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authorNorbert Preining <norbert@preining.info>2020-03-09 03:01:39 +0000
committerNorbert Preining <norbert@preining.info>2020-03-09 03:01:39 +0000
commit0561778af6f8fde0a3a1c334e2aab9a4f1142d6e (patch)
treec19fe45e4101aed56e7fb3602174c6a5e335d7d1 /macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-points.tex
parenta8880a5ee484845d9a526c5ca8d21d31214af22b (diff)
CTAN sync 202003090301
Diffstat (limited to 'macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-points.tex')
-rw-r--r--macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-points.tex603
1 files changed, 72 insertions, 531 deletions
diff --git a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-points.tex b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-points.tex
index 9cea12304c..cce1c9066f 100644
--- a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-points.tex
+++ b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-points.tex
@@ -2,25 +2,24 @@
Points can be specified in any of the following ways:
\begin{itemize}
-\item Cartesian coordinates
-\item Polar coordinates
-\item Named points
-\item Relative points
+\item Cartesian coordinates;
+\item Polar coordinates;
+\item Named points;
+\item Relative points.
\end{itemize}
Even if it's possible, I think it's a bad idea to work directly with coordinates. Preferable is to use named points.
A point is defined if it has a name linked to a unique pair of decimal numbers.
- Let $(x,y)$ or $(a:d)$ i.e. ( $x$ abscissa, $y$ ordinate) or ($a$ angle : $d$ distance ).
- This is possible because the plan has been provided with an orthonormed Cartesian coordinate system. The working axes are supposed to be (ortho)normed with unity equal to $1cm $ or something equivalent like $0.39370~in$.
- Now by default if you use a grid or axes, the rectangle used is defined by the coordinate points~: $(0,0)$ et $(10,10)$. It's the macro \tkzcname{tkzInit} of the package \tkzNamePack{tkz-base} that creates this rectangle. Look at the following two codes and the result of their compilation:
+ Let $(x,y)$ or $(a:d)$ i.e. ($x$ abscissa, $y$ ordinate) or ($a$ angle: $d$ distance).
+ This is possible because the plan has been provided with an orthonormed Cartesian coordinate system. The working axes are supposed to be (ortho)normed with unity equal to $1$~cm or something equivalent like $0.39370$~in.
+ Now by default if you use a grid or axes, the rectangle used is defined by the coordinate points: $(0,0)$ and $(10,10)$. It's the macro \tkzcname{tkzInit} of the package \tkzNamePack{tkz-base} that creates this rectangle. Look at the following two codes and the result of their compilation:
\begin{tkzexample}[latex=10cm,small]
\begin{tikzpicture}
- \tkzGrid
- \tkzDefPoint(0,0){O}
- \tkzDrawPoint[red](O)
- \tkzShowBB[line width=2pt,
- orange]
+\tkzGrid
+\tkzDefPoint(0,0){O}
+\tkzDrawPoint[red](O)
+\tkzShowBB[line width=2pt,teal]
\end{tikzpicture}
\end{tkzexample}
@@ -31,7 +30,7 @@ A point is defined if it has a name linked to a unique pair of decimal numbers.
\tkzDefPoint(5,5){A}
\tkzDrawSegment[blue](O,A)
\tkzDrawPoints[red](O,A)
- \tkzShowBB[line width=2pt,orange]
+ \tkzShowBB[line width=2pt,teal]
\end{tikzpicture}
\end{tkzexample}
@@ -40,13 +39,8 @@ A point is defined if it has a name linked to a unique pair of decimal numbers.
$y$-direction.
A point in polar coordinates requires an angle $\alpha$, in degrees,
- and distance from the origin, $d$. Unlike Cartesian coordinates, the
- distance does not have a default dimensional unit, so one must be
- supplied. The \tikz{} syntax for a point specified in polar
- coordinates is $(\alpha:r\:dim)$, where {\em dim} is a dimensional
- unit such as \texttt{cm}, \texttt{pt}, \texttt{in}, or any other
- \TeX-based unit. Other than syntax and the required dimensional unit,
- this follows usual mathematical usage.
+ and a distance $d$ from the origin with a dimensional
+ unit by default it's the \texttt{cm}.
\begin{minipage}[b]{0.5\textwidth}
@@ -59,7 +53,8 @@ A point is defined if it has a name linked to a unique pair of decimal numbers.
\tkzDefPoint(3,4){A}
\tkzDrawPoints(O,A)
\tkzLabelPoint(A){$A_1 (x_1,y_1)$}
- \tkzShowPointCoord[xlabel=$x_1$,ylabel=$y_1$](A)
+ \tkzShowPointCoord[xlabel=$x_1$,
+ ylabel=$y_1$](A)
\tkzLabelPoints(O,I)
\tkzLabelPoints[left](J)
\tkzDrawPoints[shape=cross](I,J)
@@ -74,14 +69,14 @@ A point is defined if it has a name linked to a unique pair of decimal numbers.
