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authorNorbert Preining <norbert@preining.info>2020-02-07 03:01:03 +0000
committerNorbert Preining <norbert@preining.info>2020-02-07 03:01:03 +0000
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+\section{Presentation and Overview}
+
+\subsection{Why tkz-euclide ?}
+My initial goal was to provide myself and other mathematics teachers with a tool to quickly create Euclidean geometry figures without investing too much effort in learning a new programming language.
+Of course, tkz-euclide is for math teachers who use latex and makes it possible to easily create correct drawings by means of LaTeX.
+
+It appeared that the simplest method was to reproduce the one used to obtain construction by hand.
+To describe a construction, you must of course define the objects but also the actions that you perform. It seemed to me that a syntax close to the language of mathematicians and their students would be more easily understandable; moreover, it also seemed to me that this syntax should be close to that of LaTeX.
+The objects of course are points, segments, lines, triangles, polygons and circles. As for actions, I considered five to be sufficient, namely: define, create, draw, mark and label.
+
+The syntax is perhaps too verbose but it is, I believe, easily accessible.
+As a result, the students like teachers were able to easily access this tool
+
+\subsection{tkz-euclide vs TikZ}
+
+I love programming with TikZ and without TikZ I would never have had the idea to create tkz-euclide but never forget that behind it there is TikZ and that it is always possible to insert code from TikZ. tkz-euclide doesn't prevent you from using TikZ.
+That said, I don't think mixing syntax is a good thing.
+
+There is no need to compare TikZ and tkz-euclide. The latter is not addressed to the same audience as Tikz. The first one allows you to do a lot of things, the second one only does geometry drawings. The first one can do everything the second one does, but the second one will more easily do what you want.
+
+\subsection{How it works}
+
+\subsubsection{Example Part I gold triangle}
+\begin{center}
+\begin{tikzpicture}
+
+\tkzDefPoint(0,0){C} % possible \tkzDefPoint[label=below:$C$](0,0){C} but don't do this
+\tkzDefPoint(2,6){B}
+% We get D and E with a rotation
+\tkzDefPointBy[rotation= center B angle 36](C) \tkzGetPoint{D}
+\tkzDefPointBy[rotation= center B angle 72](C) \tkzGetPoint{E}
+% Toget A we use an intersection of lines
+\tkzInterLL(B,E)(C,D) \tkzGetPoint{A}
+\tkzInterLL(C,E)(B,D) \tkzGetPoint{H}
+% drawing
+\tkzDrawArc[delta=10](B,C)(E)
+\tkzDrawPolygon(C,B,D)
+\tkzDrawSegments(D,A B,A C,E)
+% angles
+\tkzMarkAngles(C,B,D E,A,D) %this is to draw the arcs
+\tkzLabelAngles[pos=1.5](C,B,D E,A,D){$\alpha$}
+\tkzMarkRightAngle(B,H,C)
+\tkzDrawPoints(A,...,E)
+% Label only now
+\tkzLabelPoints[below left](C,A)
+\tkzLabelPoints[below right](D)
+\tkzLabelPoints[above](B,E)
+\end{tikzpicture}
+\end{center}
+
+Let's analyze the figure
+\begin{enumerate}
+ \item CBD and DBE are isosceles triangles; BC=BE and BD is a bisector of the angle CBE.From this we deduce that the CBD and DBE angles are equal and have the same measure $\alpha$.
+ \[\widehat{BAC} +\widehat{ABC} + \widehat{BCA}=180^\circ \text{in the triangle} BAC\]
+ \[3\alpha + \widehat{BCA}=180^\circ \text{in the triangle} CBD\]
+ then
+ \[\alpha + 2\widehat{BCA}=180^\circ \] soit
+ \[\widehat{BCA}=90^\circ -\alpha/2 \]
+ finally \[\widehat{CBD}=\alpha=36^\circ \]
+ the triangle CBD is a "gold" triangle
+\end{enumerate}
+
+How construct a gold triangle or an angle of $36^\circ$ ?
