sloppypar}

Another weakness of Weeks and Adkins has been shared by most modern books, 
as far as I know, 
since Descartes\apo s \emph{Geometry.} 
The weakness lies in the treatment of Thales\apo s Theorem.
The \emph{meaning} of the theorem,
the \emph{truth} of the theorem,
and the \emph{use} of the theorem to justify algebra---%
none of these are obvious.
In \emph{The Foundations of Geometry,}
Hilbert recognizes the need for work here,
and he does the work
\cite[pp.\ 24--33]{MR0116216}.
Weeks and Adkins recognize the need too,
but only to the extent of proving Thales\apo s Theorem for commensurable divisions,
then mentioning that there is an incommensurable case.
They do this in a section labelled [\textsc b] for difficulty and omissibility
\cite[pp.\ \textsc v, 212--4]{Weeks-Adkins}.
They say,
\begin{quote}
Ideas involved in proofs of theorems for incommensurable segments
are too difficult for this stage of our mathematics.
\end{quote}
Such condescension is annoying;
but in any case,
we shall establish Thales\apo s Theorem as Hilbert does,
in the sense of not using the Archimedean assumption
that underlies Euclid\apo s notion of commensurability.

First we shall use the theory of areas as developed in Book \textsc i
of the \emph{Elements.}
This relies on Common Notion 5:
the whole is greater than the part,
not only when the whole is a bounded straight line,
but also when it is a bounded region of the plane.
When point $B$ lies between $A$ and $C$ on a straight line,
then $AC$ is greater than $AB$;
and when two rectangles share a base,
the rectangle with the greater height is the greater.
Hilbert shows how to \emph{prove} the latter assertion from the former.
He does this by developing an \enquote{algebra of segments.}

We shall review this \enquote{algebra of segments};
but first we shall focus on the \enquote{algebra of areas.}
It is not really algebra,
in the sense of relying on strings of juxtaposed symbols;
it relies on an understanding of pictures.
In \emph{The Shaping of Deduction in Greek Mathematics}
\cite[p.\ 23]{MR1683176},
Reviel Netz examines how the diagram of an ancient Greek
mathematical proposition is not always recoverable from the text alone.
The diagram is an integral part of the proposition,
even its \enquote{metonym}:
it stands for the entire proposition
the way the enunciation of the proposition stands today
\cite[p.\ 38]{MR1683176}.
In the summary of Euclid called the \emph{Bones}
\cite{bones},
both the enunciations and the diagrams of the propositions of the \emph{Elements} are given.
Unlike, say, Homer\apo s \emph{Iliad,}
Euclid\apo s \emph{Elements} 
is not a work that one understands through hearing it recited by a blind poet.

The same will be true for the present essay,
if only because I have not wanted to take the trouble
to write out everything in words.
The needs of blind readers should be respected;
but this might be done best with tactile diagrams,
which could benefit sighted readers as well.
What Collingwood writes in \emph{The Principles of Art}
\cite[pp.\ 146--7]{Collingwood-PA}
about painting applies also to mathematics:
\begin{quote}
The forgotten truth about painting which was redis\-%
covered by what may be called the C\'ezanne--Berenson ap\-%
proach to it was that the spectator\apo s experience on looking 
at a picture is not a specifically visual experience at all\lips
It does not 
belong to sight alone, it belongs also (and on some occasions
even more essentially) to touch\lips
[Berenson]
is thinking, or thinking in the main, of distance and 
space and mass: not of touch sensations, but of motor 
sensations such as we experience by using our muscles and 
moving our limbs. But these are not actual motor sensations, 
they are imaginary motor sensations. In order to enjoy 
them when looking at a Masaccio we need not walk straight 
through the picture, or even stride about the gallery; what 
we are doing is to imagine ourselves as moving in these ways. 
\end{quote}
To imagine these movements, I would add,
one needs some experience of making them.
Doing mathematics requires some kind of imaginative understanding 
of what the mathematics is about.
This understanding may be engendered by drawings of triangles and circles;
but then it might just as well, if not better,
be engendered by triangles and circles that can be held and manipulated.

Descartes develops a kind of mathematics 
that might seem to require a minimum of imagination.
If you have no idea of the points that you are looking for,
you can just call them $(x,y)$ and proceed.
Pappus describes the general method
\cite[634, p.\ 82]{Jones}:
\begin{quote}
Now, analysis is the path from what one is seeking, as if it were 
established, by way of its consequences, to something that is established by 
synthesis. That is to say, in analysis we assume what is sought as if it has 
been achieved, and look for the thing from which it follows, and again what 
comes before that, until by regressing in this way we come upon some one 
of the things that are already known, or that occupy the rank of a first 
principle. We call this kind of method \senquote{analysis}, as if to say \emph{anapalin lysis}
(reduction backward). 
\end{quote}
The derivation of the nine-point conic will be by Cartesian analysis.

In Rule Four of the posthumously published 
\emph{Rules for the Direction of the Mind}
\cite[373, p.\ 17]{Descartes-Eng},
Descartes writes of a method that is
\begin{quote}
so useful\lips that without it the pursuit of learning would,
I think, be more harmful than profitable.
Hence I can readily believe that the great minds of the past
were to some extent aware of it,
guided to it even by nature alone\lips
This is our experience in the simplest of sciences,
arithmetic and geometry:
we are well aware that the geometers of antiquity
employed a sort of analysis which they went on to apply
to the solution of every problem,
though they begrudged revealing it to posterity.
At the present time a sort of arithmetic called \enquote{algebra}
is flourishing, and this is achieving for numbers
what the ancients did for figures\lips
But if one attends closely to my meaning,
one will readily see that
ordinary mathematics is far from my mind here,
that it is quite another discipline I am expounding,
and that these illustrations are more its outer garments
than its inner parts\lips
Indeed, one can even see some traces of this true mathematics, I think,
in Pappus and Diophantus who,
though not of that earliest antiquity,
lived many centuries before our time.
But I have come to think that these writers themselves,
with a kind of pernicious cunning,
later suppressed this mathematics
as, notoriously, many inventors are known to have done
where their own discoveries are concerned\lips
In the present age some very gifted men
have tried to revive this method,
for the method seems to me to be none other
than the art which goes by the outlandish name of \enquote{algebra}%
---or at least it would be
if algebra were divested 
of the multiplicity of numbers and imprehensible figures which overwhelm it 
and instead possessed that abundance of clarity and simplicity
which I believe true mathematics ought to have.
\end{quote}
Possibly Apollonius is, for Descartes, of \enquote{earliest antiquity};
but in any case he precedes Pappus and Diophantus by centuries.
He may have a secret weapon in coming up with his propositions about conic sections;
but \emph{pace} Descartes, 
I do not think it is Cartesian analysis.
One cannot have a method for finding things,
unless one already has---or \emph{somebody} has---%
a good idea of what one wants to find in the first place.
As if opening boxes to see what is inside,
Apollonius slices cones.
This is why we can now write down equations 
and call them conic sections.

Today we think of conic sections as having \emph{axes}:
one for the parabola, and two each for the ellipse and hyperbola.
The notion comes from Apollonius;
but for him, an axis is just a special case of a \emph{diameter.}
A diameter of a conic section bisects certain chords of the section
that are all parallel to one another.
In Book \textsc i of the \emph{Conics} \cite{Apollonius-Heiberg,MR1660991},
Apollonius shows that \emph{every} straight line 
through the center of an ellipse or hyperbola is a diameter in this sense;
and every straight line parallel to the axis is a diameter of a parabola.
One can give a proof by formal change of coordinates;
but the proof of Apollonius involves areas,
and it does not seem likely
that this is his translation of the former proof.
In any case, for comparison, 
we shall set down both proofs, for the parabola at least.

In introducing the nine-point circle 
near the beginning of his \emph{Introduction to Geometry} 
\cite[p.\ 18]{Coxeter-Geometry},
Coxeter quotes Pedoe 
on the same subject from \emph{Circles} \cite[p.\ 1]{MR1349339}:
\begin{quote}
  This [nine-point] circle is the first really exciting one to appear in any course
on elementary geometry.
\end{quote}
I am not sure whether to read this 
as encouragement to learn the nine-point circle, 
or as disparagement of the education 
that the student might have had to endure,
in order to be able to learn the circle.
In any case, all Euclidean circles are the same in isolation.
In Book \textsc{iii} of the \emph{Elements}
are the theorems that every angle in a semicircle is right (\textsc{iii}.31)
and that the parts of intersecting chords of a circle 
bound equal rectangles (\textsc{iii}.35).
The former theorem is elsewhere attributed to Thales.
Do not both theorems count as exciting?  
The nine-point circle is exciting 
for combining the triangles of Book~\textsc i
with the circles of Book \textsc{iii}.

\chapter{The Nine-point Circle}

\section{Centers of a triangle}\label{sect:ortho}

The angle bisectors of a triangle,
and the perpendicular bisectors of the sides of the triangle,
meet respectively at single points,
called today the \textbf{incenter} and \textbf{circumcenter} 
of the triangle \cite[pp.\ 187--8]{Weeks-Adkins}.
This is an implicit consequence of \emph{Elements} \textsc{iv}.4 and 5,
where circles are respectively inscribed in,
and circumscribed about, a triangle;
the centers of these circles are the points just mentioned.

The concurrence of the altitudes of a triangle
is used in the \emph{Book of Lemmas.}
The book is attributed ultimately to Archimedes,
and Heiberg includes a Latin rendition in his own edition of Archimedes
\cite[p.\ 427]{Heiberg-Archimedes-II}.
However, the book comes down to us originally in Arabic.
Its Proposition 4 
\cite[p.\ 304--5]{Heath-Archimedes}
concerns a semicircle with two semicircles removed,
as in Figure \ref{fig:arbelos};
\begin{figure}
\centering
\psset{unit=3mm}
\begin{pspicture}*(-2,-2)(27,14.5)%\psgrid
\psset{PosAngle=-90,PointSymbol=none}
\pstGeonode(0,0)A(9,0)B(25,0)C
\ncline%[linestyle=dotted]
AC
{\psset{PointName=none}
\pstMiddleAB ABE
\pstMiddleAB BCF
\pstMiddleAB ACG
\pstArcOAB EBA
\pstArcOAB FCB
\pstArcOAB GCA
\pstBissectBAC GBAH
}
\pstInterLC[linestyle=none,PosAngle=90] BHGAKD
\ncline%[linestyle=dotted] 
BD
\pstCircleAB BD
\pstRightAngle[RightAngleSize=2] CBD
\end{pspicture}
\caption{The arbelos}\label{fig:arbelos}
\end{figure}
the text quotes Archimedes as having called the shape an \textbf{arbelos,}
or shoemaker\apo s knife.
In the only such instance that I know of,
the big Liddell--Scott lexicon \cite[p.\ 235]{LSJ}
illustrates the \gr{>'arbhlos}
entry with a picture of the shape.
The term and the shape came to my attention,
before the high-school course that I mentioned,
in a \enquote{Mathematical Games} column of Martin Gardner
\cite[ch.\ 10, p.\ 149]{Gardner-Fractal}.
The second theorem that Gardner mentions
is Proposition 4 of the \emph{Book of Lemmas}:
the arbelos $ABCD$
is equal to the circle whose diameter is $BD$.
Proposition 5 is that the circles inscribed 
in the two parts into which 
the arbelos is cut by $BD$ are equal;
the proof appeals to the theorem
that the altitudes of a triangle concur at a point.

Today that point is the \textbf{orthocenter}
of the triangle,
and its existence follows from that of the circumcenter.
In Figure \ref{fig:perp},
\begin{figure}
\centering
\psset{unit=1.5cm}
\begin{pspicture}*(-0.5,-0.5)(6.5,4)%\psgrid
\psset{PointSymbol=none}
\pstTriangle[linestyle=dashed](0,2)H(3.5,0)K(6,2)G
\pstMiddleAB KHA
\pstMiddleAB GKB
\pstMiddleAB[PosAngle=65] HGC
\ncline AB
\ncline BC
\ncline CA
\pstProjection ABC[F]
\pstProjection[PosAngle=150] BCA[D]
\pstProjection[PosAngle=30] CAB[E]
\pstInterLL[PosAngle=90] ADBEL
\psset{RightAngleSize=0.14} % default is 0.28 unit
\pstRightAngle ADB
\pstRightAngle BEA
\psset{linestyle=dotted}
\ncline AL
\ncline BL
\ncline FL
\ncline CD
\ncline CE
\pstCircleOA[linestyle=dashed] LH
\end{pspicture}
\caption{Concurrence of altitudes}\label{fig:perp}
\end{figure}
where the sides of triangle $GHK$ 
are parallel to the respective sides of $ABC$,
the altitudes $AD$, $BE$, and $CF$ of $ABC$ \emph{are} 
the perpendicular bisectors of the sides of $GHK$.
Since these perpendicular bisectors concur at $L$,
so do the altitudes of $ABC$.