\tkzDefPoints{0/0/O,1/0/I,0/1/J}
\tkzDefPoint(40:4){P}
\tkzDrawXY[noticks,>=triangle 45]
- \tkzDrawSegment[dim={$r$,
+ \tkzDrawSegment[dim={$d$,
16pt,above=6pt}](O,P)
\tkzDrawPoints(O,P)
\tkzMarkAngle[mark=none,->](I,O,P)
\tkzFillAngle[fill=blue!20,
opacity=.5](I,O,P)
\tkzLabelAngle[pos=1.25](I,O,P){$\alpha$}
- \tkzLabelPoint(P){$P (\alpha : r )$}
+ \tkzLabelPoint(P){$P (\alpha : d )$}
\tkzDrawPoints[shape=cross](I,J)
\tkzLabelPoints(O,I)
\tkzLabelPoints[left](J)
@@ -89,51 +84,47 @@ A point is defined if it has a name linked to a unique pair of decimal numbers.
\end{tkzexample}
\end{minipage}%
-The \tkzNameMacro{tkzDefPoint} macro is used to define a point by assigning coordinates to it. This macro is based on \tkzNameMacro{coordinate}, a macro of \TIKZ\ . It can use \TIKZ-specific options such as \IoptName{TikZ}{shift}. If calculations are required then the \tkzNamePack{xfp} package is chosen. We can use Cartesian or polar coordinates.
+The \tkzNameMacro{tkzDefPoint} macro is used to define a point by assigning coordinates to it. This macro is based on \tkzNameMacro{coordinate}, a macro of \TIKZ. It can use \TIKZ-specific options such as \tkzname{shift}. If calculations are required then the \tkzNamePack{xfp} package is chosen. We can use Cartesian or polar coordinates.
\subsection{Defining a named point \tkzcname{tkzDefPoint}}
-\begin{NewMacroBox}{tkzDefPoint}{\oarg{local options}\parg{x,y}\marg{name} ou \parg{a:r}\marg{name}}
-
-\begin{tabular}{lll}
-\toprule
-arguments & défaut & définition \\
+\begin{NewMacroBox}{tkzDefPoint}{\oarg{local options}\parg{$x,y$}\marg{name} or \parg{$\alpha$:$d$}\marg{name}}%
+\begin{tabular}{lll}%
+arguments & default & definition \\
\midrule
-\TAline{(x,y)}{no default}{x et y sont deux dimensions, par défaut en cm.}
-\TAline{(a:d)}{no default}{a est un angle en degré, d une dimension}
-\TAline{\{name\}}{no default}{Nom attribué au point : $A$, $T_a$ ,$P1$ etc ...}
+\TAline{($x,y$)}{no default}{$x$ and $y$ are two dimensions, by default in cm.}
+\TAline{($\alpha$:$d$)}{no default}{$\alpha$ is an angle in degrees, $d$ is a dimension}
+\TAline{\{name\}}{no default}{Name assigned to the point: $A$, $T_a$ ,$P1$ etc ...}
\bottomrule
\end{tabular}
\medskip
-{Les arguments obligatoires de cette macro sont deux dimensions exprimées avec des décimaux, dans le premier cas ce sont deux mesures de longueur, dans le second ce sont une mesure de longueur et la mesure d'un angle en degré}
+The obligatory arguments of this macro are two dimensions expressed with decimals, in the first case they are two measures of length, in the second case they are a measure of length and the measure of an angle in degrees.
\medskip
-\begin{tabular}{lll}
+\begin{tabular}{lll}%
\toprule
options & default & definition \\
\midrule
-\TOline{label} {no default} {permet de placer un label à une distance prédéfinie}
-\TOline{shift} {no default} {Ajoute (x,y) ou (a:d) à toutes les coordonnées}
- \bottomrule
+\TOline{label} {no default} {allows you to place a label at a predefined distance}
+\TOline{shift} {no default} {adds $(x,y)$ or $(\alpha:d)$ to all coordinates}
\end{tabular}
-
\end{NewMacroBox}
- \subsubsection{Cartesian coordinates }
+\subsubsection{Cartesian coordinates }
- \begin{tkzexample}[latex=7cm,small]
- \begin{tikzpicture}
- \tkzInit[xmax=5,ymax=5]
- \tkzDefPoint(0,0){A}
- \tkzDefPoint(4,0){B}
- \tkzDefPoint(0,3){C}
- \tkzDrawPolygon(A,B,C)
- \tkzDrawPoints(A,B,C)
- \end{tikzpicture}
- \end{tkzexample}
-
- \subsubsection{Calculations with \tkzNamePack{xfp}}
+\begin{tkzexample}[latex=7cm,small]
+ \begin{tikzpicture}
+ \tkzInit[xmax=5,ymax=5]
+ \tkzDefPoint(0,0){A}
+ \tkzDefPoint(4,0){B}
+ \tkzDefPoint(0,3){C}
+ \tkzDrawPolygon(A,B,C)
+ \tkzDrawPoints(A,B,C)
+ \end{tikzpicture}
+\end{tkzexample}
+
+\subsubsection{Calculations with \tkzNamePack{xfp}}
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=1]
@@ -159,7 +150,7 @@ options & default & definition \\
\end{tkzexample}
\subsubsection{Calculations and coordinates}
-You must follow the syntax of \tkzNamePack{fxp} here. It is always possible to go through \tkzNamePack{pgfmath} but in this case, the coordinates must be calculated before using the macro \tkzcname{tkzDefPoint}.