+
+\begin{itemize}
+ \item We place the fixed points $C$ and $D$.|\tkzDefPoint(0,0){C}| and |\tkzDefPoint(4,0){D}|.
+ \item We construct a square $CDef$ and we construct the midpoint $m$ of $Cf$.
+ We can do all of this with a compass and a rule.
+ \item Then we trace an arc with center m through e. This arc cross the line $Cf$ at $n$
+ \item Now the two arcs with center $C$ et $D$ and radius $Cn$ define the point $B$.
+\end{itemize}
+
+
+\begin{minipage}{.4\textwidth}
+ \begin{tikzpicture}
+ \tkzDefPoint(0,0){C}
+ \tkzDefPoint(4,0){D}
+ \tkzDefSquare(C,D)
+ \tkzGetPoints{e}{f}
+ \tkzDefMidPoint(C,f)
+ \tkzGetPoint{m}
+ \tkzInterLC(C,f)(m,e)
+ \tkzGetSecondPoint{n}
+ \tkzInterCC[with nodes](C,C,n)(D,C,n)
+ \tkzGetFirstPoint{B}
+ \tkzDrawSegment[brown,dashed](f,n)
+ \pgfinterruptboundingbox
+ \tkzDrawPolygon[brown,dashed](C,D,e,f)
+ \tkzDrawArc[brown,dashed](m,e)(n)
+ \tkzCompass[brown,dashed,delta=20](C,B)
+ \tkzCompass[brown,dashed,delta=20](D,B)
+ \endpgfinterruptboundingbox
+ \tkzDrawPoints(C,D,B)
+ \tkzDrawPolygon(B,...,D)
+ \end{tikzpicture}
+\end{minipage}
+\begin{minipage}{.6\textwidth}
+ \begin{tkzexample}[code only,small]
+ \begin{tikzpicture}
+ \tkzDefPoint(0,0){C}
+ \tkzDefPoint(4,0){D}
+ \tkzDefSquare(C,D)
+ \tkzGetPoints{e}{f}
+ \tkzDefMidPoint(C,f)
+ \tkzGetPoint{m}
+ \tkzInterLC(C,f)(m,e)
+ \tkzGetSecondPoint{n}
+ \tkzInterCC[with nodes](C,C,n)(D,C,n)
+ \tkzGetFirstPoint{B}
+ \tkzDrawSegment[brown,dashed](f,n)
+ \pgfinterruptboundingbox
+ \tkzDrawPolygon[brown,dashed](C,D,e,f)
+ \tkzDrawArc[brown,dashed](m,e)(n)
+ \tkzCompass[brown,dashed,delta=20](C,B)
+ \tkzCompass[brown,dashed,delta=20](D,B)
+ \endpgfinterruptboundingbox
+ \tkzDrawPoints(C,D,B)
+ \tkzDrawPolygon(B,...,D)
+ \end{tikzpicture}
+ \end{tkzexample}
+\end{minipage}
+
+
+After building the golden triangle $BCD$, we build the point $A$ by noticing that $BD=DA$. Then we get the point $E$ and finally the point $F$. This is done with already intersections of defined objects (line and circle).