In Propositions 13 and 14 of \emph{On the Equilibrium of Planes}
\cite[p.\ 198--201]{Heath-Archimedes},
Archimedes shows that the center of gravity 
of a triangle must lie on a median,
and therefore must lie at the intersection of two medians.
Implicitly then, the three medians must concur at a point,
which we call the \textbf{centroid,}
though Archimedes\apo s language suggests that this is known independently.
The existence of the centroid
follows from the special case of Thales\apo s Theorem
that we shall want anyway,
to prove the nine-point circle.

\begin{theorem}\label{thm:2}
A straight line bisecting a side of a triangle
bisects a second side if and only if
the cutting line is parallel to the third side.
\end{theorem}

\begin{proof}
In triangle $ABC$ in Figure \ref{fig:lem},
\begin{figure}
\centering
\begin{pspicture}(0,-0.5)(4,3,5)
\psset{PointSymbol=none}
\pstTriangle(1,3)A(0,0)B(4,0)C
\pstMiddleAB BAD
\pstMiddleAB ACE
\pstMiddleAB CBF
\ncline DE
\ncline DF
\end{pspicture}
\caption{Bisector of two sides of a triangle}\label{fig:lem}
\end{figure}
let $D$ be the midpoint of side $AB$
(\emph{Elements} \textsc i.10),
and let $DE$ and $DF$ 
be drawn parallel to the other sides (\emph{Elements} \textsc i.31).
Then triangles $ADE$ and $DBF$ have equal sides between equal angles
(\emph{Elements} \textsc i.29),
so the triangles are congruent
(\emph{Elements} \textsc i.26).
In particular, $AE=DF$.
But in the parallelogram $CEDF$, $DF=EC$
(\emph{Elements} \textsc i.34).
Thus $AE=EC$.
Therefore $DE$ is the bisector of two sides of $ABC$.
Conversely,
since there is only one such bisector,
it must be the parallel to the third side.
\end{proof}

For completeness,
we establish the centroid.
In Figure \ref{fig:centroid},
\begin{figure}
\centering
\begin{pspicture}(0,-2)(5,4.5)
\pstTriangle(0,0)A(5,0)B(3.5,4)C
\pstMiddleAB ACE
\pstMiddleAB CBD
\pstMiddleAB[PosAngle=-70] BAF
\pstInterLL[PosAngle=110] ADBEG
\pstSymO[PosAngle=-90] FG[H]
\ncline AD\ncline BE\ncline CH\ncline AH\ncline BH
\pstSegmentMark[SegmentSymbol=MarkHash] AF
\pstSegmentMark[SegmentSymbol=MarkHash] FB
\pstSegmentMark CD
\pstSegmentMark DB
\psset{SegmentSymbol=gg}
\pstSegmentMark AD
\pstSegmentMark HB
\end{pspicture}
\caption{Concurrence of medians}\label{fig:centroid}
\end{figure}
if $AD$ and $CF$ are medians of triangle $ABC$,
and $BH$ is drawn parallel to $AD$,
then, by passing though $G$, 
$BE$ bisects $CH$ by the theorem just proved,
and the angles $FBH$ and $FAG$ are equal
by \emph{Elements} \textsc i.27.
Since the vertical angles $BFH$ and $AFG$ are equal
(by \emph{Elements} \textsc i.15,
which Proclus attributes to Thales,
as we shall discuss in Chapter \ref{ch:unity}),
and $BF=AF$,
the triangles $BFH$ and $AFG$ must be congruent
(\emph{Elements} \textsc i.26),
and in particular $BH=AG$.
Since these straight lines are also parallel,
so are $AH$ and $BE$
(\emph{Elements} \textsc i.33).
Again by Theorem \ref{thm:2},
$BE$ bisects $AC$.



\section{The angle in the semicircle}

We shall want to know that the angle in a semicircle is right.
This is Proposition \textsc{iii}.31 of Euclid;
but an attribution to Thales is passed along by
Diogenes Laertius,
the biographer of philosophers
\cite[I.24--5]{Diogenes-I}:
\begin{quote}
Pamphila says that,
having learnt geometry from the Egyptians,
he [Thales] was the first to inscribe in a circle
a right-angled triangle,
whereupon he sacrificed an ox.
Others say it was Pythagoras,
among them being Apollodorus the calculator.
\end{quote}
The theorem is easily proved
by means of \emph{Elements} \textsc i.32:
the angles of a triangle
are together equal to two right angles.
Let the side $BC$ of triangle $ABC$ in Figure \ref{fig:circ}
\begin{figure}
\centering
\psset{unit=3mm}
\begin{pspicture}(0,-2.5)(25,17.7)
\psset{PointSymbol=none}
\pstTriangle[PosAngleA=-90,PosAngleB=90,PosAngleC=-90](0,0){B}(9,12){A}(25,0){C} 
\pstMiddleAB[PosAngle=-90] BCD
\psset{linestyle=dashed}
\ncline AD
\pstArcOAB DCB
\psset{linestyle=dotted}
\pstGeonode[PosAngle=90](8.5,8)E
\ncline BE\ncline EC
\pstGeonode[PosAngle=90](9,16)F
\ncline BF\ncline FC
\end{pspicture}
\caption{The angle in the semicircle}\label{fig:circ}
\end{figure}
be bisected at $D$, and let $AD$ be drawn.
If $A$ lies on the circle with diameter $BC$,
then the triangles $ABD$ and $ACD$ are isosceles,
so their base angles are equal,
by \emph{Elements} \textsc i.5:
this is also attributed to Thales,
as we shall discuss later.
Meanwhile,
the four base angles being together equal to two right angles,
the two of them that make up angle $BAC$ must together be right.
The converse follows from \emph{Elements} \textsc i.21,
which has no other obvious use:
an angle like $BEC$ inscribed in $BAC$
is greater than $BAC$;
circumscribed, like $BFC$, less.

Thales may have established \emph{Elements} \textsc{iii}.31;
but it is hard to attribute to him the proof based on \textsc i.32
when Proclus, in his
\emph{Commentary on the First Book of Euclid\apo s \emph{Elements,}}
attributes this result to the Pythagoreans
\cite[379.2]{MR1200456}, who came after Thales.
Proclus cites the now-lost history of geometry by Eudemus,
who was apparently a student of Aristotle.

In his \emph{History of Greek Mathematics,}
Heath \cite[pp.\ 136--7]{MR654679} 
proposes an elaborate argument for \textsc{iii}.31
not using the general theorem about the sum of the angles of a triangle.
If a rectangle exists,
one can prove that the diagonals intersect at a point equidistant from the four vertices,
so that they lie on a circle whose center is that intersection point,
as in Figure \ref{fig:rect-circ}.
\begin{figure}
\subfloat[A rectangle in a circle]{\label{fig:rect-circ}
%\psset{unit=6mm}
\begin{pspicture}(-0.5,-1)(4.5,4)
\psset{PointName=none}
\pstTriangle[PosAngleA=-90](2,1.5)A(4,0)B(4,3)C
\pstSymO[PosAngle={135,-135}] A{B,C}[D,E]
\ncline AD\ncline AE
\pstCircleOA[linestyle=dashed] AB
\ncline BC\ncline CD\ncline DE\ncline EB
\end{pspicture}}
\hfill
\subfloat[A locus of right angles]{\label{fig:locus}
\psset{unit=2.5cm}
  \begin{pspicture}(-1,-1)(1,1)
\SpecialCoor
    \psarc[linestyle=dashed](0,0){1}0{180}
\psline(-1,0)(1,0)
\psline(1;0)(1;30)(1;180)
\psline(1;0)(1;60)(1;180)
\psline(1;0)(1;90)(1;180)
\psline(1;0)(1;120)(1;180)
\psline(1;0)(1;150)(1;180)
  \end{pspicture}}
\caption{Right angles and circles}
\end{figure}
In particular then,
a right angle is inscribed in a semicircle.

It seems to me one might just as well draw two diameters of a circle
and observe that their endpoints, by symmetry,
are the vertices of an equiangular quadrilateral.
This quadrilateral must then be a rectangle:
that is, the four equal angles of the quadrilateral must together make a circle.
This can be inferred from the observation that 
equiangular quadrilaterals can be used to tile floors.

Should the existence of rectangles be counted thus,
not as a theorem,
but as an observation,
if not a postulate?
In \emph{A Short History of Greek Mathematics,}
which is earlier than Heath\apo s history,
Gow \cite[p.\ 144]{Gow} 
passes along a couple of ideas of one Dr Allman about inductive reasoning.
From floor tiles, one may induce, as above,
that the angle in a semicircle is right.
By observation, 
one may find that the locus of apices of right triangles
whose bases (the hypotenuses) are all the same given segment
is a semicircle, as in Figure \ref{fig:locus}.

\section{The nine-point circle}


\begin{theorem}[Nine-point Circle]
In any triangle,
the midpoints of the sides,
the feet of the altitudes,
and the midpoints of the straight lines
drawn from the orthocenter to the vertices
lie on a single circle.
\end{theorem}

\begin{proof}
Suppose the triangle is $ABC$ in Figure \ref{fig}.
\begin{figure}
\centering
\psset{unit=15mm}
\begin{pspicture}(0,-0.5)(7,6)%\psgrid
{%
\psset{PointSymbol=none} 
\pstTriangle[PosAngleA=60,PosAngleB=-90,PosAngleC=-90](4.5,1.9){A}(0,0){B}(7,0){C} 
\pstProjection BCA[D]
\pstProjection[PosAngle=135] CAB[E]
\pstProjection[PosAngle=0] ABC[F]
\pstInterLL[PosAngle=90] BECFP
}
\pstMiddleAB[PosAngle=-90] BCG
\pstMiddleAB[PosAngle=-90] CAH
\pstMiddleAB[PosAngle=-90] ABK
\psset{RightAngleSize=0.21} % default is 0.28 unit
%\pstRightAngle ADC
\pstRightAngle AEP
\pstRightAngle CFA
\psset{linestyle=dashed}
\pstLineAB PA
\pstLineAB PB
\pstLineAB PC
\pstMiddleAB[PosAngle=45] PAL
\pstMiddleAB[PosAngle=180] PBM
\pstMiddleAB[PosAngle=45] PCN
\pstLineAB AD
\pstLineAB AE
\pstLineAB AF
\psset{linestyle=dotted}
\ncline GK
\ncline KL
\ncline LN
\ncline NG
\ncline GH
\ncline HL
\ncline LM
\ncline MG
\pstCircleAB GL
\end{pspicture}
\caption{The nine-point circle}\label{fig}
\end{figure}
Let the altitudes $BE$ and $CF$ be dropped
(\emph{Elements} \textsc i.12);
their intersection point is $P$.
Let $AP$ be drawn and extended as needed,
so as to meet $BC$ at $D$.
We already know that $P$ is the orthocenter of $ABC$,
so $AD$ must be at right angles to $BC$;
in fact we shall prove this independently.

Bisect $BC$, $CA$, and $AB$
at $G$, $H$, and $K$ respectively, 
%(\emph{Elements} \textsc i.10)
and bisect $PA$, $PB$, and $PC$
at $L$, $M$, and $N$ respectively.
By Theorem \ref{thm:2}, 
$GK$ and $LN$ are parallel to $AC$;
so they are parallel to one another,
by \emph{Elements} \textsc i.30.
Similarly $KL$ and $NG$ are parallel to one another,
being parallel to $BE$.
Then the quadrilateral $GKLN$ is a parallelogram;
it is a rectangle, 
by \emph{Elements} \textsc i.29,
since $AC$ and $BE$ are at right angles to one another.
Likewise $GHLM$ is a rectangle.
The two rectangles have common diagonal $GL$,
and so the circle with diameter $GL$
also passes through the remaining vertices of the rectangles,
by the converse of \emph{Elements} \textsc{iii}.31,
discussed above.
Similarly,
the respective diagonals $KN$ and $HM$ of the two rectangles
must be diameters of the circle,
and so $KHNM$ is a rectangle;
this yields that $AD$ is at right angles to $BC$.
The circle must pass through $E$, 
since angle $MEH$ is right, and $MH$ is a diameter;
likewise the circle passes through $F$ and $G$.
\end{proof}

The Nine-point Circle Theorem is symmetric in the vertices and orthocenter of the triangle.
These four points have the property that the straight line through any two of them
passes through neither of the other two;
moreorever, the line is at right angles to the straight line through the remaining two vertices.
In other words, the points are the vertices of a \textbf{complete quadrangle,}
and each of its three pairs of opposite sides are at right angles to one another.
The intersection of a pair of opposite sides being called a \textbf{diagonal point,}
a single circle passes through the three of these and the midpoints of the six sides.

We proceed to the case of a complete quadrangle
whose opposite sides need not be at right angles.
There will be a single \emph{conic section}
passing through the three diagonal points
and the midpoints of the six sides.
The proof will use Cartesian geometry,
as founded on Thales\apo s Theorem.

