parallel.tex

\chapter{Parallel lines}\label{chap:angle-sum}

{

\begin{wrapfigure}{o}{19mm}
\vskip-4mm
\centering
\includegraphics{mppics/pic-72}
\end{wrapfigure}

Recall that in consequence of Axiom~\ref{def:birkhoff-axioms:1},
any two distinct lines $\ell$ and $m$ have either one point in common or none; see \ref{lem:line-line}.
In the first case they are \index{intersecting lines}\emph{intersecting} (briefly $\ell\nparallel m$); 
in the second case, $\ell$ and $m$ are said to be \index{parallel!lines}\emph{parallel} (briefly, \index{36@\hskip.4mm$\parallel$\hskip.4mm, \hskip.4mm$\nparallel$\hskip.4mm}$\ell\parallel m$);
in addition, a line is always regarded as parallel to itself.

}

To emphasize that two lines in a picture are parallel we will mark them with arrows of the same type.



\begin{thm}[\abs]{Proposition}\label{prop:perp-perp} Let $\ell$, $m$, and $n$ be three lines.
Assume that $n\perp m$ and $m\perp \ell$.
Then $\ell\parallel n$. 
\end{thm}

\parit{Proof.}
Assume the contrary; 
that is, $\ell\nparallel n$.
Then there is a point, say $Z$, of intersection of $\ell$ and~$n$.
Then by Theorem~\ref{perp:ex+un},
$\ell=n$.
Since any line is parallel to itself, we have that $\ell\parallel n$ --- a contradiction.
\qeds

\begin{thm}{Theorem}\label{thm:parallel}
For any point $P$ and any line $\ell$,
there is a unique line $m$
that passes thru $P$ and is parallel to~$\ell$.
\end{thm}

The above theorem has two parts, existence and uniqueness.
In the proof of uniqueness, we will use the method of similar triangles.

\parit{Proof; existence.} 
Apply Theorem~\ref{perp:ex+un} two times,
first to construct the line $n$ thru $P$ that is perpendicular to $\ell$,
and second to construct the line $n$ thru $P$ that is perpendicular to~$m$.
Then apply Proposition~\ref{prop:perp-perp}.

\parit{Uniqueness.}
If $P\in\ell$, then $m=\ell$ by the definition of parallel lines.
Now, assume $P\notin\ell$.

Let us construct the lines $n\ni P$ and $m\ni P$ as in the proof of existence; so $n\perp \ell$, $m\perp n$, and $m\parallel \ell$.

Assume there is another line $s\ni P$ parallel to~$\ell$.
Choose a point $Q\in s$ that lies with $\ell$ on the same side from~$m$.
Let $R$ be the footpoint of $Q$ on~$n$.

\begin{figure}[!ht]
\centering
\includegraphics{mppics/pic-74}
\end{figure}

Let $D$ be the point of intersection of $n$ and~$\ell$.
According to Proposition~\ref{prop:perp-perp} $(QR)\parallel m$. 
Therefore, $Q$, $R$, and $\ell$ lie on the same side of~$m$. 
In particular, $R\in [P D)$.

Choose $Z\in [P Q)$ such that 
$$\frac{PZ}{PQ}=\frac{PD}{PR};$$
it exists by Proposition~\ref{prop:point-on-half-line}.
By SAS similarity condition (or equivalently by Axiom~\ref{def:birkhoff-axioms:4})
we have that $\triangle RPQ\sim \triangle DPZ$;
therefore $(Z D)\perp(P D)$.
It follows that $Z$ lies on $\ell$ and $s$ --- a contradiction.\qeds

\begin{thm}{Corollary}\label{cor:parallel-1}
Assume $\ell$, $m$, and $n$ are lines
such that $\ell\parallel m$ and $m\parallel n$.
Then $\ell\parallel n$.
\end{thm}

\parit{Proof.}
Assume the contrary; that is, $\ell\nparallel n$.
Then there is a point $P\in \ell\cap n$.
By Theorem~\ref{thm:parallel},
$n=\ell$ --- a contradiction.
\qeds

