axioms.tex

%\part*{Euclidean geometry}
\addtocontents{toc}{\protect\contentsline{part}{\protect\numberline{}Euclidean geometry}{}{}}
  
\chapter{Axioms}
\label{chap:axioms}

\vfill

A system of axioms appears already in Euclid's ``Elements'' --- the most successful and influential textbook ever written.

The systematic study of geometries as axiomatic systems
 was
triggered by the discovery of non-Euclidean geometry.
The branch of mathematics, emerging this way, is called ``Foundations of geometry''.

The most popular system of axioms
was proposed by David Hilbert in 1899.
This is also the first rigorous system by modern standards.
It consists of twenty axioms in five groups, six ``primitive notions'', and three ``primitive terms'';
these are not defined in terms of previously defined concepts.

Later many different systems were proposed.
It is worth mentioning
the system developed by Alexandr Alexandrov \cite{alexandrov} which is intuitive and elementary, 
the system by Friedrich Bachmann \cite{bachmann} based on the concept of symmetry,
and the system by Alfred Tarski \cite{tarski} --- a minimalist system designed for analysis using mathematical logic.

We will use another system close to the one proposed by George Birkhoff \cite{birkhoff}.
It is based on the key observations (\ref{preaxiomI})--(\ref{preaxiomV}) listed in Section~\ref{preaxioms}.
The axioms use the notions of 
metric space, 
lines, 
angles,
triangles,
equalities modulo $2\cdot\pi$ ($\equiv$), 
the continuity of maps between metric spaces,
and the congruence of triangles ($\cong$).
All of these concepts are discussed in the preliminaries.

Our system is built upon metric spaces.
In particular, we use real numbers as a building block. 
For this reason our approach is not purely axiomatic --- we build the theory upon something else;
it resembles a model-based introduction to Euclidean geometry discussed in Section~\ref{page:model}.
We used this approach to minimize the tedious parts which are unavoidable in purely axiomatic foundations.

\newpage

\section{The axioms}
\label{sec:axioms}


\begin{framed}
\begin{enumerate}[I.]
\item\label{def:birkhoff-axioms:0} The \index{plane!Euclidean plane}\index{Euclidean!plane}\emph{Euclidean plane} is a metric space with at least two points.


\item\label{def:birkhoff-axioms:1} 
There is one and only one line that contains any two given distinct points $P$ and $Q$ in the Euclidean plane.

\item\label{def:birkhoff-axioms:2} 
Any angle $AOB$ in the Euclidean plane 
defines a real number in the interval $(-\pi,\pi]$.
This number is called the \index{angle!measure}\emph{angle measure of $\angle AOB$}
and is denoted by $\measuredangle A O B$.
It satisfies the following conditions:
\begin{enumerate}[(a)]
\item\label{def:birkhoff-axioms:2a} 
Given a half-line $[O A)$ and $\alpha\in(-\pi,\pi]$, 
there is a unique  half-line $[O B)$, 
such that $\measuredangle A O B= \alpha$.
\item\label{def:birkhoff-axioms:2b} 
For any points $A$, $B$, and $C$, distinct from $O$ we have
$$\measuredangle A O B+\measuredangle B O C
\equiv\measuredangle A O C.$$
\item\label{def:birkhoff-axioms:2c} 
The function 
$$\measuredangle\:(A,O,B)\mapsto\measuredangle A O B$$
is continuous at any triple of points $(A,O,B)$,
such that $O\ne A$ and $O\ne B$ and $\measuredangle A O B\ne\pi$.