\chapter{The Nine-point Conic}

\section{Thales\apo s Theorem}\label{sect:TT}

A rudimentary form of Thales\apo s Theorem is mentioned
in the fanciful dialogue by Plutarch called
\emph{Dinner of the Seven Wise Men} \cite[\S2, pp.\ 351--3]{Plutarch-7}.
Here the character of Neiloxenus says \emph{of} and \emph{to} Thales,
\begin{quote}
he does not try to avoid, as the rest of you do, 
being a friend of kings and being called such. 
In your case, for instance, 
the king [of Egypt] finds much to admire in you, 
and in particular he was immensely pleased 
with your method of measuring the pyramid, 
because, without making any ado or asking for any instrument, 
you simply set your walking-stick upright 
at the edge of the shadow which the pyramid cast, 
and, two triangles being formed 
by the intercepting of the sun\apo s rays, 
\textbf{you demonstrated that the height of the pyramid 
bore the same relation to the length of the stick 
as the one shadow to the other.}
\end{quote}
The word translated as \enquote{relation} here,
in the sentence that I have emboldened,
is \gr{l'ogos}. 
This is usually translated as \enquote{ratio} in mathematics.
The whole sentence is
\begin{quote}
\centering
%ἔδειξας ὃν ἡ σκιὰ πρὸς τὴν σκιὰν λόγον εἶχε τὴν πυραμίδα πρὸς τὴν βακτηρίαν ἔχουσαν
  \gr{>'edeixas}\\
	\gr{<`on <h ski`a pr`os t`hn ski`an l'ogon e>~iqe}\\
	\gr{t`hn puram'ida pr`os t`hn bakthr'ian >'eqousan} \cite{Plutarch-7-Perseus};
\end{quote}
stretching the bounds of English style,
one might render this literally as
\begin{quote}
\centering
You showed,\\
what ratio the shadow had to the shadow,\\
the pyramid [as] having to the staff.
\end{quote}
It would be clearer to reverse the order of the last two lines.
If the pyramid\apo s height and shadow have lengths $P$ and $L$,
the shadow and height being measured from the center of the base,
while the lengths of Thales\apo s height and shadow are $p$ and $\ell$, 
then we may write the claim as
\begin{equation}\label{eqn:PpLl}
  P:p\as L:\ell.
\end{equation}
If not theoretically,
this must mean practically
\begin{equation}\label{eqn:PlLp}
P\cdot\ell=L\cdot p,
\end{equation}
that is, the rectangle of dimensions $P$ and $\ell$
is equal to the rectangle of dimensions $L$ and $p$.
Then what is being attributed to Thales
is something like the rectangular case of \emph{Elements} \textsc i.43:
\begin{quote}
In any parallelogram 
the complements of the parallelograms about the diameter 
are equal to one another.
\end{quote}
Thus in the parallelogram $ABCD$ in Figure \ref{fig:I.43},
\begin{figure}
\centering
\begin{pspicture}(0,-0.5)(6,4.5)
\psset{PointSymbol=none}
\pstTriangle(1,4)A(0,0)B(5,0)C
\pstMiddleAB[PointName=none] ACO
\pstSymO[PosAngle=45] OB[D]
\ncline CD\ncline DA
\pstHomO[HomCoef=0.4,PosAngle={180,40,90}] A{B,C,D}[E,G,H]
\pstInterLL EGCDF
\pstInterLL[PosAngle=-90] GHBCK
\ncline EF\ncline HK
\end{pspicture}
\caption{\emph{Elements} \textsc i.43}\label{fig:I.43}
\end{figure}
where $AGC$ is a straight line,
the parallelograms $BG$ and $GD$ are equal.
I pause to note that
Euclid\apo s two-letter notation for parallelograms here
is not at all ambiguous in Euclid\apo s Greek,
where a diagonal of a parallelogram may be \gr{<h AB},
while the parallelogram itself is \gr{t`o AB};
the articles \gr{<h} and \gr{t'o} are feminine and neuter respectively.
%\emph{Elements} \textsc i.43 concerns \emph{complementary} parallelograms, and 
Euclid\apo s word \gr{parapl'hrwma} for complement is neuter, 
like \gr{parallhl'ogrammon} \enquote{parallelogram} itself,
while \gr{gramm'h} \enquote{line} is feminine.
This observation, made in \cite{MR3312989},
is based on similar observations by Reviel Netz \cite{MR1683176}.

In Figure \ref{fig:I.43},
the parallelograms $BG$ and $GD$ are equal
because they are the result of subtracting equal triangles from equal triangles;
the equalities of the triangles
($AEG=GHA$ and so forth) are by \emph{Elements} \textsc i.34.
The theorem that Plutarch attributes to Thales
may now simply be the following.

\begin{theorem}\label{thm:Thales-right}\sloppy
In equiangular right triangles,
the rectangles bounded by alternate legs are equal.
\end{theorem}

\begin{proof}
Let the equiangular right triangles be $ABC$ and $AEF$
in Figure \ref{fig:rect}.
\begin{figure}
%\mbox{}\hfill
\psset{unit=9.5mm}
\psset{PointSymbol=none}
\begin{pspicture}(-0.3,-0.6)(5,4.6)%\psgrid
\pstTriangle[PosAngleA=-90,PosAngleB=-90,PosAngleC=90](0,0)A(5,0)B(0,4)C
\pstMiddleAB[PointName=none]BCO
\pstSymO[PointName=none] OA[D]
\pstHomO[HomCoef=0.4,PosAngle={-90,45}] A{B,C}[E, F]
\ncline EF
\pstTranslation[PosAngle=90] AE C[G]
\pstTranslation[PosAngle=90] AF B[H]
\ncline EG\ncline GC\ncline FH\ncline HB
\pstInterLL[PosAngle=135] EGFHL
\end{pspicture}
\hfill
\begin{pspicture}(0,-0.6)(5.3,4.6)
\pstTriangle[PosAngleA=-90,PosAngleB=-90,PosAngleC=90](0,0)A(5,0)B(0,4)C
\pstMiddleAB[PointName=none]BCO
\pstSymO[PosAngle=90] OA[D]
\pstHomO[HomCoef=0.4,PosAngle={-90,45}] A{B,C}[E, F]
\ncline BD\ncline CD
\ncline EF
\pstTranslation[PosAngle=90] AE C[G]
\pstTranslation[PosAngle=135] AF B[H]
\ncline EG\ncline FH
\pstInterLL[PosAngle=135] EGFHL
\ncline AL
\ncline[linestyle=dotted]LD
\end{pspicture}
%\hfill\mbox{}
\caption{The rectangular case of Thales\apo s Theorem}\label{fig:rect}
\end{figure}
We shall show
\begin{equation}\label{eqn:rect}
AF\cdot AB=AE\cdot AC.
\end{equation}
It will be enough to show 
$AF\cdot EB=AE\cdot FC$,
since we can then add the common rectangle $AE\cdot AF$ to either side.
Let rectangles $AG$ and $AH$ be completed,
as by the method whereby Euclid constructs a square
in \emph{Elements} \textsc i.46;
this gives us also the rectangle $AELF$.
Let $BH$ and $CG$ be extended to meet at $D$;
by \emph{Elements} \textsc i.43,
it will be enough to show that the diagonal $AL$,
extended, passes through $D$,
or in other words,
$L$ lies on the diagonal $AD$ of the rectangle $ABDC$.
But triangles $AFE$ and $FAL$ are congruent by \emph{Elements} \textsc i.4,
and likewise triangles $ACB$ and $CAD$ are congruent.
Thus
\begin{equation*}
\angle FAL=\angle AFE=\angle ACB=\angle CAD,
\end{equation*}
which is what we wanted to show.
\end{proof}

The foregoing is a special case of the following theorem,
which does not require the special case in its proof.

\begin{theorem}\label{thm:Thales-int}
In triangles that share an angle,
the parallelograms in this angle 
that are bounded by alternate sides of the angle
are equal
if and only if the triangles are equiangular.
\end{theorem}

\begin{proof}
Let the triangles be $ABC$ and $AEF$
in Figure \ref{fig:Thales},
\begin{figure}
\centering
\begin{pspicture}(0,-0.5)(6,4.5)
\psset{PointSymbol=none} 
\pstTriangle(0,0)A(5,0)B(1,4)C
\pstHomO[HomCoef=0.4,PosAngle={-90,180}] A{B,C}[E, F]
\ncline EF
\pstTranslation[PosAngle=90] AE C[G]
\pstTranslation AF B[H]
\ncline EG\ncline GC\ncline FH\ncline HB
\psset{linestyle=dotted}
\ncline EC\ncline FB
\end{pspicture}
\caption{Intermediate form of Thales\apo s Theorem}\label{fig:Thales}
\end{figure}
and let the parallelograms $AG$ and $AH$ be completed.
Let the diagonals $CE$ and $BF$ be drawn.
If the triangles are equiangular,
then we have
\begin{equation}\label{eqn:FE}
FE\parallel CB
\end{equation}
by \emph{Elements} \textsc i.27.
In this case,
as in Euclid\apo s proof of \textsc{vi}.2,
the triangles $FEC$ and $EFB$ are equal by \textsc i.37.
Adding triangle $AEF$ in common,
we obtain the equality of the triangles $AEC$ and $AFB$.
This gives us the equality of their doubles;
and these, by \textsc i.34, are just the parallelograms $AG$ and $AH$.

Conversely, if these parallelograms are equal,
then so are their halves,
the triangles $AEC$ and $AFB$;
hence 
the triangles $FEC$ and $EFB$ are equal,
so \eqref{eqn:FE} holds by \emph{Elements} \textsc i.39,
and then the original triangles $ABC$ and $AEF$
are equiangular by \emph{Elements} \textsc i.29.
\end{proof}

Proclus 
\cite[352.15]{MR1200456}
inadvertently gives evidence
that Thales could \emph{use} Theorem \ref{thm:Thales-right},
if not Theorem \ref{thm:Thales-int}.
Discussing \emph{Elements} \textsc i.26,
which is the triangle-congruence theorem whose two parts
are now abbreviated as A.S.A. and A.A.S. \cite[p.\ 62]{Weeks-Adkins},
and which we used earlier to prove the concurrence of the medians of a triangle,
Proclus says,
\begin{quote}
Eudemus in his history of geometry attributes the theorem itself to Thales,
saying that the method by which he is reported to have determined the distance of ships at sea
shows that he must have used it.
\end{quote}
If Thales really used Euclid\apo s \textsc i.26 for measuring distances of ships,
it may indeed have been by the method that 
Heath suggests \cite[pp.\ 132--3]{MR654679}:
climb a tower, note the angle of depression of the ship,
then find an object on land at the same angle.
The object\apo s distance is that of the ship.
This obviates any need to know the height of the tower,
or to know proportions.
Supposedly one of Napoleon\apo s engineers measured the width of a river this way.

Nonetheless, 
Gow 
\cite[p.\ 141]{Gow}
observes plausibly that the method that Heath will propose 
is not generally practical.
Thales must have had the more refined method of similar triangles:
\begin{quote}
It is hardly credible that, 
in order to ascertain the distance of the ship, the observer should 
have thought it necessary to reproduce and measure on land, in 
the horizontal plane, the enormous triangle which he constructed 
in imagination in a perpendicular plane over the sea.  Such an 
undertaking would have been so inconvenient and wearisome 
as to deprive Thales\apo\ discovery of its practical value. It is 
therefore probable that Thales knew another geometrical pro\-% 
position: viz.\ \senquote{that the sides of equiangular triangles are 
proportional.} (Euc.\ VI.\ 4.) 
\end{quote}
But Proposition \textsc{vi}.4
is overkill for measuring distances.
All one needs is the case of right triangles,
in the form of Theorem \ref{thm:Thales-right} above.
This must be the real theorem that Gow goes on to discuss:
\begin{quote}
And here no doubt we have the 
real import of those Egyptian calculations of \emph{seqt,} which Ahmes 
introduces as exercises in arithmetic. The \emph{seqt} or ratio, be\-% 
tween the distance of the ship and the height of the watch-% 
tower is the same as that between the corresponding sides of 
any small but similar triangle. The discovery, therefore, attri\-% 
buted to Thales is probably of Egyptian origin, for it is difficult 
to see what other use the Egyptians could have made of their 
\emph{seqt,} when found. It may nevertheless be true that the pro\-% 
position, Euc.\ VI.\ 4, was not known, as now stated, either to 
the Egyptians or to Thales. It would have been sufficient for 
their purposes to know, inductively, that the \emph{seqts} of equi\-% 
angular triangles were the same.
\end{quote}
Gow is right that Euclid\apo s Proposition \textsc{vi}.4 need not have been known.
But what he seems to mean is that the Egyptians and Thales
need only have had a knack for applying the theorem,
without having stepped back to recognize the theorem as such.
This may be so;
but there is no reason to think 
they had a knack for applying the theorem in full generality.

To establish that theorem, Thales\apo s Theorem,
in full generality,
we shall prove that, in the proof of Theorem \ref{thm:Thales-right},
equation \eqref{eqn:rect} still holds,
even when applied to Figure \ref{fig:Thales} in the proof of Theorem \ref{thm:Thales-int}.
To do this, we shall rely on the converse of \emph{Elements} \textsc i.43:
in Figure \ref{fig:I.43}, if the parallelograms $BG$ and $GD$ are equal,
then the point $G$ must lie on the diagonal $AD$.
We can prove this by contradiction,
or by contraposition.
If $G$ did not lie on the diagonal,
then we should be in the situation of Figure \ref{fig:I.43-conv},
\begin{figure}
\centering
\begin{pspicture}(0,-0.5)(6,4.5)
\psset{PointSymbol=none}
\pstTriangle(1,4)A(0,0)B(5,0)C
\pstMiddleAB[PointName=none] ACO
\pstSymO[PosAngle=45] OB[D]
\ncline CD\ncline DA
\pstHomO[HomCoef=0.4,PosAngle={180,40,90}] A{B,C,D}[L,N,H]
\pstInterLL LNCDM
\pstInterLL[PosAngle=-90] NHBCK
\ncline LM\ncline HK
\pstTranslation[DistCoef=0.25,PosAngle={180,0,-135}] LB{L,M,N}[E,F,G]
\ncline EF
\end{pspicture}
\caption{Converse of \emph{Elements} \textsc i.43}\label{fig:I.43-conv}
\end{figure}
where now parallelograms $BN$ and $ND$ are equal,
but $BG$ is part of $BN$, and $ND$ is part of $GD$,
so $BG$ is less than $GD$, by Euclid\apo s Common Notion 5.