By the definition, we have that $\ell\parallel m$ if and only if $m\z\parallel \ell$.
Therefore, according to the above corollary, ``$\parallel$'' is an 
\index{equivalence relation}\emph{equivalence relation}.
That is, for any lines $\ell$, $m$, and $n$ the following conditions hold:
\begin{enumerate}[(i)]
\item $\ell\parallel \ell$;
\item if $\ell\parallel m$, then $m\parallel \ell$;
\item if $\ell\parallel m$ and $m\parallel n$, then 
$\ell\parallel n$.
\end{enumerate}

\begin{thm}{Exercise}\label{ex:perp-perp}
Let $k$, $\ell$, $m$, and $n$ be lines such that $k\perp \ell$, $\ell\perp m$, and $m\perp n$.
Show that $k\nparallel n$.
\end{thm}

\begin{thm}{Exercise}\label{ex:construction-parallel}
Perform a ruler-and-compass construction of a line thru a given point that is parallel to a given line.
\end{thm}

\section{Reflection across a point}

Recall that if $O$ is the midpoint of the line segment $[XX']$,
then we say that $X'$ is a \index{reflection across a point}\emph{reflection} of $X$ across a point $O$.
In addition, we assume that $O'=O$; that is, $O$ is a reflection of itself across itself.

The following statement is a refinement of Proposition~\ref{prop:point-reflection}.

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

\parit{Proof.}
Choose a point $O$.
By Proposition~\ref{prop:point-reflection}, the reflection across $O$ is a motion.
Note that any angle $\angle XOY$ is vertical to its reflection $\angle X'OY'$.
By \ref{prop:vert}, $\measuredangle XOY\z=\measuredangle X'OY'$.
Therefore, the reflection cannot be indirect.

By \ref{prop:direct-indirect}, any motion is either direct or indirect.
Therefore, the reflection across $O$ must be direct.
\qeds

\begin{thm}{Exercise}
Suppose $\angle AOB$ is right.
Show that the composition of reflections across the lines $(OA)$ and $(OB)$ is a reflection across $O$.

Use this statement and Corollary~\ref{cor:reflection+angle} to build another proof of~\ref{prop:point-reflection+}.
\end{thm}

{

\begin{wrapfigure}{r}{23mm}
\vskip-6mm
\centering
\includegraphics{mppics/pic-78}
\end{wrapfigure}


\begin{thm}{Theorem}\label{thm:parallel-point-reflection}
Let $\ell$ be a line, $Q\in \ell$, and $P$ an arbitrary point.
Suppose $O$ is the midpoint of $[PQ]$.

Then a line $m$ passing thru $P$ is parallel to $\ell$ if and only if $m$ is a reflection of $\ell$ across $O$.
\end{thm}

}

\parit{Proof; ``if'' part.}
Assume $m$ is a reflection of $\ell$ across $O$.
Suppose $\ell\nparallel m$; that is $\ell$ and $m$ intersect at a single point $Z$.
Denote by $Z'$ be the reflection of $Z$ across $O$.

\begin{figure}[!ht]
\centering
\includegraphics{mppics/pic-80}
\end{figure}

The point $Z'$ lies on both lines $\ell$ and $m$.
It follows that $Z'=Z$ or equivalently $Z=O$.
In this case, $O\in \ell$ and therefore the reflection of $\ell$ across $O$ is $\ell$ itself;
that is, $\ell=m$ and in particular $\ell\parallel m$ --- a contradiction. 

\parit{``Only-if'' part.}
Let $\ell'$ be the reflection of $\ell$ across $O$.
According to the ``if'' part of the theorem, $\ell'\parallel \ell$.
Both lines $\ell'$ and $m$ pass thru $P$.
By uniqueness of parallel lines (\ref{thm:parallel}), if $m\parallel \ell$, then $\ell'=m$; whence the statement follows.
\qeds

\pagebreak%???

\section{Transversal property}

{

\begin{wrapfigure}{r}{25mm}
\centering
\vskip-10mm
\includegraphics{mppics/pic-82}
\end{wrapfigure}

If the line $t$ intersects each line $\ell$ and $m$ at one point, then we say that $t$ is a \index{transversal}\emph{transversal} to $\ell$ and~$m$.
For example, in the picture, line $(CB)$ is a transversal
to $(AB)$ and~$(CD)$.