+
+
+\begin{center}
+ \begin{tikzpicture}
+ \tkzDefPoint(0,0){C}
+ \tkzDefPoint(4,0){D}
+ \tkzDefSquare(C,D)
+ \tkzGetPoints{e}{f}
+ \tkzDefMidPoint(C,f)
+ \tkzGetPoint{m}
+ \tkzInterLC(C,f)(m,e)
+ \tkzGetSecondPoint{n}
+ \tkzInterCC[with nodes](C,C,n)(D,C,n)
+ \tkzGetFirstPoint{B}
+ \tkzInterLC(C,D)(D,B) \tkzGetSecondPoint{A}
+ \tkzInterLC(B,A)(B,D) \tkzGetSecondPoint{E}
+ \tkzInterLL(B,D)(C,E) \tkzGetPoint{F}
+ \tkzDrawPoints(C,D,B)
+ \tkzDrawPolygon(B,...,D)
+ \tkzDrawPolygon(B,C,D)
+ \tkzDrawSegments(D,A A,B C,E)
+ \tkzDrawArc[delta=10](B,C)(E)
+ \tkzDrawPoints(A,...,F)
+ \tkzMarkRightAngle[fill=blue!20](B,F,C)
+ \tkzFillAngles[fill=blue!10](C,B,D E,A,D)
+ \tkzMarkAngles(C,B,D E,A,D)
+ \tkzLabelAngles[pos=1.5](C,B,D E,A,D){$\alpha$}
+ \tkzLabelPoints[below](A,C,D,E)
+ \tkzLabelPoints[above right](B,F)
+ \end{tikzpicture}
+\end{center}
+
+
+
+\begin{tkzexample}[code only,small]
+ \begin{tikzpicture}
+ \tkzDefPoint(0,0){C}
+ \tkzDefPoint(4,0){D}
+ \tkzDefSquare(C,D)
+ \tkzGetPoints{e}{f}
+ \tkzDefMidPoint(C,f)
+ \tkzGetPoint{m}
+ \tkzInterLC(C,f)(m,e)
+ \tkzGetSecondPoint{n}
+ \tkzInterCC[with nodes](C,C,n)(D,C,n)
+ \tkzGetFirstPoint{B}
+ \tkzInterLC(C,D)(D,B) \tkzGetSecondPoint{A}
+ \tkzInterLC(B,A)(B,D) \tkzGetSecondPoint{E}
+ \tkzInterLL(B,D)(C,E) \tkzGetPoint{F}
+ \tkzDrawPoints(C,D,B)
+ \tkzDrawPolygon(B,...,D)
+ \tkzDrawPolygon(B,C,D)
+ \tkzDrawSegments(D,A A,B C,E)
+ \tkzDrawArc[delta=10](B,C)(E)
+ \tkzDrawPoints(A,...,F)
+ \tkzMarkRightAngle[fill=blue!20](B,F,C)
+ \tkzFillAngles[fill=blue!10](C,B,D E,A,D)
+ \tkzMarkAngles(C,B,D E,A,D)
+ \tkzLabelAngles[pos=1.5](C,B,D E,A,D){$\alpha$}
+ \tkzLabelPoints[below](A,C,D,E)
+ \tkzLabelPoints[above right](B,F)
+ \end{tikzpicture}
+\end{tkzexample}
+
+\subsubsection{Example Part II two others methods gold and euclide triangle}
+
+tkz-euclide knows how to define a "gold" or "euclide" triangle. We can define BCD and BCA comme des triangles d'or
+
+
+ \begin{center}
+ \begin{tkzexample}[code only,small]
+ \begin{tikzpicture}
+ \tkzDefPoint(0,0){C}
+ \tkzDefPoint(4,0){D}
+ \tkzDefTriangle[gold](C,D)
+ \tkzGetPoint{B}
+ \tkzDefTriangle[gold](B,C)
+ \tkzGetPoint{A}
+ \tkzInterLC(B,A)(B,D) \tkzGetSecondPoint{E}
+ \tkzInterLL(B,D)(C,E) \tkzGetPoint{F}
+ \tkzDrawPoints(C,D,B)
+ \tkzDrawPolygon(B,...,D)
+ \tkzDrawPolygon(B,C,D)
+ \tkzDrawSegments(D,A A,B C,E)
+ \tkzDrawArc[delta=10](B,C)(E)
+ \tkzDrawPoints(A,...,F)
+ \tkzMarkRightAngle[fill=blue!20](B,F,C)
+ \tkzFillAngles[fill=blue!10](C,B,D E,A,D)
+ \tkzMarkAngles(C,B,D E,A,D)
+ \tkzLabelAngles[pos=1.5](C,B,D E,A,D){$\alpha$}