In Euclid\apo s proof of the Pythagorean Theorem,
\emph{Elements} \textsc i.47,
three auxiliary straight lines concur.
Heath 
\cite[Vol.\ 1, p.\ 367]{MR17:814b}
passes along Hero\apo s proof of this,
including, as a lemma, the converse of \textsc i.43.
Hero\apo s proof is direct,
but relies on \emph{Elements} \textsc i.39:
equal triangles lying on the same side of the same base 
are in the same parallels.
We used this in proving Theorem \ref{thm:Thales-int},
and it is the converse of \textsc i.37;
Euclid proves it by contradiction, using Common Notion~5.

\begin{theorem}\label{thm:Thales}
If two parallelograms share an angle,
the parallelograms are equal if and only if
the rectangles bounded by the same sides are equal.
\end{theorem}

\begin{proof}
Let the parallelograms be $AG$ and $AH$
in Figure \ref{fig:Thales-full}.
\begin{figure}
\centering
\begin{pspicture}(-3.3,-2.1)(6.9,3.8)%\psgrid
  \psset{PointSymbol=none}
\pstTriangle[PosAngleA=45,linecolor=white](0,0)A(5,0)B(1.4,3.3)C % C originally (1,4)
\pstMiddleAB[PointSymbol=none,PointName=none]BCO
\pstSymO[PosAngle=45] OA[D]
\pstHomO[HomCoef=0.4,PosAngle={-45,-150}] A{B,C}[E, F]
\ncline BD\ncline CD
\pstTranslation[PosAngle=90] AE C[G]
\pstTranslation AF B[H]
\ncline EG\ncline FH
\pstInterLL[PosAngle=135] EGFHL
\pstInterLC[PointSymbol=none, PointName=none]AEAF{F1}{F2}
\pstRotation[RotAngle=-90,PosAngle=-90]A{F2}[M]
\pstTranslation[PointSymbol=none, PointName=none]AMB
\pstInterLL[PosAngle=-90]FA{B'}MN
\pstInterLL[PosAngle=-90]LE{B'}MP
\pstInterLL[PosAngle=-90]HB{B'}MQ
\pstInterLC[PointSymbol=none, PointName=none]AFAE{E1}{E2}
\pstRotation[RotAngle=90,PosAngle=180]A{E2}[R]
\pstTranslation[PointSymbol=none, PointName=none]ARC
\pstInterLL[PosAngle=180]EA{C'}RS
\pstInterLL[PosAngle=180]LF{C'}RT
\pstInterLL[PosAngle=180]GC{C'}RU
\pstInterLL[PosAngle=-135] USQNX
\ncline CU
\ncline UX
\ncline XQ
\ncline QB
\ncline FT
\ncline BS
\ncline CN
\ncline EP
\ncline AR
\ncline AM
\pstRightAngle FAR
\pstRightAngle EAM
\ncline[linestyle=dotted]XD
\ncline FE\ncline CB
\pstArcOAB[linestyle=dashed] AMF
\pstArcOAB[linestyle=dashed] AER
%\psset{MarkAngle=90}
%\pstSegmentMark AE
%\pstSegmentMark AR
%\pstSegmentMark[SegmentSymbol=MarkHash] FA
%\pstSegmentMark[SegmentSymbol=MarkHash] MA
\end{pspicture}
\caption{Thales\apo s Theorem}\label{fig:Thales-full}
\end{figure}
Supposing them equal,
we prove \eqref{eqn:rect}, namely
$AF\cdot AB=AE\cdot AC$.
Let the parallelogram $ABDC$ be completed.
By the converse of \emph{Elements} \textsc i.43,
the point $L$ lies on the diagonal $AD$.
Now erect the perpendiculars $AR$ and $AM$
(using \emph{Elements} \textsc i.10),
and make them equal to $AE$ and $AF$ respectively,
as by drawing circles.
Each of the parallelograms $NE$ and $SF$ 
is equal to a rectangle of sides equal to $AE$ and $AF$,
by \emph{Elements} \textsc i.35.
Therefore $A$ lies on the diagonal $LX$,
again by the converse of \emph{Elements} \textsc i.43,
so $A$ and $L$ both lie on the diagonal $DX$ of the large parallelogram.
Consequently, the parallelograms
$SC$ and $NB$ are equal;
but they are also equal to $AE\cdot AC$ and $AF\cdot AB$ respectively.
The converse is similar.
\end{proof}

Theorems \ref{thm:Thales-int} and \ref{thm:Thales} together
are what we are calling Thales\apo s Theorem,
provided we can establish \emph{Elements} \textsc{vi}.16:
\begin{quote}
If four straight lines be proportional,
the rectangle contained by the extremes
is equal to the rectangle contained by the means;
and [conversely].
\end{quote}
To do this, we need a proper theoretical definition of proportion.

\section{Thales and Desargues}\label{sect:Th-D}

I suggested that the equivalence of the proportion \eqref{eqn:PpLl}
with the equation \eqref{eqn:PlLp} was \enquote{practical.}
To read the colons in the expression
\begin{equation}\label{eqn:abcd}
a:b\as c:d,
\end{equation}
we can say any of the following:
\begin{compactitem}
\item
$a$, $b$, $c$, and $d$ are proportional;
\item
$a$ is to $b$ as $c$ is to $d$;
\item
the ratio of $a$ to $b$ is the same as the ratio of $c$ to $d$.
\end{compactitem}
From the last clause, we can delete the phrase \enquote{the same as}:
this is effectively what Plutarch does in the passage quoted above,
although the translator puts the phrase back in.
The equation
\begin{equation}\label{eqn:adbc}
a\cdot d=b\cdot c.
\end{equation}  
does not in itself express a property of the ordered pair $(a,b)$
that is the same as the corresponding property of $(c,d)$;
it expresses only a relation between the pairs.
Immediately we have $a\cdot b=b\cdot a$,
and if \eqref{eqn:adbc} holds, so does $c\cdot b=d\cdot a$;
so the relation expressed by \eqref{eqn:adbc} 
between $(a,b)$ and $(c,d)$
is reflexive and symmetric.
The relation is not obviously transitive: if \eqref{eqn:adbc} holds,
and $c\cdot f=d\cdot e$,
it is not obvious that $a\cdot f=b\cdot e$.
It would be \emph{true,} for example, 
if we allowed passage to a third dimension,
obtaining from the hypotheses
\begin{align*}
  a\cdot d\cdot f&=b\cdot c\cdot f,&
  b\cdot c\cdot f&=b\cdot d\cdot e,
\end{align*}
whence, presumably, the desired conclusion would follow;
but this would be a theorem.
Therefore \eqref{eqn:adbc} alone cannot constitute the 
\emph{definition} of \eqref{eqn:abcd}.
I have argued this elsewhere in the context of Euclid\apo s number theory
\cite{Pierce-Eu-nth}.

Suppose now that each of the letters in \eqref{eqn:abcd}
stands for a \textbf{length:} not a number, 
but the class of Euclidean bounded straight lines 
that are equal to a particular straight line.
I propose to define \eqref{eqn:abcd}
to mean that for all lengths $x$ and $y$,
\begin{equation}\label{eqn:ef}
b\cdot x=a\cdot y\iff d\cdot x=c\cdot y.
\end{equation}
In other words, \eqref{eqn:abcd} means that the sets
\begin{align*}
&\{(x,y)\colon b\cdot x=a\cdot y\},&
&\{(x,y)\colon d\cdot x=c\cdot y\}
\end{align*}
are the same.
This definition ensures \emph{logically} 
that the relation of having the same ratio is transitive,
as any relation described as a sameness should be.
The definition also ensures that \eqref{eqn:abcd} \emph{implies} \eqref{eqn:adbc}:
let $(x,y)=(c,d)$ in \eqref{eqn:ef}.
The definition avoids the \enquote{Archimedean} assumption
required by the definition attributed to Eudoxus,
found in Book \textsc v of the \emph{Elements.}
However,
we need to prove that \eqref{eqn:adbc} implies \eqref{eqn:abcd}.
This implication is \emph{Elements} \textsc{vi}.16 
for the new definition of proportion.

\begin{theorem}\label{thm:rect-prop}
If, of four lengths,
the rectangle bounded by the extremes 
is equal to the rectangle bounded by the means,
then the lengths are in proportion.
\end{theorem}
 
\begin{proof}
Supposing we are given four lengths $a$, $b$, $c$, $d$
such that \eqref{eqn:adbc} holds,
we want to show \eqref{eqn:abcd},
as defined by \eqref{eqn:ef}.
It is enough to show that,
if two lengths $e$ and $f$ are such that
\begin{equation}\label{eqn:afbe}
	a\cdot f=b\cdot e,
\end{equation}
then
\begin{equation}\label{eqn:cfde}
c\cdot f=d\cdot e.
\end{equation}
In Figure \ref{fig:trans},
\begin{figure}
\centering
\begin{pspicture}(1,-0.6)(8.8,4.6)%\psgrid
\pstGeonode[PosAngle={90,90,-90}](3,4)A(8,4)B(3.2,0)C
\pstMiddleAB[PointSymbol=none,PointName=none] BCO
\pstSymO OA[D]
\ncline BD\ncline DC
\pstTranslation[DistCoef=0.35,PosAngle={-90,-90}]BA{C,D}[E,F]
\ncline BF
\pstHomO[HomCoef=0.55,PosAngle={90,45,45,135,-75}]A{B,C,D,E,F}[G,H,K,L,M]
\ncline AG\taput{$a$}
\ncline GB\taput{$b$}
\ncline AH\trput{$c$}
\ncline HC\trput{$d$}
\ncline AL\tlput{$e$}
\ncline LE\tlput{$g$}
\pstInterLL[PosAngle=-90] GKCDN
\ncline GN
\pstInterLL HKBDP
\ncline LP
\pstInterLL[PosAngle=-90] GMEFQ
\ncline GQ
\pstInterLL[PosAngle=-75] LMBFR
\ncline EF
\psset{linestyle=dotted}
\ncline CB\ncline BE
\ncline HG\ncline GL
\end{pspicture}
\caption{Transitivity}\label{fig:trans}
\end{figure}
let $AH$ and $AL$ have lengths $c$ and $e$ respectively.
Draw $AG$ parallel to $HL$ (\emph{Elements} \textsc i.31),
and let $AG$ have length $a$.
Complete the parallelogram $ABDC$
so that $GB$ and $HC$ have lengths $b$ and $d$ respectively.
By \eqref{eqn:adbc} and Theorem \ref{thm:Thales},
the parallelograms $GP$ and $HN$ are equal.
Now complete the parallelogram $ABFE$,
and denote by $g$ the length of $LE$.
The parallelograms $GR$ and $GP$ are equal,
by \emph{Elements} \textsc i.35.
Both $LM$ and $HK$ have length $a$ (\emph{Elements} \textsc i.34),
so parallelograms $LQ$ and $HN$ are equal, 
by \emph{Elements} \textsc i.36.
Thus $GR$ and $LQ$ are equal.
Hence, by Theorem \ref{thm:Thales}, we have
\begin{equation*}
a\cdot g=b\cdot e,
\end{equation*}
so $a\cdot g=a\cdot f$ by \eqref{eqn:afbe}, 
and therefore $g=f$ by Common Notion 5.
Since $LH\parallel EC$,
we have \eqref{eqn:cfde} by Theorem \ref{thm:Thales}.
\end{proof}

There are easy consequences,
corresponding to Propositions 16 and 22 of \emph{Elements} Book \textsc v;
but these propositions are not themselves so easy to prove
with the Eudoxan definition of proportion.

\begin{corollary}[Alternation]\label{cor:alt}
\begin{equation*}
a:b\as c:d\implies a:c\as b:d.
\end{equation*}
\end{corollary}

\begin{proof}
Each proportion is equivalent to $a\cdot d=b\cdot c$.
\end{proof}

In a note on \textsc v.16,
Heath \cite[Vol.\ 2, pp.\ 165--6]{MR17:814b}
observes that the proposition is easier to prove when---as for us---%
the magnitudes being considered are lengths.
He quotes the textbook of Smith and Bryant \cite[pp.\ 298--9]{Smith-Bryant},
which derives the special case from \textsc{vi}.1:
parallelograms and triangles under the same height
are to one another as their bases.
We might write this as
\begin{equation*}
a:b\as a\cdot c:b\cdot c.
\end{equation*}
One could take this as a \emph{definition} of proportion;
but then one has the problem of transitivity, as before.
If 
\begin{equation*}
a:b\as c\cdot e:d\cdot e
\end{equation*}
for some $e$,
meaning $c\cdot e=a\cdot f$ and $d\cdot e=b\cdot f$ for some $f$,
one has to show the same for arbitrary $e$.

\begin{corollary}[Cancellation]\label{cor:can}
\begin{equation*}
\left.\begin{gathered}
a:b\as d:e\\
b:c\as e:f
\end{gathered}\right\}\implies a:c\as d:f.
\end{equation*}
\end{corollary}

\begin{proof}
Under the hypothesis, by alternation,
$a:d\as b:e$ and $b:e\as c:f$.
Then $a:d\as c:f$,
since sameness of ratio is transitive by definition.
\end{proof}

Theorems \ref{thm:Thales-int}, \ref{thm:Thales}, and \ref{thm:rect-prop} together
constitute Thales\apo s Theorem.
In proving Theorem \ref{thm:rect-prop},
we effectively showed in Figure \ref{fig:trans}
\begin{equation}\label{eqn:BCGH}
BC\parallel GH\And CE\parallel HL\implies
BE\parallel GL,
\end{equation}
provided also
\begin{equation}\label{eqn:CEAB}
CE\parallel AB.
\end{equation}
Then by Thales\apo s Theorem itself,
since sameness of ratio is transitive,
\eqref{eqn:BCGH} holds,
without need for \eqref{eqn:CEAB}.
This is Desargues\apo s Theorem:
if the straight lines through corresponding vertices of two triangles concur,
and two pairs of corresponding sides of the triangles are parallel,
then the third pair must be parallel,
as in Figure \ref{fig:Des}.
\begin{figure}
\centering
\begin{pspicture}(6,4)
\psset{PointName=none}
\pstTriangle(0,0)A(3,1)B(0,4)C
\pstGeonode(6,1)O
\pstHomO[HomCoef=0.7] O{A,B,C}[D,E,F]
\ncline AB\ncline BC\ncline CA
\ncline DE\ncline EF\ncline FD
\psset{linestyle=dotted}
\ncline OA\ncline OB\ncline OC
\end{pspicture}
\caption{Desargues\apo s Theorem}\label{fig:Des}
\end{figure}
We have proved this from Book \textsc i of Euclid,
without the Archimedean assumption of Book \textsc v,
though \emph{with} Common Notion 5 for areas.