\end{enumerate}

\item\label{def:birkhoff-axioms:3}  
In the Euclidean plane, we have
$\triangle A B C\cong\triangle A' B' C'$
if and only if 
\begin{align*}
A' B'&=A B, & A' C'&= A C, &&\text{and}
&\measuredangle C' A' B'&=\pm\measuredangle C A B.
\end{align*}
\item\label{def:birkhoff-axioms:4}
If for two triangles $ABC$, $AB'C'$ in the Euclidean plane
and for $k>0$ we have
\begin{align*}
B'&\in [AB),
& C'&\in [AC),
\\
AB'&=k\cdot AB,&
AC'&=k\cdot AC,
\end{align*}
then
\begin{align*}
B'C'&=k\cdot BC,&
\measuredangle ABC&=\measuredangle AB'C',
&
\measuredangle ACB&=\measuredangle AC'B'.
\end{align*}
\end{enumerate}
\end{framed}

From now on,  
we can use no information about the Euclidean plane that does not follow from the five axioms above.

\begin{thm}{Exercise}\label{ex:infinite}
Show that there are (a) an infinite set of points,
(b) an infinite set of lines on the plane.
\end{thm}

\section{Lines and half-lines}

\begin{thm}[\abs]{Proposition}\label{lem:line-line}
\let\thefootnote\relax\footnotetext{${}^\a$ The mark ``$\a$'' indicates that Axiom~\ref{def:birkhoff-axioms:4} is not used in the proof.
Please ignore this mark for now; it will become important in Chapter~\ref{chap:non-euclid}.}
Any two distinct lines intersect at most at one point.
\end{thm}

\parit{Proof.}
Assume that two lines $\ell$ and $m$ intersect at two distinct points $P$ and~$Q$.
Applying Axiom~\ref{def:birkhoff-axioms:1}, we get that $\ell=m$.
\qeds

\begin{thm}{Exercise}\label{ex:[OA)=[OA')}
Suppose $A'\in[OA)$ and $A'\not=O$. 
Show that 
\[[O A)\z=[O A').\]

\end{thm}

\begin{thm}[\abs]{Proposition}\label{prop:point-on-half-line}
Given $r\ge 0$ and a half-line $[O A)$ there is a unique $A'\in [O A)$  such that $O A'=r$.
\end{thm}

\parit{Proof.}
According to the definition of a half-line, 
there is an isometry 
$$f\:[O A)\to [0,\infty)$$
such that $f(O)=0$.
By the definition of an isometry, $O A'=f(A')$ for any $A'\z\in [O A)$.
Thus, $O A'=r$ if and only if $f(A')=r$.

Since an isometry must be bijective, the statement follows.
\qeds

\section{Zero angle}

\begin{thm}[\abs]{Proposition}\label{lem:AOA=0}
$\measuredangle A O A= 0$ for any $A\not=O$.
\end{thm}

\parit{Proof.}
According to Axiom~\ref{def:birkhoff-axioms:2b},
$$\measuredangle A O A
+
\measuredangle A O A 
\equiv
\measuredangle A O A.$$
Subtract  $\measuredangle A O A$ from both sides, we get that
$\measuredangle A O A \equiv 0$.

By Axiom~\ref{def:birkhoff-axioms:2}, $-\pi<\measuredangle A O A\le \pi$;
therefore $\measuredangle A O A \z= 0$.
\qeds

\begin{thm}{Exercise}\label{ex:2.4} 
Assume $\measuredangle A O B= 0$.
Show that $[OA)=[OB)$.
\end{thm}

\begin{thm}[\abs]{Proposition}\label{lem:AOB+BOA=0}
For any $A$ and $B$ distinct from $O$,
we have 
$$\measuredangle A O B\equiv-\measuredangle B O A.$$

\end{thm}

\parit{Proof.}
According to Axiom~\ref{def:birkhoff-axioms:2b},
$$\measuredangle A O B+\measuredangle B O A \equiv\measuredangle A O A.$$
By Proposition~\ref{lem:AOA=0}, $\measuredangle A O A=0$.
Hence the result.
\qeds

\section{Straight angle}

If $\measuredangle A O B=\pi$,
we say that $\angle A O B$ is a 
\index{angle!straight angle}\emph{straight angle}.
By Proposition~\ref{lem:AOB+BOA=0}, 
if $\angle A O B$ is straight,
then so is $\angle B O A$.