+ \tkzLabelPoints[below](A,C,D,E)
+ \tkzLabelPoints[above right](B,F)
+ \end{tikzpicture}
+ \end{tkzexample}
+ \end{center}
+
+Voici une dernière méthode qui utilise des rotations
+
+\begin{center}
+ \begin{tkzexample}[code only,small]
+ \begin{tikzpicture}
+ \tkzDefPoint(0,0){C} % possible
+ % \tkzDefPoint[label=below:$C$](0,0){C}
+ % but don't do this
+ \tkzDefPoint(2,6){B}
+ % We get D and E with a rotation
+ \tkzDefPointBy[rotation= center B angle 36](C) \tkzGetPoint{D}
+ \tkzDefPointBy[rotation= center B angle 72](C) \tkzGetPoint{E}
+ % To get A we use an intersection of lines
+ \tkzInterLL(B,E)(C,D) \tkzGetPoint{A}
+ \tkzInterLL(C,E)(B,D) \tkzGetPoint{H}
+ % drawing
+ \tkzDrawArc[delta=10](B,C)(E)
+ \tkzDrawPolygon(C,B,D)
+ \tkzDrawSegments(D,A B,A C,E)
+ % angles
+ \tkzMarkAngles(C,B,D E,A,D) %this is to draw the arcs
+ \tkzLabelAngles[pos=1.5](C,B,D E,A,D){$\alpha$}
+ \tkzMarkRightAngle(B,H,C)
+ \tkzDrawPoints(A,...,E)
+ % Label only now
+ \tkzLabelPoints[below left](C,A)
+ \tkzLabelPoints[below right](D)
+ \tkzLabelPoints[above](B,E)
+ \end{tikzpicture}
+ \end{tkzexample}
+\end{center}
+
+
+\subsubsection{Complete but minimal example}
+
+
+A unit of length being chosen, the example shows how to obtain a segment of length $\sqrt{a}$ from a segment of length $a$, using a ruler and a compass.
+
+$IB=a$, $AI=1$
+
+\vspace{12pt}
+\hypertarget{firstex}{}
+
+\begin{tikzpicture}[scale=1,ra/.style={fill=gray!20}]
+ % fixed points
+ \tkzDefPoint(0,0){A}
+ \tkzDefPoint(1,0){I}
+ % calculation
+ \tkzDefPointBy[homothety=center A ratio 10 ](I) \tkzGetPoint{B}
+ \tkzDefMidPoint(A,B) \tkzGetPoint{M}
+ \tkzDefPointWith[orthogonal](I,M) \tkzGetPoint{i}
+ \tkzInterLC(I,i)(M,B) \tkzGetSecondPoint{C}
+ \tkzDrawSegment[style=orange](I,C)
+ \tkzDrawArc(M,B)(A)
+ \tkzDrawSegment[dim={$1$,-16pt,}](A,I)
+ \tkzDrawSegment[dim={$a/2$,-10pt,}](I,M)
+ \tkzDrawSegment[dim={$a/2$,-16pt,}](M,B)
+ \tkzMarkRightAngle[ra](A,I,C)
+ \tkzDrawPoints(I,A,B,C,M)
+ \tkzLabelPoint[left](A){$A(0,0)$}
+ \tkzLabelPoints[above right](I,M)
+ \tkzLabelPoints[above left](C)
+ \tkzLabelPoint[right](B){$B(10,0)$}
+ \tkzLabelSegment[right=4pt](I,C){$IC=\sqrt{a}$}
+\end{tikzpicture}
+
+\emph{Commentaires}
+
+\begin{itemize}
+\item The Preamble
+
+
+ Let us first look at the preamble. If you need it, you have to load \tkzname{xcolor} before \tkzname{tkz-euclide}, that is, before \TIKZ\ . \TIKZ\ may cause problems with the active characters , but...
+ provides a library in its latest version that's supposed to solve these problems \NameLib{babel}.