}

\begin{thm}{Transversal property}\label{thm:parallel-2} 
$(AB)\parallel(C D)$ if and only if 
$$2\cdot(\measuredangle A B C+\measuredangle B C D)\equiv 0.
\eqlbl{A B C + B C D}$$ 
Equivalently, 
$$\measuredangle A B C+\measuredangle B C D
\equiv 
0
\quad
\text{or}
\quad
\measuredangle A B C+\measuredangle B C D
\equiv
\pi.$$

Moreover, if $(AB)\ne(C D)$, then in the first case, 
$A$ and $D$ lie on opposite sides of $(BC)$;
in the second case,
$A$ and $D$ lie on the same sides of~$(BC)$.
\end{thm}



\parit{Proof; ``only-if'' part.}
Denote by $O$ the midpoint of $[BC]$.

Assume $(AB)\parallel(C D)$.
According to Theorem~\ref{thm:parallel-point-reflection},
$(CD)$ is a reflection of $(AB)$ across $O$.

\begin{wrapfigure}{r}{31mm}
\vskip-4mm
\centering
\includegraphics{mppics/pic-84}
\end{wrapfigure}

Let $A'$ be the reflection of $A$ across $O$.
Then $A'\in (CD)$ and by Proposition~\ref{prop:point-reflection+} we have that
\[\measuredangle ABO=\measuredangle A'CO.\eqlbl{A B O = A' C O}\]
Note that 
\[\measuredangle ABO\equiv\measuredangle ABC,
\qquad
\measuredangle A'CO\equiv-\measuredangle BCA'.\eqlbl{eq:A B O= A B C}
\]
Since $A'$, $C$, and $D$ lie on one line, Exercise~\ref{ex:ABCO-line} implies that 
\[2\cdot \measuredangle BCD\equiv 2\cdot \measuredangle BCA'.\eqlbl{eq:2BCD= 2BCA'}\]
Finally, \ref{A B O = A' C O}, \ref{eq:A B O= A B C}, and \ref{eq:2BCD= 2BCA'} imply \ref{A B C + B C D}.
\qeds

\parit{``If''-part.}
By Theorem~\ref{thm:parallel} there is a unique line $(CD)$ thru $C$ that is parallel to $(AB)$.
From the ``only-if'' part we know that  \ref{A B C + B C D} holds.

On the other hand, there is a \textit{unique} line $(CD)$ such that \ref{A B C + B C D} holds.
Indeed, suppose there are two such lines $(CD)$ and $(CD')$, then
$$2\cdot(\measuredangle A B C+\measuredangle B C D)\equiv 2\cdot(\measuredangle A B C+\measuredangle B C D')\equiv0.
$$ 
Therefore 
$2\cdot\measuredangle B C D\equiv 2\cdot\measuredangle B C D'$
and by Exercise~\ref{ex:ABCO-line},  $D'\in (CD)$, or equivalently the line $(CD)$ coincides with $(CD')$.

Therefore if \ref{A B C + B C D} holds, then $(CD)\parallel (AB)$.


\parit{Last statement.}
If $(AB)\ne(C D)$ and $A$ and $D$ lie on the opposite sides of $(BC)$, then $\angle ABC$ and $\angle BCD$ have opposite signs.
Therefore
\[-\pi\z<\measuredangle A B C+\measuredangle B C D<\pi.\]
Applying \ref{A B C + B C D}, we get $\measuredangle A B C+\measuredangle B C D=0$.