+You must follow the syntax of \tkzNamePack{xfp} here. It is always possible to go through \tkzNamePack{pgfmath} but in this case, the coordinates must be calculated before using the macro \tkzcname{tkzDefPoint}.
\begin{tkzexample}[latex=6cm,small]
\begin{tikzpicture}[scale=.5]
@@ -171,8 +162,7 @@ You must follow the syntax of \tkzNamePack{fxp} here. It is always possible to g
\subsubsection{Relative points}
-
-First, we can use the \tkzNameEnv{scope} environment from \TIKZ\ ..
+First, we can use the \tkzNameEnv{scope} environment from \TIKZ.
In the following example, we have a way to define an equilateral triangle.
\begin{tkzexample}[latex=7cm,small]
@@ -193,26 +183,24 @@ In the following example, we have a way to define an equilateral triangle.
\end{tkzexample}
%<--------------------------------------------------------------------------->
-\subsection{Point relative to another : \tkzcname{tkzDefShiftPoint}}
-\begin{NewMacroBox}{tkzDefShiftPoint}{\oarg{Point}\parg{x,y}\marg{name} ou \parg{a:d}\marg{name}}
-\begin{tabular}{lll}
+\subsection{Point relative to another: \tkzcname{tkzDefShiftPoint}}
+\begin{NewMacroBox}{tkzDefShiftPoint}{\oarg{Point}\parg{$x,y$}\marg{name} or \parg{$\alpha$:$d$}\marg{name}}%
+\begin{tabular}{lll}%
arguments & default & definition \\
\midrule
-\TAline{(x,y)}{no default}{x and y are two dimensions, by default in cm.}
-\TAline{(a:d)}{no default}{a is an angle in degrees, d is a dimension}
+\TAline{($x,y$)}{no default}{$x$ and $y$ are two dimensions, by default in cm.}
+\TAline{($\alpha$:$d$)}{no default}{$\alpha$ is an angle in degrees, $d$ is a dimension}
\midrule
options & default & definition \\
\midrule
\TOline{[pt]} {no default} {\tkzcname{tkzDefShiftPoint}[A](0:4)\{B\}}
-\bottomrule
\end{tabular}
-
\end{NewMacroBox}
\subsubsection{Isosceles triangle with \tkzcname{tkzDefShiftPoint}}
-This macro allows you to place one point relative to another. This is equivalent to a translation. Here is how to construct an isosceles triangle with main vertex A and angle at vertex of $30^{\circ} $.
+This macro allows you to place one point relative to another. This is equivalent to a translation. Here is how to construct an isosceles triangle with main vertex $A$ and angle at vertex of $30^{\circ} $.
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[rotate=-30]
@@ -220,29 +208,26 @@ This macro allows you to place one point relative to another. This is equivalent
\tkzDefShiftPoint[A](0:4){B}
\tkzDefShiftPoint[A](30:4){C}
\tkzDrawSegments(A,B B,C C,A)
- \tkzMarkSegments[mark=|,
- color=red](A,B A,C)
+ \tkzMarkSegments[mark=|,color=red](A,B A,C)
\tkzDrawPoints(A,B,C)
\tkzLabelPoints(B,C)
\tkzLabelPoints[above left](A)
\end{tikzpicture}
\end{tkzexample}
-
-
\subsubsection{Equilateral triangle}
Let's see how to get an equilateral triangle (there is much simpler)
-
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=1]
\tkzDefPoint(2,3){A}
- \tkzDefShiftPoint[A](30:4){B}
- \tkzDefShiftPoint[A](-30:4){C}
+ \tkzDefShiftPoint[A](30:3){B}
+ \tkzDefShiftPoint[A](-30:3){C}
\tkzDrawPolygon(A,B,C)
\tkzDrawPoints(A,B,C)
\tkzLabelPoints(B,C)
\tkzLabelPoints[above left](A)
+ \tkzMarkSegments[mark=|,color=red](A,B A,C B,C)
\end{tikzpicture}
\end{tkzexample}
@@ -251,22 +236,21 @@ There's a simpler way
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}
\tkzDefPoint(0,0){A}
- \tkzDefPoint(60:3){B}
- \tkzDefShiftPointCoord[B](30:4){C}
- \tkzDefShiftPointCoord[A](30:4){D}
+ \tkzDefPoint(30:3){B}
+ \tkzDefShiftPointCoord[B](10:2){C}
+ \tkzDefShiftPointCoord[A](10:2){D}
\tkzDrawPolygon(A,...,D)
\tkzDrawPoints(A,...,D)