+
+\begin{tkzltxexample}[]
+\documentclass{standalone} % or another class
+ % \usepackage{xcolor} % before tikz or tkz-euclide if necessary
+\usepackage{tkz-euclide} % no need to load TikZ
+ % \usetkzobj{all} is no longer necessary
+ % \usetikzlibrary{babel} if there are problems with the active characters
+\end{tkzltxexample}
+
+The following code consists of several parts:
+
+ \item Definition of fixed points: the first part includes the definitions of the points necessary for the construction, these are the fixed points. The macros \tkzcname{tkkzInit} and \tkzcname{tkkzClip} in most cases are not necessary.
+
+\begin{tkzltxexample}[]
+ \tkzDefPoint(0,0){O}
+ \tkzDefPoint(1,0){I}
+ \tkzDefPoint(10,0){B}
+\end{tkzltxexample}
+
+ \item The second part is dedicated to the creation of new points from the fixed points;
+ a $B$ point is placed at $10 cm$ from $A$. The middle of $[AB]$ is defined by $M$ and then the orthogonal line to the $(AB)$ line is searched for at the $I$ point. Then we look for the intersection of this line with the semi-circle of center $M$ passing through $A$.
+
+\begin{tkzltxexample}[]
+ \tkzDefPointBy[homothety=center A ratio 10 ](I)
+ \tkzGetPoint{B}
+ \tkzDefMidPoint(A,B)
+ \tkzGetPoint{M}
+ \tkzDefPointWith[orthogonal](I,M)
+ \tkzGetPoint{H}
+ \tkzInterLC(I,H)(M,A)
+ \tkzGetSecondPoint{B}
+ \end{tkzltxexample}
+
+
+ \item The third one includes the different drawings;
+ \begin{tkzltxexample}[]
+ \tkzDrawSegment[style=dashed](I,H)
+ \tkzDrawPoints(O,I,A,B,M)
+ \tkzDrawArc(M,A)(O)
+ \tkzDrawSegment[dim={$1$,-16pt,}](O,I) % voir la documentation pour l'usage de dim
+ \tkzDrawSegment[dim={$a/2$,-10pt,}](I,M)
+ \tkzDrawSegment[dim={$a/2$,-16pt,}](M,A)
+ \end{tkzltxexample}
+
+\item Marking: the fourth is devoted to marking;
+
+
+\begin{tkzltxexample}[]
+ \tkzMarkRightAngle(A,I,B)
+ \end{tkzltxexample}
+
+ \item Labelling: the latter only deals with the placement of labels.
+\begin{tkzltxexample}[]
+ \tkzLabelPoint[left](O){$A(0,0)$}
+ \tkzLabelPoint[right](A){$B(10,0)$}
+ \tkzLabelSegment[right=4pt](I,B){$\sqrt{a^2}=a \ (a>0)$}
+\end{tkzltxexample}
+
+
+\item The full code:
+
+
+\begin{tkzexample}[code only]
+ \begin{tikzpicture}[scale=1,ra/.style={fill=gray!20}]
+ % fixed points
+ \tkzDefPoint(0,0){A}
+ \tkzDefPoint(1,0){I}
+ % calculation
+ \tkzDefPointBy[homothety=center A ratio 10 ](I) \tkzGetPoint{B}
+ \tkzDefMidPoint(A,B) \tkzGetPoint{M}
+ \tkzDefPointWith[orthogonal](I,M) \tkzGetPoint{i}
+ \tkzInterLC(I,i)(M,B) \tkzGetSecondPoint{C}
+
+ \tkzDrawSegment[style=orange](I,C)
+ \tkzDrawArc(M,B)(A)
+ \tkzDrawSegment[dim={$1$,-16pt,}](A,I)
+ \tkzDrawSegment[dim={$a/2$,-10pt,}](I,M)
+ \tkzDrawSegment[dim={$a/2$,-16pt,}](M,B)
+ \tkzMarkRightAngle[ra](A,I,C)
+ \tkzDrawPoints(I,A,B,C,M)
+ \tkzLabelPoint[left](A){$A(0,0)$}
+ \tkzLabelPoints[above right](I,M)
+ \tkzLabelPoints[above left](C)
+ \tkzLabelPoint[right](B){$B(10,0)$}
+ \tkzLabelSegment[right=4pt](I,C){$IC=\sqrt{a}$}
+ \end{tikzpicture}
+\end{tkzexample}
+\end{itemize}
+
+\newpage
+\subsection{The Elements of tkz code}
+In this paragraph, we start looking at the "rules" and "symbols" used to create a figure with tkz-euclide.