Desargues\apo s Theorem has other cases in the Euclidean plane.
When we add to this plane the \enquote{line at infinity,}
thus obtaining the projective plane,
then the three pairs of parallel lines in the theorem
intersect on the new line.
But then any line of the projective plane
can serve as a line at infinity added to a Euclidean plane.
A way to show this is by the kind of coordinatization 
that we are in the process of developing.

\section{Locus problems}\label{sect:L-p}

Fixing a unit length,
Descartes \cite[p.\ 5]{Descartes-Geometry}
defines the product $ab$ of lengths $a$ and $b$
as another length,
given by the rule that we may express as
\begin{equation*}
1:a\as b:ab.
\end{equation*}
\sloppy
Denoting Decartes\apo s product thus, by juxtaposition alone,
while continuing to denote with a dot the area of a rectangle
with given dimensions,
by Theorem \ref{thm:rect-prop} we have
\begin{equation*}
ab\cdot 1=a\cdot b.
\end{equation*}
\fussy
In particular,
Cartesian multiplication is commutative,
and it distributes over addition,
since
\begin{align*}
	a\cdot b&=b\cdot a,&
	a\cdot(b+c)&=a\cdot b+a\cdot c,
\end{align*}
and from Common Notion 5 we have
\begin{equation*}
d\cdot 1=e\cdot1\implies d=e.
\end{equation*}
That Cartesian multiplication is associative 
can be seen from the related operation of composition of ratios,
given by the rule
\begin{equation*}%\label{eqn:abc}
(a:b)\comp(b:c)\as a:c.
\end{equation*}
One may prefer to use the sign $=$ of equality here, 
rather than the sign $\as$ of sameness of ratio,
if one judges it not to be immediate that the compound ratio
$(a:b)\comp(b:c)$ depends not merely on the ratios $a:b$ and $b:c$,
but on their given representations in terms of $a$, $b$, and $c$.
Nonetheless, it does depend only on the ratios,
by Corollary \ref{cor:can}.
Then composition of ratios is immediately associative.
We have generally
\begin{equation*}
(a:b)\comp(c:d)\as e:d,
\end{equation*}
provided $a:b\as e:c$;
and such $e$ can be found,
by the method of \emph{Elements} \textsc i.44 and 45.
Then
\begin{equation*}
(a:1)\comp(b:1)\as ab:1,
\end{equation*}
and from this we can derive $a(bc)=(ab)c$.
Also
\begin{equation*}
(a:1)\comp(1:a)\as 1:1,
\end{equation*}
so multiplication is invertible.

One can define the sum of two ratios as one defines the sum of fractions in school,
by finding a common denominator.
In particular,
\begin{equation*}
(a:1)+(b:1)=(a+b):1,
\end{equation*}
where now it does seem appropriate to start using the sign of equality.
Now both lengths and ratios compose fields,
in fact ordered fields,
which are isomorphic under $x\mapsto x:1$.
Descartes shows this implicitly, 
in order to solve ancient problems.
One may object that we have not introduced additive inverses,
whether of lengths or of ratios.
We can do this by assigning to each class of parallel straight lines a direction,
so that the signed length of $BA$ is the additive inverse of the signed length of $AB$.

Descartes
\cite[p.\ 40, n.\ 59]{Descartes-Geometry}
alludes to a passage in the \emph{Collection} where
Pappus \cite[pp.~346--53]{MR13:419a} describes three kinds of geometry problem:
\textbf{plane,} as being solved by means of straight lines and circles, 
which lie in a plane;
\textbf{solid,} as requiring also the use of conic sections, 
which are sections of a solid figure, the cone; and
\textbf{linear,} as involving more complicated \emph{lines,} that is, curves.
An example of a linear problem then would be the quadrature or squaring of the circle,
achieved by means of the \textbf{quadratrix}
or \enquote{tetragonizer}
(\gr{tetragwn'izousa}),
which Pappus \cite[pp.~336--47]{MR13:419a} defines
as being traced in a square,
such as $ABGD$ in Figure \ref{fig:quad},
\begin{figure}[ht]
\centering
\psset{unit=5cm}
\begin{pspicture}(-0.1,-0.1)(1.1,1.1)
\pspolygon(0,0)(1,0)(1,1)(0,1)
\parametricplot{0.001}{1}{t 90 t mul tan div t}
\psarc[linestyle=dotted](0,0){1}{0}{90}
\psset{linestyle=dashed}
\psline(0,0)(0.577,1)
\psline(0,0.667)(1,0.667)
\uput[d](0.193,0.667){$x$}
\uput[l](0,0.333){$y$}
\uput{10pt}[30](0,0){$\theta$}
\uput[dl](0,0){$A$}
\uput[ul](0,1){$B$}
\uput[ur](1,1){$G$}
\uput[dr](1,0){$D$}
\uput[d](0.637,0){$H$}
\end{pspicture}
\caption{The quadratrix}\label{fig:quad}
\end{figure}
by the intersection of two straight lines,
one horizontal and moving from the top edge $BG$ to the bottom edge $AD$,
the other swinging about the lower left corner $A$ 
from the left edge $AB$ to the bottom edge $AD$.
If there is a point $H$ where the quadratrix meets the lower edge of the square,
then, as Pappus shows,
\begin{equation*}
BD:AB\as AB:AH,
\end{equation*}
where $BD$ is the circular arc centered at $A$.
In modern terms,
with variables as in the figure,
$AB$ being taken as a unit,
\begin{align*}
\frac{\theta}y&=\frac{\uppi}2,&
\tan\theta=\frac yx,
\end{align*}
so
\begin{equation*}
x=\frac2{\uppi}\cdot\frac{\theta}{\tan\theta}.
\end{equation*}
As $\theta$ vanishes, $x$ goes to $2/\uppi$.
This then is the length of $AH$.
Pappus points out
that we have no way to construct the quadratrix
without knowing where the point $H$ is in the first place.
He attributes this criticism to one Sporus,
about whom we apparently have no source 
but Pappus himself \cite[p.~285, n.~78]{MR2093668}.


A solid problem that Pappus describes \cite[pp.~486--9]{MR13:419a} 
is the \textbf{four-line locus problem:}
find the locus of points such that 
the rectangle whose dimensions are the distances to two given straight lines 
bears a given ratio to 
the rectangle whose dimensions are the distances to two more given straight lines.
According to Pappus, theorems of Apollonius were needed to solve this problem;
but it is not clear whether Pappus thinks 
Apollonius actually did work out a full solution.
By the last three propositions, namely 54--6, 
of Book III of the \emph{Conics} of Apollonius,
it is implied that the conic sections are three-line loci, that is,
solutions to the four-line locus problem when two of the lines are identical.
Taliaferro \cite[pp.~267--75]{MR1660991} works out the details
and derives the theorem that the conic sections are four-line loci.

Descartes works out a full solution to the four-line locus problem
\cite[pp.\ 59--80]{Descartes-Geometry}.
He also solves a particular \emph{five}-line locus problem,
where four of the straight lines---%
say $\ell_0$, $\ell_1$, $\ell_2$, and $\ell_3$---%
are parallel to one another,
each a distance $a$ from the previous,
while the fifth line---$\ell_4$---is perpendicular to them
\cite[pp.\ 83--4]{Descartes-Geometry}.
What is the locus of points such that the product of their distances
to $\ell_0$, $\ell_1$, and $\ell_3$
is equal to the product of $a$
with the distances to $\ell_2$ and $\ell_4$?
One can write down an \emph{equation} for the locus,
and Descartes does.
%Effectively 
Letting 
distances from $\ell_4$ and $\ell_2$ be $x$ and $y$ respectively,
%the $x$- and $y$-axes be $\ell_2$ and $\ell_4$,
%the positive direction of the latter being from $\ell_3$ towards $\ell_0$,
Descartes obtains
\begin{equation}\label{eqn:5}
y^3-2ay^2-a^2y+2a^3=axy.
\end{equation}
This may allow us to plot points on the desired locus,
obtaining the bold solid curve in Figure \ref{fig:5-line-sol};
\begin{figure}
\centering
\psset{unit=2cm}
\begin{pspicture}*(-3,-2)(2,5)%\psgrid
\psline(-2,-2)(-2,5)
\psline(-1,-2)(-1,5)
\psline( 0,-2)( 0,5)
\psline( 1,-2)( 1,5)
\psline(-3, 0)( 2,0)
\uput[ur](-2,-2){$\ell_0$}
\uput[ur](-1,-2){$\ell_1$}
\uput[ur](-0,-2){$\ell_2$}
\uput[ur]( 1,-2){$\ell_3$}
\uput[ur](-3, 0){$\ell_4$}
\psset{linewidth=1.6pt,linestyle=solid}
\psplot{-3}{-0.25}{x dup dup dup mul exch 2 mul add 1 sub exch 2 exch div sub}
\psplot{0.25}{2}{x dup dup dup mul exch 2 mul add 1 sub exch 2 exch div sub}
\psset{linestyle=dashed}
\psplot{-3}{2}{x dup mul 0.5 add}
\psline(-3,-0.75)(2,3)
\psdots[dotstyle=o](-2,0)(0,1.5)
\end{pspicture}
\caption{Solution of a five-line locus problem}\label{fig:5-line-sol}
\end{figure}
but we could already do that.
The equation is thus not a solution to the locus problem,
since it does not tell us what the locus \emph{is.}
But Descartes shows that the locus is traced
by the intersection of a moving parabola
with a straight line passing through one fixed point
and one point that moves with the parabola,
as suggested by the dashed lines in Figure \ref{fig:5-line-sol}.
The parabola
has axis sliding along $\ell_2$,
and the latus rectum of the parabola is $a$,
so the parabola is given by $ax=y^2$
when its vertex is on $\ell_4$.
The straight line passes through the intersection of $\ell_0$ and $\ell_4$
and through the point on the axis of the parabola whose distance from the vertex is $a$.

We shall be looking at latera recta again later; meanwhile,
one may consult my article \enquote{Abscissas and Ordinates} \cite{MR3312989},
to learn more than one ever imagined wanting to know about the terminology.

Descartes\apo s solution of a five-line locus problem
is apparently one that Pappus would recognize as such.
Thus Descartes\apo s algebraic methods would seem to represent an advance,
and not just a different way of doing mathematics.
As Descartes knows \cite[p.\ 22, n.\ 34]{Descartes-Geometry}, 
Pappus \cite[pp.~600-3]{MR13:419b} could \emph{formulate} 
the $2n$- and $(2n+1)$-line locus problems for arbitrary $n$.
If $n>3$, the ratio of the product of $n$ segments with the product of $n$ segments
can be understood as the ratio compounded of the respective ratios of segment to segment.
Given $2n$ lengths $a_1$, \lips, $a_n$, $b_1$, \lips, $b_n$,
we can understand the ratio of the product of the $a_k$
to the product of the $b_k$ as the composite ratio
\begin{equation*}
(a_1:b_1)\comp\cdots\comp(a_n:b_n).
\end{equation*}
Pappus recognizes this.
Descartes expresses the solution of the $2n$-line locus problem
as an $n$th-degree polynomial equation in $x$ and $y$, where
$y$ is the distance from a point of the locus 
to one of the given straight lines,
and $x$ is the distance from a given point on that line
to the foot of the perpendicular from the point of the locus.
Today we call the line the \textbf{$x$-axis,}
and the perpendicular through the given point on it the \textbf{$y$-axis;}
but Descartes does not seem to have done this expressly.

Descartes does effectively allow oblique axes. 
The original locus problems literally involve
not distances to the given lines,
but lengths of straight lines drawn at given angles to the given lines.
%For the methods of Descartes, the distinction is trivial.
%For ancients like Apollonius, the distinction would seem not to be trivial.

\section{The nine-point conic}\label{sect:9-p-c}

The nine-point conic is the solution of a locus problem.
The solution had been known earlier;
but apparently the solution was first remarked on 
in 1892 by Maxime B\^ocher \cite{Bocher},
who says,
\begin{quotation}
It does not seem to have been noticed that a few well-known facts, 
when properly stated, 
yield the following direct generalization of the famous nine point circle theorem:---

\begingroup
\itshape Given a triangle $ABC$ and a point $P$ in its plane, 
a conic can be drawn through the following nine points:

(1) The middle points of the sides of the triangle; 

(2) The middle points of the lines joining $P$ to the vertices of the triangle; 

(3) The points where these last named lines cut the sides of the triangle.\endgroup

The conic possessing these properties 
is simply the locus of the centre 
of the conics passing through the four points $A$, $B$, $C$, $P$
(cf.\ Salmon\apo s Conic Sections, p.\ 153, Ex.\ 3, and p.\ 302, Ex.\ 15).
\end{quotation}
B\^ocher\apo s references fit the sixth and tenth editions of Salmon\apo s
\emph{Treatise on Conic Sections,}
dated 1879 and 1896 respectively \cite{Salmon-1879,Salmon-1896};
but in the \enquote{Third Edition, revised and enlarged,}
dated 1855
\cite{Salmon-1855},
the references would be p.\ 137, Ex.\ 4, and p.\ 284, Art.\ 339.


Following Salmon, we start with a general equation
\begin{equation}\label{eqn}
ax^2+2hxy+by^2+2gx+2fy+c=0
\end{equation}
of the second degree.
We do not assume that $x$ and $y$ are measured at right angles to one another:
we allow oblique axes.
We do assume $ab\neq0$.
If we wish, we can eliminate the $xy$ term
by redrawing the $x$-axis,
and changing its scale,
so that the point now designated by $(x,y)$
is the one that was called $(x+hy/a,y)$ before.
The curve that was defined by \eqref{eqn}
is now defined by
\begin{equation}\label{eqn:no-xy}
ax^2+\left(b-\frac{h^2}a\right)y^2+2gx+2\left(f-\frac{gh}a\right)y+c=0.
\end{equation}
If the linear terms in \eqref{eqn} are absent,
then they are absent from \eqref{eqn:no-xy} as well.