\end{tikzpicture}
\end{tkzexample}
%<--------------------------------------------------------------------------->
+\subsection{Definition of multiple points: \tkzcname{tkzDefPoints}}
-\subsection{Definition of multiple points : \tkzcname{tkzDefPoints}}
-
-\begin{NewMacroBox}{tkzDefPoints}{\oarg{local options}\marg{$x_1/y_1/n_1,x_2/y_2/n_2$, ...}}
-$x_i$ et $y_i$ are the coordinates of a referenced point $n_i$
+\begin{NewMacroBox}{tkzDefPoints}{\oarg{local options}\marg{$x_1/y_1/n_1,x_2/y_2/n_2$, ...}}%
+$x_i$ and $y_i$ are the coordinates of a referenced point $n_i$
-\begin{tabular}{lll}
+\begin{tabular}{lll}%
\toprule
arguments & default & example \\
\midrule
@@ -274,19 +258,14 @@ arguments & default & example \\
\end{tabular}
\medskip
-\begin{tabular}{lll}
-\toprule
+\begin{tabular}{lll}%
options & default & definition \\
\midrule
-\TOline{label} {no default} {allows you to place a label at a predefined distance}
-\TOline{shift} {no default} {Adds (x,y) or (a:d) to all coordinates}
- \bottomrule
+\TOline{shift} {no default} {Adds $(x,y)$ or $(\alpha:d)$ to all coordinates}
\end{tabular}
-
\end{NewMacroBox}
\subsection{Create a triangle}
-
\begin{tkzexample}[latex=6cm,small]
\begin{tikzpicture}[scale=1]
\tkzDefPoints{0/0/A,4/0/B,4/3/C}
@@ -305,19 +284,17 @@ Note here the syntax for drawing the polygon.
\end{tikzpicture}
\end{tkzexample}
-\newpage
\section{Special points}
The introduction of the dots was done in \tkzname{tkz-base}, the most important macro being \tkzcname{tkzDefPoint}. Here are some special points.
-
%<--------------------------------------------------------------------------->
\subsection{Middle of a segment \tkzcname{tkzDefMidPoint}}
It is a question of determining the middle of a segment.
-\begin{NewMacroBox}{tkzDefMidPoint}{\parg{pt1,pt2}}
+\begin{NewMacroBox}{tkzDefMidPoint}{\parg{pt1,pt2}}%
The result is in \tkzname{tkzPointResult}. We can access it with \tkzcname{tkzGetPoint}.
\medskip
-\begin{tabular}{lll}
+\begin{tabular}{lll}%
\toprule
arguments & default & definition \\
\midrule
@@ -326,7 +303,7 @@ arguments & default & definition \\
\end{NewMacroBox}
\subsubsection{Use of \tkzcname{tkzDefMidPoint}}
-Review the use of \tkzcname{tkzDefPoint} in \NamePack{tkz-base}.
+Review the use of \tkzcname{tkzDefPoint} in \tkzNamePack{tkz-base}.
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=1]
\tkzDefPoint(2,3){A}
@@ -341,18 +318,18 @@ Review the use of \tkzcname{tkzDefPoint} in \NamePack{tkz-base}.
\subsection{Barycentric coordinates }
$pt_1$, $pt_2$, \dots, $pt_n$ being $n$ points, they define $n$ vectors $\overrightarrow{v_1}$, $\overrightarrow{v_2}$, \dots, $\overrightarrow{v_n}$ with the origin of the referential as the common endpoint. $\alpha_1$, $\alpha_2$,
-\dots $\alpha_n$ is $n$ numbers, the vector obtained by :
+\dots $\alpha_n$ are $n$ numbers, the vector obtained by:
\begin{align*}
\frac{\alpha_1 \overrightarrow{v_1} + \alpha_2 \overrightarrow{v_2} + \cdots + \alpha_n \overrightarrow{v_n}}{\alpha_1
+ \alpha_2 + \cdots + \alpha_n}
\end{align*}
defines a single point.
-\begin{NewMacroBox}{tkzDefBarycentricPoint}{\parg{pt1=$\alpha_1$,pt2=$\alpha_2$,\ldots}}
-\begin{tabular}{lll}
+\begin{NewMacroBox}{tkzDefBarycentricPoint}{\parg{pt1=$\alpha_1$,pt2=$\alpha_2$,\dots}}%
+\begin{tabular}{lll}%
arguments & default & definition \\
\midrule
-\TAline{(pt1=$\alpha_1$,pt2=$\alpha_2$,\ldots)}{no default}{Each point has a assigned weight}
+\TAline{(pt1=$\alpha_1$,pt2=$\alpha_2$,\dots)}{no default}{Each point has a assigned weight}
\bottomrule
\end{tabular}
@@ -362,7 +339,7 @@ You need at least two points.