+
+ The primitive objects are points. You can refer to a point at any time using the name given when defining it. (it is possible to assign a different name later on).
+
+\medskip
+In general, tkz-euclide macros have a name beginning with tkz. There are four main categories starting with~:
+|\tkzDef...| |\tkzDraw...| |\tkzMark...| et |\tkzLabel...|
+
+Among the first category, |\tkzDefPoint| allows you to define fixed points. It will be studied in detail later. Here we will see in detail the macro DefTriangle |\tkzDefTriangle|.
+
+This macro makes it possible to associate to a pair of points a third point in order to define a certain triangle |\tkzDefTriangle(A,B)|. The obtained point is referenced |tkzPointResult| and it is possible to choose another reference with |\tkzGetPoint{C}| for example.
+Parentheses are used to pass arguments. In |(A,B)| $A$ and $B$ are the points with which a third will be defined.
+
+However, in |{C}| we use braces to retrieve the new point.
+In order to choose a certain type of triangle among the following choices :
+ |equilateral|, |halftone|, |pythagoras|, |school|, |golden or sublime|, |euclide|, |gold|, |cheops|...
+ and |two angles| you just have to choose between hooks, for example~:
+
+|\tkzDefTriangle[euclide](A,B) \tkzGetPoint{C}|
+
+\begin{minipage}{0.5\textwidth}
+ \begin{tikzpicture}[scale=.75]
+ \tkzDefPoints{0/0/A,8/0/B}
+ \foreach \tr in {equilateral,half,pythagore,%
+ school,golden,euclide, gold,cheops}
+ {\tkzDefTriangle[\tr](A,B) \tkzGetPoint{C}
+ \tkzDrawPoint(C)
+ \tkzLabelPoint[right](C){\tr}
+ \tkzDrawSegments(A,C C,B)}
+ \tkzDrawPoints(A,B)
+ \tkzDrawSegments(A,B)
+ \end{tikzpicture}
+\end{minipage}
+\begin{minipage}{0.5\textwidth}
+ \begin{tkzexample}[code only,small]
+ \begin{tikzpicture}[scale=.75]
+ \tkzDefPoints{0/0/A,8/0/B}
+ \foreach \tr in {equilateral,half,pythagore,%
+ school,golden,euclide, gold,cheops}
+ {\tkzDefTriangle[\tr](A,B) \tkzGetPoint{C}
+ \tkzDrawPoint(C)
+ \tkzLabelPoint[right](C){\tr}
+ \tkzDrawSegments(A,C C,B)}
+ \tkzDrawPoints(A,B)
+ \tkzDrawSegments(A,B)
+ \end{tikzpicture}
+ \end{tkzexample}
+
+\end{minipage}
+
+
+\subsection{Conventions}
+
+For this documentation, I used the geometric French and personal conventions for naming the points:
+\begin{itemize}
+\item $O$ is a center for a circle, a rotation, etc.;
+\item $M$ defined a midpoint;
+\item $H$ defined the foot of an altitude;
+\item $P'$ is the image of $P$ by a transformation ;
+\item $a$ defined an angle (degree), $r$ the length of a radius, $d$ a length (or dimension);
+\item ($x_1$,$y_1$) coordinates of the point $A_1$, ($x_A$,$y_A$) coordinates of the point $A$;
+\item $[AB]$ a line segment, $(AB)$ a line.