The equation \eqref{eqn:no-xy} defines an ellipse or hyperbola,
no matter what the angle is between the axes,
or what their relative scales are.
This is perhaps not, strictly speaking, high-school knowledge.
One may know from school that an equation
\begin{equation*}
\frac{x^2}{a^2}\pm\frac{y^2}{b^2}=1
\end{equation*}
defines a certain curve called ellipse or hyperbola,
depending on whether the upper or lower sign is taken.
But one assumes that $x$ and $y$ are measured orthogonally.
One does not learn why the curves are named as they are
\cite{MR3312989},
and one does not learn that,
if the appropriate oblique axes are chosen,
then the curve has an equation of the same form.
But this is just what Book I of the \emph{Conics}
of Apollonius is devoted to showing \cite{MR1660991}.

Going back to the original axes,
and the curve defined by \eqref{eqn},
we translate the axes,
so that the new origin is the point formerly called $(x',y')$.
The curve is now given by
\begin{equation*}
ax^2+2hxy+by^2+2g'x+2f'y+c'=0,
\end{equation*}
where
\begin{align}\label{eqn:g'f'}
	g'&=ax'+hy'+g,&
	f'&=by'+hx'+f,
\end{align}
and we are not interested in $c'$.
The curve given by \eqref{eqn} has center at $(x',y')$
just in case $(g',f')=(0,0)$.

For a complete quadrangle
in which one pair (at least) of opposite sides are not parallel,
those sides
determine a coordinate system.
In this system,
suppose the vertices of the complete quadrilateral are 
$(\lambda,0)$ and $(\lambda',0)$ on the $x$-axis
and $(0,\mu)$ and $(0,\mu')$ on the $y$-axis.
Let the conic given by \eqref{eqn} pass through these four points.
Setting $y=0$,
we obtain the equation
\begin{equation*}
ax^2+2gx+c=0,
\end{equation*}
which must have roots $\lambda$ and $\lambda'$,
so that the equation is
\begin{equation*}
a\bigl(x^2-(\lambda+\lambda')x+\lambda\lambda'\bigr)=0.
\end{equation*}
From this we obtain
\begin{align*}
	2g&=-a(\lambda+\lambda'),&
	c&=a\lambda\lambda'.
\end{align*}
Likewise, setting $x=0$ in \eqref{eqn}
yields the equation
\begin{equation*}
by^2+2fy+c=0,
\end{equation*}
which must be
\begin{equation*}
b\bigl(y^2-(\mu+\mu')y+\mu\mu'\bigr)=0,
\end{equation*}
from which we obtain
\begin{align*}
	2f&=-b(\mu+\mu'),&
	c&=b\mu\mu'.
\end{align*}
From the two expressions for $c$, we have
\begin{equation*}
a\lambda\lambda'=b\mu\mu'.
\end{equation*}
In \eqref{eqn}, we are free to let $a=\mu\mu'$.
Then $b=\lambda\lambda'$,
and \eqref{eqn} becomes
\begin{multline}\label{eqn:new}
\mu\mu'x^2+2hxy+\lambda\lambda'y^2\\
{}-\mu\mu'(\lambda+\lambda)x-\lambda\lambda'(\mu+\mu')y+\lambda\lambda'\mu\mu'=0.
\end{multline}
By the computations for \eqref{eqn:g'f'},
the center of the conic in \eqref{eqn:new} satisfies
\begin{align*}
2\mu    \mu'    x+2hy-\mu    \mu'    (\lambda+\lambda')&=0,\\
2\lambda\lambda'y+2hx-\lambda\lambda'(\mu    +    \mu')&=0.
\end{align*}
Eliminating $h$ yields
\begin{equation}\label{eqn:2mm}
	2\mu\mu'x^2-\mu\mu'(\lambda+\lambda')x=
2\lambda\lambda'y^2-\lambda\lambda'(\mu+\mu')y.
\end{equation}
This is the equation of an ellipse or hyperbola passing through the origin.
We can also write the equation as
\begin{equation}\label{eqn:con-sol}
	\mu\mu'x\left(x-\frac{\lambda+\lambda'}2\right)=
\lambda\lambda'y\left(y-\frac{\mu+\mu'}2\right);
\end{equation}
this shows that the conic passes through the midpoints 
of the sides of the complete quadrangle that lie along the axes.
By symmetry, the conic passes through the midpoints of all six sides of the quadrangle,
and through its three diagonal points---%
if these exist, that is, if none of the three pairs of opposite sides of the quadrangle are parallel.

I suggested earlier that for Descartes to solve a five-line locus problem,
it was not enough to find the equation \eqref{eqn:5};
he had to describe the solution geometrically.
We do have a thorough geometric understanding of solutions of second-degree equations
like \eqref{eqn:2mm} and \eqref{eqn:con-sol} or even \eqref{eqn}.
Alternatively,
there is a modern \enquote{synthetic} approach
\cite[8.71, p.\ 118]{Coxeter-RPP},
that is, an approach based not on field axioms,
but on geometric axioms:
here axioms for a projective plane,
but with a line at infinity designated,
so that midpoints of segments of other lines can be defined.

I do not know how directly the nine-point conic
can be derived from the work of Apollonius.
Here I shall just want to look briefly
at an example of how Apollonius uses areas
in a way not easily made algebraic.
David Hilbert shows how algebra is \emph{possible,}
without \emph{a priori} assumptions about areas;
but it is not clear how much is gained.

\chapter{Lengths and Areas}

\section{Algebra}

The points of an unbounded straight line 
are elements of an ordered abelian group
with respect to the obvious notion of addition,
once an origin and a direction have been selected.
If we set up two straight lines at right angles to one another,
letting their intersection point be the origin,
then,
after fixing also a unit length, 
we obtain Hilbert\apo s definition of multiplication 
as in Figure \ref{fig:Hilbert},
\begin{figure}
\centering
\begin{pspicture}(0,-0.5)(4,3.2)
\psset{PointName=none,PointSymbol=none}
\pstTriangle[PosAngleB=-90,PosAngleC=180](0,0)O(4,0)A(0,3)B
\pstHomO[HomCoef=0.6]O{A,B}[C,D]
\nput{-90}C{$1$}
\nput{-90}A{$b$}
\nput{180}D{$a$}
\nput{180}B{$a\hilmul b$}
\nput{-135}O{$0$}
\ncline CD
%\psset{RightAngleSize=0.21} % default is 0.28 unit
\pstRightAngle AOB
\end{pspicture}
\caption{Hilbert\apo s multiplication}\label{fig:Hilbert}
\end{figure}
the two oblique lines being parallel.

Not having accepted the Eudoxan definition of proportion,
or Thales\apo s Theorem,
Hilbert can still show that multiplication is commutative and associative.
He does this by means of what he calls Pascal\apo s Theorem,
although it was referred to in the Introduction above as Pappus\apo s Theorem:
if the vertices of a hexagon 
lie alternately on two straight lines in the projective plane,
then the intersection points of the three pairs of opposite sides
lie on a straight line.
Pascal announced the generalization
in which the original two straight lines can be an arbitrary conic section
\cite{Pascal-Conics,Pascal-Coniques}.

When we give Pappus\apo s Theorem in the Euclidean plane,
the simplest case occurs as in Figure \ref{fig:Pappus},
\begin{figure}
\centering
\begin{pspicture}(0,-0.5)(6,4.5)
\psset{PointName=none}
\pstTriangle(0,0)A(6,0)B(5,4)C
\pstHomO[HomCoef=0.7,PosAngle=-90] AB[D]
\pstHomO[HomCoef=0.6,PosAngle={135,-90}] A{C,D}[E,F]
\pstHomO[HomCoef=0.7,PosAngle=135] AE[G]
\ncline CD\ncline BE\ncline EF\ncline FG\ncline GD
\nput{180}A{$0$}
\nput{-90}F{$a$}
\nput{-90}D{$d\hilmul b$}
\nput{0}B{$d\hilmul(c\hilmul a)$}
\nput{135}G{$b$}
\nput{135}E{$c\hilmul a$}
\nput{0}C{$c\hilmul(d\hilmul b)$}
\end{pspicture}
\caption{Pappus\apo s Theorem}\label{fig:Pappus}
\end{figure}
labelled for proving commutativity and associativity of multiplication
(in case the angle at $0$ is right).
In general,
if there are two pairs of parallel opposite sides of the hexagon 
that is woven like a spider\apo s web or cat\apo s cradle across the angle,
then the third pair of opposite sides are parallel as well.
By the numbering in Hultsch\apo s edition of Pappus\apo s \emph{Collection} \cite{Pappus},
which is apparently the numbering made originally by Commandinus
\cite[pp.\ 62--3, 77]{Jones},
the result is Proposition 134 in Book \textsc{vii};
it is also number \textsc{viii} of Pappus\apo s lemmas for the now-lost \emph{Porisms} of Euclid.

Lemma \textsc{viii} seems to have been sadly forgotten.
The case of Pappus\apo s Theorem
where the three pairs of opposite sides all intersect
is Propositions 138 and 139, or Lemmas \textsc{xii} and \textsc{xiii}:
these cover the cases when the straight lines on which the vertices of the hexagon lie
are parallel and not, respectively.
Kline cites only Proposition 139 
as giving Pappus\apo s Theorem
\cite[p.\ 128]{MR0472307}.
In his summary of most of Pappus\apo s lemmas for Euclid\apo s \emph{Porisms,}
Heath 
\cite[p.\ 419--24]{MR654680}
lists Propositions 138, 139, 141, and 143
as constituting Pappus\apo s Theorem.
The last two are converses of the first two,
as Pappus states them.
It is not clear that Pappus recognizes a single theme
behind the several of his lemmas that constitute the theorem named for him.
He omits the case where exactly one pair of opposite sides are parallel.

\emph{Heath} omits to mention Proposition 134 at all.
This is a strange oversight for an important theorem.
Pappus proves it by means of areas,
using Euclid\apo s \textsc i.39,
as we proved Theorem \ref{thm:Thales-int}.
Without using areas,
Hilbert gives two elaborate proofs,
one of which,
in the situation of Figure \ref{fig:a=ac},
\begin{figure}
\centering
\begin{pspicture}(0,-0.5)(4,3)
\psset{PointSymbol=none,PointName=none}
\pstTriangle(0,0)A(4,0)B(4,3)C
\ncline AB\tbput{$a$}
\ncline AC\tlput{$c$}
\psset{MarkAngleRadius=0.6}
\pstMarkAngle BAC{$\alpha$}
\pstRightAngle ABC
\end{pspicture}
\caption{Hilbert\apo s trigonometry}\label{fig:a=ac}
\end{figure}
uses the notation
\begin{equation*}
a=\alpha c
\end{equation*}
for what today might be written as $a=c\cos\alpha$.

For proving associativity and commutativity of multiplication,
Hartshorne has a more streamlined approach,
in \emph{Geometry: Euclid and Beyond}
\cite[pp.\ 170--2]{MR1761093}.
In Figure \ref{fig:Hartshorne},
\begin{figure}
\centering
\begin{pspicture}(-3,-3.5)(3,3)
\psset{PointSymbol=none}
\pstGeonode[PosAngle=135](-0.5,-0.25)A
{%
\psset{PointName=none,PointSymbol=none}
\pstGeonode(0,0)O(3,0)X(0,3)Y
\pstTranslation OXA[F]
\pstTranslation OYA[G]
}
\pstCircleOA OX
\pstInterLC[PosAngleA=180] AFOXBC
\pstInterLC[PosAngleA=-90,PosAngleB=90] AGOYDE
\ncline BD\ncline DC\ncline CE\ncline EB
\ncline AB\taput{$c$}
\ncline AE\tlput{$a\hilmul c$}
\ncline AD\tlput{$b\hilmul c$}
\ncline AC\taput{$b\hilmul(a\hilmul c)$}\tbput{$a\hilmul(b\hilmul c)$}
\pstMarkAngle DBA{}
\pstMarkAngle AEC{}
\psset{MarkAngleRadius=0.5}
\pstMarkAngle ABE{}
\pstMarkAngle CDA{}
\psset{MarkAngleRadius=0.6}
\pstMarkAngle DBA{}
\pstMarkAngle AEC{}
\pstRightAngle CAE
\end{pspicture}
\caption{Associativity and commutativity}\label{fig:Hartshorne}
\end{figure}
$AC$ has the two lengths indicated, so these are the same;
that is,
\begin{equation}\label{eqn:abc}
a\hilmul(b\hilmul c)=b\hilmul(a\hilmul c).
\end{equation}
Letting $c=1$ gives commutativity;
then this with \eqref{eqn:abc} gives associativity.
Distributivity follows from Figure \ref{fig:dist},
\begin{figure}
\centering
\begin{pspicture}(0,-0.5)(5,4)
\psset{PointSymbol=none,PointName=none}
\pstGeonode (0,0)A(5,0)B(5,4)C
\pstHomO[HomCoef=0.4]CA[D]
\pstProjection ABD[E]
\ncline AE\tbput{$b$}
\ncline ED\trput{$a\hilmul b$}
\pstProjection BCD[F]
\ncline DF\tbput{$c$}
\ncline FC\trput{$a\hilmul c$}
\ncline AC
\pstRightAngle AED
\pstRightAngle DFC
\psset{linestyle=dotted}
\ncline EB\ncline BF
\end{pspicture}
\caption{Distributivity}\label{fig:dist}
\end{figure}
where
\begin{equation*}
a\hilmul b+a\hilmul c=a\hilmul(b+c).
\end{equation*}

The advantage of defining multiplication in terms of lengths alone
(and right angles, and parallelism, 
but not parallelograms or other bounded regions of the plane)
is that it allows all straight-sided regions to be linearly ordered by size,
without assuming \emph{a priori} that the whole region is greater than the part.