\subsubsection{Using \tkzcname{tkzDefBarycentricPoint} with two points}
-In the following example, we obtain the barycentre of points A and B with coefficients 1 and 2, in other words:
+In the following example, we obtain the barycentre of points $A$ and $B$ with coefficients $1$ and $2$, in other words:
\[
\overrightarrow{AI}= \frac{2}{3}\overrightarrow{AB}
\]
@@ -380,8 +357,7 @@ In the following example, we obtain the barycentre of points A and B with coeffi
\end{tkzexample}
\subsubsection{Using \tkzcname{tkzDefBarycentricPoint} with three points}
-
-This time M is simply the centre of gravity of the triangle. For reasons of simplification and homogeneity, there is also \tkzcname{tkzCentroid}
+This time $M$ is simply the centre of gravity of the triangle. For reasons of simplification and homogeneity, there is also \tkzcname{tkzCentroid}.
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=.8]
\tkzDefPoint(2,1){A}
@@ -430,441 +406,6 @@ The centres of the two homotheties in which two circles correspond are called ex
\end{tikzpicture}
\end{tkzexample}
-\clearpage \newpage
-\section{Special points relating to a triangle}
-
-\subsection{Triangle center : \tkzcname{tkzDefTriangleCenter}}
-
-This macro allows you to define the center of a triangle.
-
-
-\begin{NewMacroBox}{tkzDefTriangleCenter}{\oarg{local options}\parg{A,B,C}}
-\tkzHandBomb\ Be careful, the arguments are lists of three points. This macro is used in conjunction with \tkzcname{tkzGetPoint} to get the center you are looking for. You can use \tkzname{tkzPointResult} if it is not necessary to keep the results.
-
-\medskip
-\begin{tabular}{lll}
-\toprule
-arguments & default & definition \\
-
-\midrule
-\TAline{(pt1,pt2,pt3)}{no default}{three points}
-\midrule
-options & default & definition \\
-\midrule
-\TOline{ortho} {circum}{Intersection of the altitudes of a triangle}
-\TOline{centroid} {circum}{centre of gravity. Intersection of the medians }
-\TOline{circum}{circum}{circle center circumscribed}
-\TOline{in} {circum}{centre du cercle inscrit dans à un triangle }
-\TOline{ex} {circum}{center of a circle exinscribed to a triangle }
-\TOline{euler}{circum}{centre of Euler's circle }
-\TOline{symmedian} {circum}{Lemoine's point or symmedian centre or Grebe's point }
-\TOline{spieker} {circum}{Spieker Circle Center}
-\TOline{nagel}{circum}{Nagel Centre}
-\TOline{mittenpunkt} {circum}{or else MiddlePoint center}
-\TOline{feuerbach}{circum}{Feuerbach Point}
-
-\end{tabular}
-\end{NewMacroBox}
-
-\subsubsection{\IoptName{tkzDefTriangleCenter}{ortho}}
- The intersection H of the three altitudes of a triangle is called the orthocenter.
-
-
-\begin{tkzexample}[latex=5cm,small]
-\begin{tikzpicture}
- \tkzDefPoint(0,0){A}
- \tkzDefPoint(5,1){B}
- \tkzDefPoint(1,4){C}
- \tkzClipPolygon(A,B,C)
- \tkzDefTriangleCenter[ortho](B,C,A)
- \tkzGetPoint{H}
- \tkzDefSpcTriangle[orthic,name=H](A,B,C){a,b,c}
- \tkzDrawPolygon[color=blue](A,B,C)
- \tkzDrawPoints(A,B,C,H)
- \tkzDrawLines[add=0 and 1](A,Ha B,Hb C,Hc)
- \tkzLabelPoint(H){$H$}
- \tkzAutoLabelPoints[center=H](A,B,C)
- \tkzMarkRightAngles(A,Ha,B B,Hb,C C,Hc,A)
-\end{tikzpicture}
-\end{tkzexample}
-
-\subsubsection{\IoptName{tkzDefTriangleCenter}{centroid}}
-
-\begin{tkzexample}[latex=5cm,small]
-\begin{tikzpicture}[scale=.75]
- \tkzDefPoints{-1/1/A,5/1/B}
- \tkzDefEquilateral(A,B)
- \tkzGetPoint{C}
- \tkzDefTriangleCenter[centroid](A,B,C)
- \tkzGetPoint{G}
- \tkzDrawPolygon[color=brown](A,B,C)
- \tkzDrawPoints(A,B,C,G)
- \tkzDrawLines[add = 0 and 2/3](A,G B,G C,G)
-\end{tikzpicture}
-\end{tkzexample}
-
-\subsubsection{\IoptName{tkzDefTriangleCenter}{circum}}
-
-\begin{tkzexample}[latex=6cm,small]
- \begin{tikzpicture}
- \tkzDefPoints{0/1/A,3/2/B,1/4/C}
- \tkzDefTriangleCenter[circum](A,B,C)
- \tkzGetPoint{G}
- \tkzDrawPolygon[color=brown](A,B,C)
- \tkzDrawCircle(G,A)
- \tkzDrawPoints(A,B,C,G)
- \end{tikzpicture}
-\end{tkzexample}
-
-
-\subsubsection{\IoptName{tkzDefTriangleCenter}{in}}
- In geometry, the incircle or inscribed circle of a triangle is the largest circle contained in the triangle; it touches (is tangent to) the three sides. The center of the incircle is a triangle center called the triangle's incenter.