Similarly, if $A$ and $D$ lie on the same side of $(BC)$,
then $\angle ABC$ and $\angle BCD$ have the same sign.
Therefore
\[0<|\measuredangle A B C+\measuredangle B C D|<2\cdot\pi,\]
and \ref{A B C + B C D} implies that $\measuredangle A B C+\measuredangle B C D\z\equiv\pi$.
\qeds

\begin{thm}{Exercise}\label{ex:smililar+parallel}
Let $\triangle ABC$ be a nondegenerate triangle, and $P$ lies between $A$ and $B$.
Suppose that a line $\ell$ passes thru $P$ and is parallel to $(AC)$.
Show that $\ell$ crosses the side $[BC]$ at another point, say $Q$, and 
\[\triangle ABC\sim\triangle PBQ.\]
In particular, 
\[\frac{PB}{AB}=\frac{QB}{CB}.\]

\end{thm} 

\begin{thm}{Exercise}\label{ex:trisection}
Trisect a given segment with a ruler and a compass.
\end{thm}

\section{Angles of triangles}

\begin{thm}{Theorem}\label{thm:3sum}
In any $\triangle A B C$, we have
$$\measuredangle A B C+ \measuredangle B C A + \measuredangle C A B \equiv \pi.$$

\end{thm}

\parit{Proof.} 
If $\triangle A B C$ is degenerate, then the equality follows from Corollary~\ref{cor:degenerate=pi}.
Now, assume that $\triangle A B C$ is nondegenerate.

\begin{wrapfigure}{o}{23mm}
\centering
\includegraphics{mppics/pic-86}
\end{wrapfigure} 

Let $X$ be the reflection of $C$ across the midpoint $M$ of $[AB]$.
By Proposition~\ref{prop:point-reflection+}
$\measuredangle BAX\z=\measuredangle ABC$.
Note that $(AX)$ is a reflection of $(CB)$ across $M$;
therefore by Theorem~\ref{thm:parallel-point-reflection}, $(AX)\z\parallel (CB)$.

Since $[BM]$ and $[MX]$ do not intersect $(CA)$,
the points $B$, $M$, and $X$ lie on the same side of $(CA)$.
Applying the transversal property for the transversal $(CA)$ to $(AX)$ and $(CB)$, we get that 
\[\measuredangle BCA+\measuredangle CAX\equiv \pi.\eqlbl{eq:ABC+CAB}\]

Since $\measuredangle BAX=\measuredangle ABC$,
we have 
\[\measuredangle CAX\equiv\measuredangle CAB+\measuredangle ABC.\]
The latter identity and \ref{eq:ABC+CAB} imply the theorem.\qeds

\begin{thm}{Exercise}\label{ex:|3sum|}
Show that 
$$|\measuredangle A B C|+ |\measuredangle B C A| + |\measuredangle C A B| = \pi$$
for any $\triangle ABC$.
\end{thm} 

{

\begin{wrapfigure}{r}{30mm}
\vskip-0mm
\centering
\includegraphics{mppics/pic-88}
\vskip4mm
\includegraphics{mppics/pic-90}
%\vskip4mm
%\includegraphics{mppics/pic-92}
\end{wrapfigure}

\begin{thm}{Exercise}\label{ex:pent}
Let $\triangle ABC$ be a nondegenerate triangle.
Assume there is a point $D\in [BC]$ 
such that 
\[\measuredangle BAD\equiv \measuredangle DAC,
\quad
BA=AD=DC.\]
Find the angles of $\triangle ABC$. 
\end{thm}


\begin{thm}{Exercise}\label{ex:right-isos}
Let $\triangle ABC$ be an isosceles nondegenerate triangle with the base~$[AC]$.
Suppose $D$ is a reflection of $A$ across $B$.
Show that $\angle ACD$ is right.
\end{thm}



%\begin{thm}{Exercise}\label{ex:pi/4-isos}
%Let $\triangle ABC$ be an isosceles nondegenerate triangle with base~$[AC]$. 
%Assume that a circle is passing thru $A$, centered at a point on $[AB]$, and tangent to $(BC)$ at the point~$X$.
%Show that $\measuredangle CAX\z=\pm\tfrac\pi4$.
%\end{thm}

}

\begin{thm}{Exercise}\label{ex:quadrangle}
Show that for any quadrangle $ABCD$, we have
$$\measuredangle ABC+\measuredangle BCD+\measuredangle CDA+\measuredangle DAB\equiv 0.$$

\end{thm}


\section{Parallelograms}

{

\begin{wrapfigure}{r}{26mm}
\vskip-10mm
\centering
\includegraphics{mppics/pic-94}
\end{wrapfigure}

A quadrangle $ABCD$ in the Euclidean plane is called \index{quadrangle!degenerate quadrangle}\index{degenerate!quadrangle}\emph{nondegenerate} if no three points from $A,B,C,D$ lie on one line.