We say that point $O$ \index{between}\emph{lies between} points $A$ and $B$, 
if $O\not= A$, $O\not= B$, and $O\in[A B]$.

\begin{thm}[\abs]{Theorem}\label{thm:straight-angle}
An angle $A O B$ is straight 
if and only if $O$ 
\index{between}\emph{lies between} $A$ and~$B$.
\end{thm}

\begin{wrapfigure}{r}{40mm}
\centering
\vskip-8mm
\includegraphics{mppics/pic-8}
\end{wrapfigure}

\parit{Proof.}
By Proposition~\ref{prop:point-on-half-line},  we may assume that
$O A = O B = 1$.

\parit{``If'' part.}
Assume $O$  
lies between $A$ and~$B$.
Set  $\alpha=\measuredangle A O B$.

Applying Axiom~\ref{def:birkhoff-axioms:2a},
we get a half-line $[OA')$ such that $\alpha\z=\measuredangle B O A'$.
By Proposition~\ref{prop:point-on-half-line}, we can assume that $OA'=1$.
According to Axiom~\ref{def:birkhoff-axioms:3},
\[\triangle AOB\z\cong\triangle BOA'.\]
Suppose that $f$ denotes the corresponding motion of the plane;
that is, $f$ is a motion such that $f(A)=B$, $f(O)=O$, and $f(B)=A'$. 

\begin{wrapfigure}{o}{40mm}
\centering
\includegraphics{mppics/pic-10}
\end{wrapfigure}

Then 
\[(A'B)=f((AB))\ni f(O)=O.\]
Therefore, both lines $(AB)$ and $(A'B)$ contain $B$ and~$O$.
By Axiom~\ref{def:birkhoff-axioms:1}, $(AB)=(A'B)$.

By the definition of a line,
$(AB)$ contains exactly two points $A$ and $B$ at distance $1$ from~$O$.
Since $OA'=1$ and $A'\ne B$, we get that $A=A'$.

By Axiom~\ref{def:birkhoff-axioms:2b} and Proposition~\ref{lem:AOA=0}, we get that
\begin{align*}
2\cdot\alpha&=
\measuredangle AOB+\measuredangle BOA'=
\\
&=\measuredangle AOB+\measuredangle BOA\equiv
\\
&\equiv\measuredangle AOA=
\\
&= 0.
\end{align*}
Therefore, by Exercise~\ref{ex:2a=0}, $\alpha$ is either $0$ or~$\pi$.

Since $[OA)\ne [OB)$,  
we have that $\alpha\ne 0$, see Exercise~\ref{ex:2.4}.
Therefore, $\alpha=\pi$.


\parit{``Only if'' part.}
Suppose that $\measuredangle A O B= \pi$.
Consider the line $(OA)$ and choose a point $B'$ on $(OA)$ so that $O$ lies between $A$ and~$B'$.

From above, we have that $\measuredangle AOB'=\pi$.
Applying Axiom~\ref{def:birkhoff-axioms:2a}, 
we get that $[O B)\z=[O B')$.
In particular, $O$ lies between $A$ and~$B$.
\qeds 

A triangle $ABC$ is called 
\index{triangle!degenerate triangle}\index{degenerate! triangle}\emph{degenerate}
if $A$, $B$, and $C$ lie on one line.
The following corollary is just a reformulation of Theorem~\ref{thm:straight-angle}.

\begin{thm}[\abs]{Corollary}\label{cor:degenerate=pi}
A triangle is degenerate if and only if one of its angles is equal to $\pi$ or~$0$.
Moreover, in a degenerate triangle, the angle measures are $0$, $0$, and $\pi$.
\end{thm}

\begin{thm}{Exercise}\label{ex:lineAOB}
Show that three distinct points $A$, $O$, and $B$ lie on one line if and only if 
$$2\cdot \measuredangle AOB\equiv 0.$$ 

\end{thm}

\begin{thm}{Exercise}\label{ex:ABCO-line}
Let $A$, $B$, and $C$ be three points distinct from~$O$.
Show that $B$, $O$, and $C$ lie on one line if and only if
$$2\cdot \measuredangle AOB\equiv 2\cdot \measuredangle AOC.$$ 