- The center of the incircle, called the incenter, can be found as the intersection of the three internal angle bisectors. The center of an excircle is the intersection of the internal bisector of one angle (at vertex A, for example) and the external bisectors of the other two. The center of this excircle is called the excenter relative to the vertex A, or the excenter of A.[3] Because the internal bisector of an angle is perpendicular to its external bisector, it follows that the center of the incircle together with the three excircle centers form an orthocentric system.(\url{https://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle})
-
- \medskip
- We get the centre of the inscribed circle of the triangle. The result is of course in \tkzname{tkzPointResult}. We can retrieve it with \tkzcname{tkzGetPoint}.
-
-\begin{tkzexample}[latex=6cm,small]
-\begin{tikzpicture}
- \tkzDefPoints{0/1/A,3/2/B,1/4/C}
- \tkzDefTriangleCenter[in](A,B,C)\tkzGetPoint{I}
- \tkzDefPointBy[projection=onto A--C](I)
- \tkzGetPoint{Ib}
- \tkzDrawPolygon[color=blue](A,B,C)
- \tkzDrawPoints(A,B,C,I)
- \tkzDrawLines[add = 0 and 2/3](A,I B,I C,I)
- \tkzDrawCircle(I,Ib)
-\end{tikzpicture}
-\end{tkzexample}
-
-\subsubsection{\IoptName{tkzDefTriangleCenter}{ex}}
-
-
-An excircle or escribed circle of the triangle is a circle lying outside the triangle, tangent to one of its sides and tangent to the extensions of the other two. Every triangle has three distinct excircles, each tangent to one of the triangle's sides.
-(\url{https://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle})
-
-
- We get the centre of an inscribed circle of the triangle. The result is of course in \tkzname{tkzPointResult}. We can retrieve it with \tkzcname{tkzGetPoint}.
-
-\begin{tkzexample}[latex=8cm,small]
-\begin{tikzpicture}[scale=.5]
- \tkzDefPoints{0/1/A,3/2/B,1/4/C}
- \tkzDefCircle[ex](B,C,A)
- \tkzGetFirstPoint{J_c}
- \tkzGetSecondPoint{Tc}
- \tkzDrawPolygon[color=blue](A,B,C)
- \tkzDrawPoints(A,B,C,J_c)
- \tkzDrawCircle[red](J_c,Tc)
- \tkzDrawLines[add=1.5 and 0](A,C B,C)
- \tkzLabelPoints(J_c)
-\end{tikzpicture}
-\end{tkzexample}
-
-\subsubsection{Utilisation de \IoptName{tkzDefTriangleCenter}{euler} }
-This macro allows to obtain the center of the circle of the nine points or euler's circle or Feuerbach's circle.
-The nine-point circle, also called Euler's circle or the Feuerbach circle, is the circle that passes through the perpendicular feet $H_A$, $H_B$, and $H_C$ dropped from the vertices of any reference triangle ABC on the sides opposite them. Euler showed in 1765 that it also passes through the midpoints $M_A$, $M_B$, $M_C$ of the sides of ABC. By Feuerbach's theorem, the nine-point circle also passes through the midpoints $E_A$, $E_B$, and $E_C$ of the segments that join the vertices and the orthocenter H. These points are commonly referred to as the Euler points. (\url{http://mathworld.wolfram.com/Nine-PointCircle.html})
-
-\begin{tkzexample}[latex=7cm,small]
-\begin{tikzpicture}[scale=1]
- \tkzDefPoints{0/0/A,6/0/B,0.8/4/C}
- \tkzDefSpcTriangle[medial,
- name=M](A,B,C){_A,_B,_C}
- \tkzDefTriangleCenter[euler](A,B,C)
- \tkzGetPoint{N} % I= N nine points
- \tkzDefTriangleCenter[ortho](A,B,C)
- \tkzGetPoint{H}
- \tkzDefMidPoint(A,H) \tkzGetPoint{E_A}
- \tkzDefMidPoint(C,H) \tkzGetPoint{E_C}
- \tkzDefMidPoint(B,H) \tkzGetPoint{E_B}
- \tkzDefSpcTriangle[ortho,name=H](A,B,C){_A,_B,_C}
- \tkzDrawPolygon[color=blue](A,B,C)
- \tkzDrawCircle(N,E_A)
- \tkzDrawSegments[blue](A,H_A B,H_B C,H_C)
- \tkzDrawPoints(A,B,C,N,H)
- \tkzDrawPoints[red](M_A,M_B,M_C)
- \tkzDrawPoints[blue]( H_A,H_B,H_C)
- \tkzDrawPoints[green](E_A,E_B,E_C)
- \tkzAutoLabelPoints[center=N,