}

A nondegenerate quadrangle  is called a \index{parallelogram}\emph{parallelogram}
if its opposite sides are parallel.



\begin{thm}{Lemma}\label{lem:parallelogram}
Any parallelogram is \index{central symmetry}\emph{centrally symmetric} with respect to a midpoint of one of its diagonals;
that is, the reflection across the midpoint maps the parallelogram to itself.

In particular, if $\square A B C D$ is a parallelogram, then
\begin{enumerate}[(a)]
\item\label{lem:parallelogram:midpoint} its diagonals $[AC]$ and $[BD]$ intersect each other at their midpoints;
\item $\measuredangle A B C= \measuredangle C D A$;
\item $AB=CD$.
\end{enumerate}
\end{thm}

{

\begin{wrapfigure}{r}{33mm}
\centering
\includegraphics{mppics/pic-96}
\end{wrapfigure}

\parit{Proof.} Let $\square A B C D$ be a parallelogram.
Denote by $M$ the midpoint of $[AC]$.

Since $(AB)\parallel (CD)$, Theorem~\ref{thm:parallel-point-reflection} implies that $(CD)$ is a reflection of $(AB)$ across $M$.
In the same way, $(BC)$ is a reflection of $(DA)$ across $M$.
Since $\square A B C D$ is nondegenerate, it follows that $D$ is a reflection of $B$ across $M$; in other words, $M$ is the midpoint of $[BD]$.

The remaining statements follow since reflection across $M$ is a direct motion of the plane (see \ref{prop:point-reflection+}).
\qeds

}

\begin{thm}{Exercise}\label{ex:4parallels}
Assume that point $P$ is distinct from vertices of a parallelogram $ABCD$.
Let $a$, $b$, $c$ and $d$ be lines thru points $A$, $B$, $C$ and $D$ such that 
$a\parallel (CP)$,
$b\parallel (DP)$,
$c\parallel (AP)$, and
$d\parallel (BP)$.
Show that the lines $a$, $b$, $c$ and $d$ intersect at one point.
\end{thm}


\begin{thm}{Exercise}\label{ex:romb}
Assume $ABCD$ is a quadrangle such that
\[AB=CD=BC=DA.\]
Show that $ABCD$ is a parallelogram.
\end{thm}

A quadrangle as in the exercise above is called a \index{rhombus}\emph{rhombus}.

A quadrangle $ABCD$ is called a \index{rectangle}\emph{rectangle} if the angles $ABC$, $BCD$, $CDA$, and $DAB$ are right.
According to the transversal property (\ref{thm:parallel-2}),
any rectangle is a parallelogram.

A rectangle with equal sides is called a \index{square}\emph{square}.

\begin{thm}{Exercise}\label{ex:rectangle}
Show that a parallelogram $ABCD$ is a rectangle
if and only if $AC=BD$.
\end{thm}

\begin{thm}{Exercise}\label{ex:romb2}
Show that a parallelogram $ABCD$ is a rhombus
if and only if $(AC)\perp (BD)$.
\end{thm}

Assume $\ell\parallel m$, and $X,Y\in m$.
Let $X'$ and $Y'$ denote the footpoints of $X$ and $Y$ on~$\ell$.
Note that $\square XYY'X'$ is a rectangle.
By Lemma~\ref{lem:parallelogram}, $XX'=YY'$.
That is, any point on $m$ lies at the same distance from $\ell$.
This distance is called the \index{distance!between parallel lines}\emph{distance between} $\ell$ and~$m$.