Euclid provides for an ordering in \emph{Elements} \textsc i.44 and 45,
which show that every straight-sided region is equal to a rectangle on a given base.
Finding the rectangle involves \textsc i.43.
Showing the rectangle unique requires the converse of \textsc i.43,
which in turn requires Common Notion 5 for areas.
The triangle may be equal to the rectangle on the same base with half the height;
but there are three choices of base,
and so three rectangles result that are equal to the triangle.
When they are all made equal to rectangles on the same base,
why should they have the same height?
Common Notion 5 is one reason; but Hilbert doesn\apo t need it.

The triangle is equal to the rectangle whose base is half the perimeter of the triangle
and whose height is the radius of the inscribed circle:
see Figure \ref{fig:incircle},
\begin{figure}
\centering
\begin{pspicture}(0,-0.5)(5,3.5)
\psset{PointSymbol=none,RightAngleSize=0.21}
\pstTriangle(0,0)A(4,0)B(5,3)C
{\psset{PointName=none}
\pstBissectBAC[linestyle=none] BACX
\pstBissectBAC[linestyle=none] CBAY
}
\pstInterLL[PosAngle=90] AXBYD
\ncline AD\ncline BD\ncline CD
\pstProjection BCD[E]
\pstProjection CAD[F]
\pstProjection ABD[G]
\pstCircleOA DE
\ncline DE\ncline DF\ncline DG
\pstRightAngle AGD
\pstRightAngle CED
\pstRightAngle AFD
\end{pspicture}
\caption{Circle inscribed in triangle}\label{fig:incircle}
\end{figure}
where the triangle $ABC$ is equal to the sum of
$AG\cdot GD$, $BG\cdot GD$, and $CE\cdot ED$,
but $ED=GD$.
However, if $ABC$ is cut into two triangles,
and rectangles equal to the two triangles are found,
added together, and made equal to a rectangle whose base is again half the perimeter of $ABC$,
we need to know that the height is equal to $GD$.

In the notation of Figure \ref{fig:area},
\begin{figure}
\centering
\begin{pspicture}(0,-0.5)(5,3.5)
\psset{PointSymbol=none,RightAngleSize=0.21}
\pstTriangle[PosAngleB=-90](0,0)A(4,0)B(5,3)C
\psset{PointName=none}
\pstProjection ABC[H]
\pstProjection ACB[K]
\pstRightAngle AHC
\pstRightAngle AKB
\psset{linestyle=dotted}
\ncline BH\ncline HC\trput{$h$}
\ncline BK\tlput{$k$}
\psset{linestyle=none}
\ncline AC\taput{$c$}
\ncline AB\tbput{$b$}
\pstMarkAngle BAC{$\alpha$}
\end{pspicture}
\caption{Area of triangle in two ways}\label{fig:area}
\end{figure}
we have
\begin{align*}
	h&=c\sin\alpha,&k&=b\sin\alpha,
\end{align*}
where $\sin\alpha$ stands for the appropriate length,
and the multiplication is Hilbert\apo s; 
and then
\begin{equation*}
bh=bc\sin\alpha=cb\sin\alpha=ck.
\end{equation*}
Thus we can define the area of $ABC$ unambiguously as the \emph{length} $bh/2$.
With this definition, when the triangle is divided into two parts,
or indeed into many parts, as Hilbert shows,
the area of the whole is the sum of the areas of the parts.

Moreover, if two areas become equal when equal areas are added,
then the two original areas are themselves equal:
this is Euclid\apo s Common Notion 3,
but Hilbert makes it a theorem.
One might classify this theorem with the one
that almost every human being learns in childhood,
but almost nobody ever recognizes \emph{as} a theorem:
no matter how you count a finite set,
you always get the same number.
It \emph{can} be valuable to become clear about the basics,
as I have argued concerning the little-recognized distinction
between induction and recursion \cite{Pierce-IR}.

\section{Apollonius}\label{sect:A}

\begin{sloppypar}
We look at a proof by Apollonius,
in order to consider Descartes\apo s idea,
quoted in the Introduction,
that ancient mathematicians had a secret method.
\end{sloppypar}

We first set the stage;
this is done in more detail in \cite{MR3312989}.
A \textbf{cone} is determined by a \textbf{base,}
which is a circle, and a \textbf{vertex,}
not in the plane of the base,
but not necessarily hovering right over the center of the base either:
the cone may be oblique.
The surface of the cone is traced out
by the straight lines that pass through the vertex
and the circumference of the base.
A diameter of the base is also the base of an
\textbf{axial triangle,}
whose apex is the vertex of the cone.
If a chord of the base of the cone 
cuts the base of the axial triangle at right angles,
then a plane containing the chord
and parallel to a side of the axial triangle
cuts the surface of the cone in a \textbf{parabola.}
The cutting plane cuts the axial triangle
in a straight line that is called a \textbf{diameter} of the parabola
because the line bisects the chords of the parabola that are parallel to the base of the cone.
Half of such a chord is an \textbf{ordinate;}
it cuts off from the diameter the corresponding \textbf{abscissa,}
the other endpoint of this being the \textbf{vertex} of the parabola.
There is some bounded straight line,
the \textbf{latus rectum,}
such that, when the square on any ordinate
is made equal to a rectangle on the abscissa,
the other side of the rectangle is precisely the latus rectum:
this is Proposition \textsc i.11 of Apollonius,
and it is the reason for the term \emph{parabola,}
meaning application.

The tangent to the parabola at the vertex
is parallel to the ordinates.
We are going to show that
every straight line parallel to the diameter
is another diameter,
with a corresponding latus rectum;
and latera recta are to one another
as the squares on the straight lines,
each of which is drawn
drawn tangent to the parabola
from vertex to other diameter.

In Cartesian terms,
we may start with a diameter that is an \textbf{axis}
in the sense of being at right angles to its ordinates.
If the latus rectum is $\ell$, then the parabola can be given by
\begin{equation}\label{eqn:ay}
\ell y=x^2.
\end{equation}
As in Figure \ref{fig:ay=x2},
\begin{figure}
\centering
\psset{unit=48mm}
\begin{pspicture}(-0.5,-1.5)(1.2,1.66)
%  \psgrid
{\psset{linestyle=dashed}
\psline(0,-1.5)(1,0)(1,1)(0,1)
\psline(0.75,0.5625)(0,0.5625)
}
\uput[d](1,0){$\ell $}
\uput[l](0,1){$\ell $}
\uput[d](0.75,0){$a$}
\uput[l](0,0.5625){$a^2/\ell $}
\uput[r](0,-0.5625){$-a^2/\ell $}
\uput[l](0,-1.5){$-2a$}
\psplot[plotpoints=200]{-0.5}{1.25}{x dup mul}
{\psset{linestyle=dotted}
         \psline(-0.25,0.0625)(0,0.0625)
\psline(-0.25,-0.9375)(-0.25,0.0625)(0.75,1.5625)
}
\uput[r](0,0.0625){$y$}
\uput[dl](-0.25,0){$x$}
\uput[r](0.75,1.5625){$y'$}
\uput[dr](-0.25,-0.9375){$x'$}
\psset{linewidth=1.6pt}
\psline{->}(0,-1.5)(0,1.66)
\psline{->}(-0.5,0)(1.25,0)
\psset{linestyle=dashed}
\psline{->}(0.75,0)(0.75,1.66)
\psline{->}(-0.5,-1.3125)(0,-0.5625)(1.25,1.3125)
\end{pspicture}
\caption{Change of coordinates}\label{fig:ay=x2}
\end{figure}
the tangent to the parabola at $(a,a^2/\ell )$
cuts the $y$-axis at $-a^2/\ell$:
this can be shown with calculus,
but it is Proposition \textsc i.35 of Apollonius.
The tangent then is
\begin{equation*}
\ell y=2ax-a^2.
\end{equation*}
We shall take this and $x=a$ as new axes,
say $x'$- and $y'$-axes.
If $d$ is the distance between
the new origin and the intersection of the $x'$-axis with the $y$-axis,
then, since the $x$- and $y$-axes are orthogonal,
by the Pythagorean Theorem (\emph{Elements} \textsc i.47) we have
\begin{equation*}
d^2=a^2+\left(\frac{2a^2}\ell \right)^2=\frac{a^2}{\ell^2}(\ell^2+4a^2).
\end{equation*}
Let 
$b=\sqrt{\ell^2+4a^2}$,
so $d=ba/\ell$.
Then
\begin{align*}
	\ell x'&=\frac{\ell d}a(x-a)=bx-ba,     &\ell y'&=\ell y-2ax+a^2,
\end{align*}
so
\begin{align*}
	bx&=\ell x'+ba,&	\ell y&=\ell y'+\frac{2a}b(\ell x'+ba)-a^2\\
	  &        &    &=\ell y'+\frac{2\ell a}bx'+a^2.
\end{align*}
Plugging into \eqref{eqn:ay} yields
\begin{gather*}
	\ell y'+\frac{2\ell a}bx'+a^2
	=\left(\frac \ell bx'+a\right)^2,\\
b^2y'=\ell (x')^2.
\end{gather*}
In particular, the new latus rectum is $b^2/\ell$,
which is as claimed, since $b^2/\ell^2=d^2/a^2$.

For his own proof of this,
Apollonius uses a lemma,
Proposition \textsc i.42 
\cite[p.\ 128--31]{Apollonius-Heiberg}.
In Figure~\ref{fig:A.42}, 
\begin{figure}
\centering
\psset{unit=6cm}
\begin{pspicture}(-0.55,-0.3)(0.675,0.6)%\psgrid
\parametricplot{-0.25}{0.5}{t dup mul t}
\psline(-0.5,-0.25)(0.25,0.5)
\uput[u](0.25,0.5){$\grm G$}
\uput[d](-0.5,-0.25){$\grm A$}
\psline(0.0625,0.25)(-0.4375,-0.25)
\uput[d](-0.4375,-0.25){$\grm E$}
\uput[r](0.0625,0.25){$\grm D$}
\psline(0.0625,-0.25)(-0.3125,0.5)
\uput[d](0.0625,-0.25){$\grm B$}
\uput[u](-0.3125,0.5){$\grm H$}
\psline(0.25,0.5)(-0.3125,0.5)
\psline(0.3125,-0.25)(-0.0625,0.5)
\uput[d](0.3125,-0.25){$\grm Z$}
\psline(0.625,-0.25)(0.25,0.5)
\uput[d](0.625,-0.25){$\grm J$}
\psline(0.625,-0.25)(-0.5,-0.25)
\end{pspicture}
\caption{Proposition \textsc i.42 of Apollonius}\label{fig:A.42}
\end{figure}
we have a parabola $\grm{GDB}$ with diameter $\grm{BJ}$.
Here $\grm{DZ}$ and $\grm{GJ}$ are ordinates,
and $\grm{BH}$ is parallel to these,
so it is tangent to the parabola at $\grm B$.
The straight line $\grm{AG}$ is tangent to the parabola at $\grm G$,
which means,
by \textsc i.35,
\begin{equation}\label{eqn:ABBJ}
\grm{AB}=\grm{BJ}.
\end{equation}
Then
\begin{equation}\label{eqn:AGJ}
\grm{AGJ}=\grm{HJ},
\end{equation}
the latter being the parallelogram with those opposite vertices.
The straight line $\grm{DE}$ is drawn parallel to $\grm{GA}$.
Then triangles $\grm{AGJ}$ and $\grm{EDZ}$ are similar,
so their ratio is that of the squares on their bases,
those bases being the ordinates mentioned.
Then the abscissas $\grm{BJ}$ and $\grm{BZ}$ are in that ratio,
and hence the parallelograms $\grm{HJ}$ and $\grm{HZ}$ are in that ratio.
By \eqref{eqn:AGJ} then,
$\grm{AGJ}$ has the same ratio to $\grm{HZ}$ that it does to $\grm{EDZ}$.
Therefore
\begin{equation*}
\grm{EDZ}=\grm{HZ}.
\end{equation*}
The relative positions of $\grm D$ and $\grm G$ on the parabola are irrelevant to the argument:
this will matter for the next theorem.

In Figure~\ref{fig:A.49} now,
\begin{figure}
\centering
\psset{unit=6cm}
\begin{pspicture}(-0.55,-0.3)(1.1125,0.85)
\parametricplot{-0.25}{0.75}{t dup mul t}
\psline(0.625,-0.25)(0.25,0.5)(-0.5,-0.25)(1.0625,-0.25)(0.5625,0.75)(-0.4375,-0.25)
\uput[d](0.625,-0.25){$\grm X$}
\uput[ul](0.25,0.5){$\grm D$}
\uput[d](-0.5,-0.25){$\grm G$}
\uput[d](1.0625,-0.25){$\grm M$}
\uput[u](0.5625,0.75){$\grm K$}
\uput[d](-0.4375,-0.25){$\grm P$}
\psline(0.6825,0.5)(-0.3175,0.5)(0.0625,-0.25)
\uput[r](0.6825,0.5){$\grm N$}
\uput[u](-0.3175,0.5){$\grm Z$}
\uput[d](0.0625,-0.25){$\grm B$}
\uput[dr](0.3125,0.5){$\grm L$}
\uput[l](-0.1275,0.125){$\grm E$}
\end{pspicture}
\caption{Proposition \textsc i.49 of Apollonius}\label{fig:A.49}
\end{figure}
$\grm{KDB}$ is a parabola with diameter $\grm{BM}$,
and 
$\grm{GD}$ is tangent to the parabola, and
through $\grm D$, parallel to the diameter, 
straight line $\grm{DN}$ is drawn
and extended to $\grm Z$ so that
$\grm{ZB}$ is parallel to the ordinate $\grm{DX}$.
A length $\grm H$ (not depicted) is taken such that
\begin{equation}\label{eqn:H}
\grm{ED}:\grm{DZ}\as\grm H:2\grm{GD}.
\end{equation}
Through a random point $\grm K$ on the parabola, 
$\grm{KL}$ is drawn parallel to the tangent $\grm{GD}$.
We shall show
\begin{equation}\label{eqn:KL}
\grm{KL}^2=\grm H\cdot\grm{DL},
\end{equation}
so that $\grm{DN}$ serves as a new diameter of the parabola,
with corresponding latus rectum $\grm H$.
This is Proposition \textsc i.49 of Apollonius.