- font=\scriptsize](A,B,C,%
- M_A,M_B,M_C,%
- H_A,H_B,H_C,%
- E_A,E_B,E_C)
- \tkzLabelPoints[font=\scriptsize](H,N)
- \tkzMarkSegments[mark=s|,size=3pt,
- color=blue,line width=1pt](B,E_B E_B,H)
-\end{tikzpicture}
-\end{tkzexample}
-
-
-\subsubsection{Using option \IoptName{tkzDefTriangleCenter}{symmedian}}
-
-\begin{tkzexample}[latex=6cm,small]
-\begin{tikzpicture}
- \tkzDefPoint(0,0){A}
- \tkzDefPoint(5,0){B}
- \tkzDefPoint(1,4){C}
- \tkzDefTriangleCenter[symmedian](A,B,C)\tkzGetPoint{K}
- \tkzDefTriangleCenter[median](A,B,C)\tkzGetPoint{G}
- \tkzDefTriangleCenter[in](A,B,C)\tkzGetPoint{I}
- \tkzDefSpcTriangle[centroid,name=M](A,B,C){a,b,c}
- \tkzDefSpcTriangle[incentral,name=I](A,B,C){a,b,c}
- \tkzDrawPolygon[color=blue](A,B,C)
- \tkzDrawPoints(A,B,C,K)
- \tkzDrawLines[add = 0 and 2/3,blue](A,K B,K C,K)
- \tkzDrawSegments[red,dashed](A,Ma B,Mb C,Mc)
- \tkzDrawSegments[orange,dashed](A,Ia B,Ib C,Ic)
- \tkzDrawLine(G,I)
-\end{tikzpicture}
-\end{tkzexample}
-
-\subsubsection{Using option \IoptName{tkzDefTriangleCenter}{nagel}}
-
-Let$ Ta$ be the point at which the $Ja$ excircle meets the side BC of a triangle $ABC$, and define Tband Tc similarly. Then the lines $ATa$, $BTb$, and $CTc$ concur in the Nagel point $Na$.
-\href{http://mathworld.wolfram.com/NagelPoint.html}{Weisstein, Eric W. "Nagel point." From MathWorld--A Wolfram Web Resource. }
-
-
-\begin{tkzexample}[latex=8cm,small]
- \begin{tikzpicture}[scale=.5]
- \tkzDefPoints{0/0/A,6/0/B,4/6/C}
- \tkzDefSpcTriangle[ex](A,B,C){Ja,Jb,Jc}
- \tkzDefSpcTriangle[extouch](A,B,C){Ta,Tb,Tc}
- \tkzDrawPoints(Ja,Jb,Jc,Ta,Tb,Tc)
- \tkzLabelPoints(Ja,Jb,Jc,Ta,Tb,Tc)
- \tkzDrawPolygon[blue](A,B,C)
- \tkzDefTriangleCenter[nagel](A,B,C) \tkzGetPoint{Na}
- \tkzDrawPoints[blue](B,C,A)
- \tkzDrawPoints[red](Na)
- \tkzLabelPoints[blue](B,C,A)
- \tkzLabelPoints[red](Na)
- \tkzDrawLines[add=0 and 1](A,Ta B,Tb C,Tc)
- \tkzShowBB\tkzClipBB
- \tkzDrawLines[add=1 and 1,dashed](A,B B,C C,A)
- \tkzDrawCircles[ex,gray](A,B,C C,A,B B,C,A)
- \tkzDrawSegments[dashed](Ja,Ta Jb,Tb Jc,Tc)
- \tkzMarkRightAngles[fill=gray!20](Ja,Ta,C Jb,Tb,A Jc,Tc,B)
- \end{tikzpicture}
-\end{tkzexample}
-
-
-\subsubsection{Option Triangle "mittenpunkt"}
-\begin{tkzexample}[latex=8cm,small]
-\begin{tikzpicture}[scale=.5]
- \tkzDefPoints{0/0/A,6/0/B,4/6/C}
- \tkzDefSpcTriangle[centroid](A,B,C){Ma,Mb,Mc}
- \tkzDefSpcTriangle[ex](A,B,C){Ja,Jb,Jc}
- \tkzDefSpcTriangle[extouch](A,B,C){Ta,Tb,Tc}
- \tkzDefTriangleCenter[mittenpunkt](A,B,C)
- \tkzGetPoint{Mi}
- \tkzDrawPoints(Ma,Mb,Mc,Ja,Jb,Jc)
- \tkzClipBB
- \tkzDrawPolygon[blue](A,B,C)
- \tkzDrawLines[add=0 and 1](Ja,Ma
- Jb,Mb Jc,Mc)
- \tkzDrawLines[add=1 and 1](A,B A,C B,C)
- \tkzDrawCircles[gray](Ja,Ta Jb,Tb Jc,Tc)
- \tkzDrawPoints[blue](B,C,A)
- \tkzDrawPoints[red](Mi)
- \tkzLabelPoints[red](Mi)
- \tkzLabelPoints[left](Mb)
- \tkzLabelPoints(Ma,Mc,Jb,Jc)
- \tkzLabelPoints[above left](Ja,Jc)
- \tkzShowBB
-\end{tikzpicture}
-\end{tkzexample}
-%<--------------------------------------------------------------------------->
-%<--------------------------------------------------------------------------->
-\clearpage \newpage
-\section{Draw a point}
-\subsubsection{Drawing points \tkzcname{tkzDrawPoint}} \hypertarget{tdrp}{}
-
-\begin{NewMacroBox}{tkzDrawPoint}{\oarg{local options}\parg{name}}
-\begin{tabular}{lll}
-arguments & default & definition \\
-\midrule
-\TAline{name of point} {no default} {Only one point name is accepted}
-\bottomrule
-\end{tabular}
-
-\medskip
-The argument is required. The disc takes the color of the circle, but lighter. It is possible to change everything. The point is a node and therefore it is invariant if the drawing is modified by scaling.