+\end{itemize}
+
+\subsection{How to use the tkz-euclide package ?}
+\subsubsection{Let's look at a classic example}
+In order to show the right way, we will see how to build an equilateral triangle. Several possibilities are open to us, we are going to follow the steps of Euclid.
+
+\begin{itemize}
+\item First of all you have to use a document class. The best choice to test your code is to create a single figure with the class \tkzname{standalone}\index{standalone}.
+\begin{verbatim}
+\documentclass{standalone}
+\end{verbatim}
+\item Then load the tkz-euclide package:
+\begin{verbatim}
+\usepackage{tkz-euclide}
+\end{verbatim}
+
+ You don't need to load \TIKZ\ because the tkz-euclide package works on top of TikZ and loads it.
+ \item {\color{red} \bomb \sout{|\BS usetkzobj{all}| }}
+ With the new version 3.02 you don't need this line anymore. All objects are now loaded.
+ \item Start the document and open a TikZ picture environment:
+\begin{verbatim}
+\begin{document}
+\begin{tikzpicture}
+\end{verbatim}
+
+\item Now we define two fixed points:
+\begin{verbatim}
+\tkzDefPoint(O,O){A}
+\tkzDefPoint(5,2){B}
+\end{verbatim}
+
+\item Two points define two circles, let's use these circles :
+
+ circle with center $A$ through $B$ and circle with center $B$ through $A$. These two circles have two points in common.
+\begin{verbatim}
+\tkzInterCC(A,B)(B,A)
+\end{verbatim}
+we can get the points of intersection with
+\begin{verbatim}
+\tkzGetPoints{C}{D}
+\end{verbatim}
+
+\item All the necessary points are obtained, we can move on to the final steps including the plots.
+\begin{verbatim}
+\tkzDrawPolygon(A,B,C)% The triangle
+\end{verbatim}
+\item Draw all points A,B,C and D :
+\begin{verbatim}
+\tkzDrawPoints(A,...,D)
+\end{verbatim}
+
+\item The final step, we print labels to the points and use options for positioning:\\
+\begin{verbatim}
+\tkzLabelPoints[below left](A)
+\tkzLabelPoints(B,D)
+\tkzLabelPoint (above] (C){$C$}
+\end{verbatim}
+\item We finally close both environments
+\begin{verbatim}
+\end{tikzpicture}
+\end{document}
+\end{verbatim}
+
+\item The complete code
+
+\begin{tkzexample}[latex=8cm,small]
+ \begin{tikzpicture}[scale=.5]
+ % fixed points
+ \tkzDefPoint(0,0){A}
+ \tkzDefPoint(5,2){B}
+ % calculus
+ \tkzInterCC(A,B)(B,A)
+ \tkzGetPoints{C}{D}
+ % drawings
+ \tkzDrawCircles[gray,dashed](A,B B,A)
+ \tkzDrawPolygon(A,B,C)
+ \tkzDrawPoints(A,...,D)
+ % marking
+ \tkzMarkSegments[mark=s||](A,B B,C C,A)
+ % labelling
+ \tkzLabelSegments[swap](A,B){$c$}
+ \tkzLabelPoints(A,B,D)
+ \tkzLabelPoints[above](C)
+\end{tikzpicture}
+\end{tkzexample}
+
+ \end{itemize}
+
+\subsubsection{"Set, Calculate, Draw, Mark, Label"}
+The title could have been : \texttt{Separation of Calculus and Drawings}
+
+When a document is prepared using the LaTeX system, the source code of the document can be divided into two parts: the document body and the preamble.
+Under this methodology, publications can be structured, styled and typeset with minimal effort.
+I propose a similar methodology for creating figures with tkz-euclide.
+
+The first part defines the fixed points, the second part allows the creation of new points. These are the two main parts. All that is left to do is to draw, mark and label.
+
+
+
+
+\endinput
+
+
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+