Since, as before, $\grm{GB}=\grm{BX}$,
we have
\begin{equation}\label{eqn:EZD}
\grm{EZD}=\grm{EGB}.
\end{equation}
Let ordinate $\grm{KNM}$ be drawn.  
Adding to either side of \eqref{eqn:EZD} the pentagon $\grm{DEBMN}$, 
we have
\begin{equation*}
\grm{ZM}=\grm{DGMN}.
\end{equation*}
Let $\grm{KL}$ be extended to $\grm P$.  
By the lemma that we proved above, 
$\grm{KPM}=\grm{ZM}$.  Thus
\begin{equation*}
\grm{KPM}=\grm{DGMN}.
\end{equation*}
Subtracting the trapezoid $\grm{LPMN}$ gives
\begin{equation*}
\grm{KLN}=\grm{LG}.
\end{equation*}
From this, by Theorem \ref{thm:Thales} above, we have
\begin{equation}\label{eqn:KL:}
\grm{KL}\cdot\grm{LN}=2\grm{LD}\cdot\grm{DG}.
\end{equation}
We now compute
\begin{align*}
	\grm{KL}^2:\grm{KL}\cdot\grm{LN}
	&\as	\grm{KL}:\grm{LN}&&\\
&\as\grm{ED}:\grm{DZ}&&\\
&\as\grm H:2\grm{GD}&&\text{[by \eqref{eqn:H}]}\\
&\as\grm H\cdot\grm{LD}:2\grm{LD}\cdot\grm{GD}&&\\
&\as\grm H\cdot\grm{LD}:\grm{KL}\cdot\grm{LN},&&\text{[by \eqref{eqn:KL:}]}
\end{align*}
which yields \eqref{eqn:KL}.
We have assumed $\grm K$ 
to be on the other side of $\grm{DN}$ from $\grm{BM}$.  
The argument can be adapted to the other case.  
Then, as a corollary, 
we have that $\grm{DN}$ bisects all chords parallel to $\grm{DG}$.  
In fact Apollonius proves this independently, in Proposition \textsc i.46.

Could Apollonius have created the proof of \textsc i.49
for pedagogical or ideological reasons,
after verifying the theorem itself
by Cartesian methods, such we we employed?
I have not found any reason to think so.
Before Apollonius,
it seems that only the \emph{right} cone was studied,
and the only sections considered were made by planes
that were orthogonal to straight lines in the surface of the cone
\cite[p.\ xxiv]{Heath-Apollonius}.
Whether an ellipse, parabola, or hyperbola was obtained
depended on the angle at the vertex of the cone.
The recognition that the cone can be oblique,
and every section can be obtained from every cone,
seems to be due to Apollonius.
That our Cartesian argument started with orthogonal axes
corresponds to starting with a right cone.
On the other hand, this feature was not essential to the argument;
we did not really need the parameter $b$.


\chapter{Unity}\label{ch:unity}

In addition to \emph{Elements} \textsc i.26, 
which is A.S.A. and A.A.S. as discussed earlier,
Proclus 
\cite[157.11, 250.20, 299.4]{MR1200456}
attributes to Thales three more of Euclid\apo s propositions,
depicted in Figure \ref{fig:3}:
\begin{figure}
\mbox{}\hfill
\psset{unit=16mm}
\begin{pspicture}(-1,-1)(1,1)
\pscircle(0,0)1
\psline(-1,0)(1,0)
\psline(0,-1)(0,1)
\psline(-0.86,-0.5)(0.86,0.5)
\psline(-0.5,-0.86)(0.5,0.86)
\psline(-0.86,0.5)(0.86,-0.5)
\psline(-0.5,0.86)(0.5,-0.86)
\end{pspicture}
\hfill
\begin{pspicture}(-1,-1)(1,1)
\pspolygon(-1,-1)(1,-1)(0,1)
\end{pspicture}
\hfill
\begin{pspicture}(-0.5,-1)(0.5,1)
\psline(-0.5,-1)(0.5,1)
\psline(-0.5,1)(0.5,-1)
\end{pspicture}
\hfill\mbox{}
\caption{Symmetries}\label{fig:3}
\end{figure}
\begin{compactenum}[1)]
\item
the diameter bisects the circle,
as in the remark on, or addendum to, the definition of diameter
given at the head of the \emph{Elements}%
%(the definition is number 17 in Heiberg\apo s and Heath\apo s reckoning)
;
\item
the base angles of an isosceles triangle are equal
(\textsc i.5);
\item
vertical angles are equal
(\textsc i.15).
\end{compactenum}
Kant alludes to the second of these theorems
in the Preface to the B Edition of the \emph{Critique of Pure Reason,} 
in a purple passage of praise for the person who discovered the theorem
\cite[\textsc b x--xi, p.\ 107--8]{Kant}:
\begin{quote}
\textbf{Mathematics} has,
from the earliest times to which the history of human reason reaches,
in that admirable people the Greeks,
traveled the secure path of a science.
Yet it must not be thought that it was as easy for it as for logic---%
in which reason has to do only with itself---%
to find that royal path, or rather itself to open it up;
rather, I believe that mathematics was left groping about for a long time
(chiefly among the Egyptians),
and that its transformation is to be ascribed to a \textbf{revolution,}
brought about by the happy inspiration of a single man in an attempt
from which the road to be taken onward could no longer be missed,
and the secure course of a science was entered on and prescribed for all time
and to an infinite extent.
The history of this revolution in the way of thinking---%
which was far more important than the discovery of the way around the famous Cape---%
and of the lucky one who brought it about,
has not been preserved for us.
But the legend handed down to us by Diogenes Laertius---%
who names the reputed inventor of the smallest elements of geometrical demonstration,
even of those that, according to common judgment,
stand in no need of proof---%
proves that the memory of the alteration wrought by the discovery of this new path
in its earliest footsteps
must have seemed exceedingly important to mathematicians,
and was thereby rendered unforgettable.
A new light broke upon the person who demonstrated the isosceles triangle
(whether he was called \enquote{Thales} or had some other name).
\end{quote}
The boldface is Kant\apo s.
The editors cite a letter 
in which Kant confirms the allusion to \emph{Elements} \textsc i.5.

In apparent disagreement with Kant,
I would suggest that revolutions in thought need not persist;
they must be made afresh by each new thinker.
The student need not realize her or his potential for new thought,
no matter how open the royal path may seem to the teacher.

Kant refers to the discovery of the southern route around Africa:
does he mean the discovery by Bartolomeu Dias in 1488,
or the discovery by the Phoenicians,
sailing the other directions,
two thousand years earlier,
described by Herodotus \cite[\textsc{iv}.42, pp.\ 297--9]{Strassler}?
The account of Herodotus is made plausible by his disbelief that,
in sailing west around the Cape of Good Hope,
the Phoenicians could have found the sun on their right.
Their route was not maintained,
which is why Dias can be hailed as its discoverer.

Likewise may routes to mathematical understanding
not be maintained.
Much of ancient mathematics has been lost to us,
in the slow catastrophe alluded to by the title of
\emph{The Forgotten Revolution}
\cite[p.\ 8]{MR2038833}.
Here Lucio Russo points out 
that the last of the eight books of Apollonius on conic sections no longer exists,
while Books \textsc{v--vii} survive only in Arabic translation,
and we have only Books \textsc{i--iv} in the original Greek.
Presumably this is because the later books of Apollonius 
were found too difficult
by anybody who could afford to have copies made.

We may be able to recover the achievement of Thales,
if Thales be his name.
There is no reason in principle why we cannot understand him
as well as we understand anybody;
but time and loss present great obstacles.
Kant continues with his own interpretation of Thales\apo s thought:
\begin{quote}
For he found that what he had to do
was not to trace what he saw in this figure,
or even trace its mere concept,
and read off, as it were, from the properties of the figure;
but rather that he had to produce the latter 
from what he himself thought into the object and presented
(through construction) according to \emph{a priori} concepts,
and that in order to know something securely \emph{a priori}
he had to ascribe to the thing nothing except what followed necessarily
from what he himself had put into it in accordance with its concept.
\end{quote}
Kant is right,
if he means that the equality of the base angles of an isosceles triangle
does not follow merely from a figure of an isosceles triangle.
The figure itself represents only one instance of the general claim.
One has to recognize that the figure is constructed according to a \emph{principle,}
which in this case can be understood as symmetry.
The three shapes in Figure \ref{fig:3} embody a symmetry
that justifies the corresponding theorems,
even though Euclid does not appeal to this symmetry in his proofs. 

As we discussed,
Thales is also thought to have discovered that the angle in a semicircle is right,
and he may have done this by recognizing that
any two diameters of a circle
are diagonals of an equiangular quadrilateral.
Such quadrilaterals are rectangles,
but this is not so fundamental an observation
as the equality of the base angles of an isosceles triangle;
in fact the \enquote{observation} can be disputed,
as it was by Lobachevski \cite{Lobachevski}.


Thales is held to be the founder of the Ionian school of philosophy.
In \emph{Before Philosophy} \cite[p.\ 251]{Frankfort},
the Frankforts say of the Ionian school that its members
\begin{quote}
proceeded,
with preposterous boldness,
on an entirely unproved assumption.
They held that the universe is an intelligible whole.
In other words,
they presumed that a single order underlies the chaos of our perceptions and,
furthermore, that we are able to comprehend that order.
\end{quote}
For Thales, 
the order of the world was apparently to be explained through the medium of water.
However, this information is third-hand at best.
Aristotle says in \emph{De Caelo}
\cite[II.13, pp.\ 430]{Aristo-Heavens},
\begin{quote}
By these considerations some have been led to assert 
that the earth below us is infinite, 
saying, with Xenophanes of Colophon, 
that it has \senquote{pushed its roots to infinity},---in order to save the trouble of seeking for the cause\lips
Others say the earth rests upon water. 
This, indeed, is the oldest theory that has been preserved, 
and is attributed to Thales of Miletus. 
It was supposed to stay still 
because it floated like wood and other similar substances, 
which are so constituted as to rest upon water but not upon air. 
As if the same account had not to be given of the water which carries the earth as of the earth itself! 
\end{quote}
Aristotle has only second-hand information on Thales,
who seems not to have written any books.
It may be that Aristotle does not understand
what questions Thales was trying to answer.
Aristotle has his own questions,
and criticizes Thales for not answering them.

Aristotle may do a little better by Thales in the \emph{Metaphysics} 
\cite[983\textsuperscript a24, pp.\ 693--5]{Aristo-Metaphysics-Ross},
where he says,
\begin{quote}
Of the first philosophers, then, 
most thought the principles which were of the nature of matter were the only principles of all things\lips
Yet they do not all agree as to the number and the nature of these principles. 
Thales, the founder of this type of philosophy, says the principle is water 
(for which reason he declared that the earth rests on water),
getting the notion perhaps from seeing that the nutriment of all things is moist, 
and that heat itself is generated from the moist and kept alive by it 
(and that from which they come to be is a principle of all things). 
He got his notion from this fact, 
and from the fact that the seeds [\gr{t`a sp'ermata}]
of all things have a moist nature, 
and that water is the origin of the nature of moist things.
\end{quote}
In \emph{The Idea of Nature} \cite[pp.\ 31--2]{Collingwood-IN},
Collingwood has a poetic interpretation of this:
\begin{quote}
The point to be noticed here is not what Aristotle says but what it presupposes, 
namely that Thales conceived the world of nature as an organism: 
in fact, as an animal\lips
he may possibly have conceived the earth as grazing, so to speak, 
on the water in which it floats,
thus repairing its own tissues and the tissues of everything in it 
by taking in water from this ocean and transforming it, 
by processes akin to respiration and digestion, 
into the various parts of its own body\lips
This animal lived in the medium out of which it was made, 
as a cow lives in a meadow. 
But now the question arose, How did the cow get there?\lips
The world was not born, it was made; 
made by the only maker that dare frame its fearful symmetry: God.
\end{quote}
Collingwood is presumably alluding to 
the sixth and final stanza of William Blake\apo s poem
\enquote{The Tyger} \cite{Blake-Experience}:
\begin{quote}
\settowidth{\versewidth}{Dare frame thy fearful symmetry?}
\begin{verse}[\versewidth]
Tyger Tyger burning bright,\\
In the forests of the night,\\
What immortal hand or eye,\\
Dare frame thy fearful symmetry?
\end{verse}
\end{quote}
Like most animals,
the tiger exhibits bilateral symmetry,
the kind of symmetry shown in Figure \ref{fig:3},
though this may not be what Blake is referring to:
his illustration for the poem
shows a tiger from the side, not the front.
For Euclid, \gr{summetr'ia} is what we now call commensurability;
symmetry may also be balance and harmony in a non-mathematical sense
\cite{Pierce-on-comm-symm}.
But this would seem to be what Thales saw, or sought, in the world,
and it is akin to his recognition
of the unity underlying all isosceles triangles,
a unity whereby the equality of the base angles of each of them
can be established once for all.
What made Thales a philosopher
made him a mathematician.

\AfterBibliographyPreamble{\smaller}

%\bibliographystyle{plain}
%\bibliography{../references}
%\bibliography{../../references}

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\end{document}