\section{Method of coordinates}

The following exercise is important;
it shows that our axiomatic definition agrees with the model described in Section~\ref{page:model}.


\begin{thm}{Exercise}\label{ex:coordinates} 
Let $\ell$ and $m$ be perpendicular lines in the Euclidean plane.
Given a point $P$, let $P_\ell$ and $P_m$ denote the footpoints of $P$ on $\ell$ and $m$ respectively.

\begin{enumerate}[(a)]
\item Show that for any $X\in \ell$ and $Y\in m$ there is a unique point $P$ such that $P_\ell=X$ and $P_m=Y$.
\end{enumerate}

\begin{enumerate}[(a)]\addtocounter{enumi}{1}
\item
Show that 
$PQ^2=P_\ell Q_\ell^2+P_mQ_m^2$
for any pair of points $P$ and~$Q$.
\end{enumerate}

\begin{enumerate}[(a)]\addtocounter{enumi}{2}
\item Conclude that the plane is isometric to $(\mathbb{R}^2,d_2)$; see \ref{ex:dist-square}.
\end{enumerate}

\end{thm}

\begin{wrapfigure}{r}{37mm}
\centering
\includegraphics{mppics/pic-98}
\end{wrapfigure}

Once this exercise is solved, we can apply 
the method of coordinates
to solve any problem in Euclidean plane geometry.
This method is powerful and universal;
it will be developed further in Chapter~\ref{chap:complex}.

\begin{thm}{Exercise}\label{ex:abc}
Use Exercise~\ref{ex:coordinates}
to give an alternative proof of Theorem~\ref{thm:abc} in the Euclidean plane.

That is, prove that given the real numbers $a$, $b$, and $c$ such that 
 $$0<a\le b\le c\le a+b,$$
there is a triangle $ABC$
such that $a=BC$, $b=CA$, and $c=AB$.
\end{thm} 

\begin{thm}{Exercise}\label{ex:line-coord}
Consider two distinct points $A=(x_A,y_A)$ and $B\z=(x_B,y_B)$ on the coordinate plane.
Show that the perpendicular bisector of $[AB]$ is described by the equation
\[2\cdot (x_B-x_A)\cdot x+2\cdot (y_B-y_A)\cdot y=x_B^2+y_B^2-x_A^2-y_A^2.\]

Conclude that line can be defined as a subset of the coordinate plane of the following type:
\begin{enumerate}[(a)]
\item  Solutions of an equation $a\cdot x+b\cdot y=c$
for constants $a$, $b$, and $c$ such that $a\ne 0$ or $b\ne0$.
\item\label{ex:line-coord:parameter} The set of points $(a\cdot t+c,b\cdot t+d)$ for constants $a$, $b$, $c$, and $d$ such that $a\ne 0$ or $b\ne0$ and all $t\in \mathbb{R}$.
\end{enumerate}

\end{thm}

\section{Apollonian circle}\label{sec:Apollonian circle}
The exercises in this section illustrate the method of coordinates; they will not be used further in the sequel.

\begin{thm}{Exercise}\label{ex:circle-coord}
Show that for fixed real values $a$, $b$, and $c$ the equation 
\[x^2+y^2+a\cdot x+b\cdot y+c=0\]
describes a circle, a point, or an empty set.

Show that if it is a circle, then $(-\tfrac a2,-\tfrac b2)$ is its center,
and $r\z=\tfrac12\cdot \sqrt{a^2+b^2-4\cdot c}$ is its radius.
\end{thm}

\begin{thm}{Exercise}\label{ex:apolonnius}
Use the previous exercise to show that given a positive real number $k\ne1$,
the locus of points $M$ such that $AM=k\cdot BM$ 
for distinct points $A$ and $B$
is a circle. 
\end{thm}

\begin{figure}[!ht]
\centering
\includegraphics{mppics/pic-100}
\end{figure}

The circle in the exercise above is an example of the so-called \index{Apollonian circle}\emph{Apollonian circle with focuses $A$ and $B$}.
A few of these circles for different values of $k$ are shown in the picture;
for $k=1$, it becomes the perpendicular bisector of $[AB]$.


\begin{thm}{Exercise}\label{ex:apolonnius-construction}
Construct an Apollonian circle with given focuses $A$ and $B$ thru a given point $M$ using ruler and compass.
\end{thm}