\end{thm}

\begin{thm}{Exercise}\label{ex:infinite-number-of-lines} 
Show that there is a nondegenerate triangle.
\end{thm}

\section{Vertical angles}

A pair of angles $AOB$ and $A'OB'$ 
is called \index{angle!vertical angles}\index{vertical angles}\emph{vertical}
if point $O$ 
lies between $A$ and $A'$ 
and between $B$ and $B'$ at the same time.


\begin{thm}[\abs]{Proposition}\label{prop:vert}
Vertical angles have equal measures.
\end{thm}


\parit{Proof.}
Assume that the angles $AOB$ and $A'OB'$ are vertical.
Note that $\angle AOA'$ and $\angle BOB'$ are straight.
Therefore, $\measuredangle AOA'\z=\measuredangle BOB'=\pi$.

{

\begin{wrapfigure}[4]{o}{24mm}
\vskip-2mm
\centering
\includegraphics{mppics/pic-12}
\end{wrapfigure}

It follows that
\begin{align*}
0&=\measuredangle AOA'-\measuredangle BOB'\equiv
\\
&\equiv 
\measuredangle AOB+\measuredangle BOA'-\measuredangle BOA'-\measuredangle A'OB'
\equiv
\\
&\equiv\measuredangle AOB-\measuredangle A'OB'.
\end{align*}
Since $-\pi<\measuredangle AOB\le \pi$ and $-\pi<\measuredangle A'OB'\le \pi$, we get that $\measuredangle AOB\z=\measuredangle A'OB'$.
\qeds

}

\section{Reflection across a point}

\begin{thm}[\abs]{Lemma}\label{lem:point-reflection}
Given distinct points $X$ and $O$, there is a unique point $X'$ such that $O$ is the midpoint of a line segment $[XX']$.
\end{thm}

\begin{wrapfigure}{o}{33mm}
\vskip-0mm
\centering
\includegraphics{mppics/pic-915}
\end{wrapfigure}

\parit{Proof.}
The point $O$ is the midpoint of a line segment $[XX']$ if and only if $X'\in [XO)$ and $XX'=2\cdot XO$.

By Axiom~\ref{def:birkhoff-axioms:1}, the line $(XO)$ and therefore half-line $[XO)$ are uniquely defined.
Further, by Proposition~\ref{prop:point-on-half-line} there is a unique point $X'\in [XO)$ such that $XX'=2\cdot XO$.
\qeds

The point $X'$ provided by the lemma is called \index{reflection across a point}\emph{reflection} of point $X$ across $O$.
We also assume that $O'=O$; that is, $O$ is a reflection of itself across itself.

The point $O$ is called the \index{center!of reflection}\emph{center of reflection}.

\begin{thm}[\abs]{Proposition}\label{prop:point-reflection}
A reflection across a point is a motion of the plane.
\end{thm}

\parit{Proof.}
Let $O$ be the center of reflection.
Observe that if $X'$ is a reflection of $X$ across $O$,
then $X$ is a reflection of $X'$.
In other words, the composition of the reflection with itself is the identity map.
In particular, the reflection is a bijection.

\begin{wrapfigure}{o}{33mm}
\vskip-6mm
\centering
\includegraphics{mppics/pic-76}
\end{wrapfigure}

Now choose two points $X$ and $Y$;
let $X'$ and $Y'$ be their reflections across $O$.
To check that the reflection is distance preserving, we need to show that $X'Y'=XY$.

We may assume that $X$, $Y$, and $O$ are distinct; otherwise, the statement is trivial.
By the definition of the reflection, we have that $OX=OX'$, $OY=OY'$.
Note also that the angles $XOY$ and $X'OY'$ are vertical.
By \ref{prop:vert}, $\measuredangle XOY\z=\measuredangle X'OY'$.

Now Axiom~\ref{def:birkhoff-axioms:3} implies that $\triangle XOY\cong\triangle X'OY'$ and $X'Y'=XY$.
\qeds

\begin{thm}{Exercise}\label{ex:refelection-of-line}
Show that the reflection across a point maps any line to a line.
\end{thm}