-
-\medskip
-\begin{tabular}{lll}
-\toprule
-options & default & definition \\
-\midrule
-\TOline{shape} {circle}{Possible \tkzname{cross} ou \tkzname{cross out}}
-\TOline{size} {6}{$6 \times$ \tkzcname{pgflinewidth}}
-\TOline{color} {black}{the default color can be changed }
-\bottomrule
-\end{tabular}
-
-\medskip
-{We can create other forms such as \tkzname{cross}}
-\end{NewMacroBox}
-
-\subsubsection{Example of point drawings}
-Note that \tkzname{scale} does not affect the shape of the dots. Which is normal. Most of the time, we are satisfied with a single point shape that we can define from the beginning, either with a macro or by modifying a configuration file.
-
-
-\begin{tkzexample}[latex=5cm,small]
- \begin{tikzpicture}[scale=.5]
- \tkzDefPoint(1,3){A}
- \tkzDefPoint(4,1){B}
- \tkzDefPoint(0,0){O}
- \tkzDrawPoint[color=red](A)
- \tkzDrawPoint[fill=blue!20,draw=blue](B)
- \tkzDrawPoint[color=green](O)
- \end{tikzpicture}
-\end{tkzexample}
-
-It is possible to draw several points at once but this macro is a little slower than the previous one. Moreover, we have to make do with the same options for all the points.
-
-\hypertarget{tdrps}{}
-\begin{NewMacroBox}{tkzDrawPoints}{\oarg{local options}\parg{liste}}
-\begin{tabular}{lll}
-arguments & default & definition \\
-\midrule
-\TAline{points list}{no default}{example \tkzcname{tkzDrawPoints(A,B,C)}}
-\bottomrule
-\end{tabular}
-
-\medskip
-\begin{tabular}{lll}
-\toprule
-options & default & definition \\
-\midrule
-\TOline{shape} {circle}{Possible \tkzname{cross} ou \tkzname{cross out}}
-\TOline{size} {6}{$6 \times$ \tkzcname{pgflinewidth}}
-\TOline{color} {black}{the default color can be changed }
-\bottomrule
-\end{tabular}
-
-\medskip
-\tkzHandBomb\ Beware of the final "s", an oversight leads to cascading errors if you try to draw multiple points. The options are the same as for the previous macro.
-\end{NewMacroBox}
-
-\subsubsection{First example}
-
-\begin{tkzexample}[latex=7cm,small]
-\begin{tikzpicture}
- \tkzDefPoint(1,3){A}
- \tkzDefPoint(4,1){B}
- \tkzDefPoint(0,0){C}
- \tkzDrawPoints[size=6,color=red,
- fill=red!50](A,B,C)
-\end{tikzpicture}
-\end{tkzexample}
-
-\subsubsection{Second example}
-
-\begin{tkzexample}[latex=7cm,small]
-\begin{tikzpicture}[scale=.5]
- \tkzDefPoint(2,3){A} \tkzDefPoint(5,-1){B}
- \tkzDefPoint[label=below:$\mathcal{C}$,
- shift={(2,3)}](-30:5.5){E}
- \begin{scope}[shift=(A)]
- \tkzDefPoint(30:5){C}
- \end{scope}
- \tkzCalcLength[cm](A,B)\tkzGetLength{rAB}
- \tkzDrawCircle[R](A,\rAB cm)
- \tkzDrawSegment(A,B)
- \tkzDrawPoints(A,B,C)
- \tkzLabelPoints(B,C)
- \tkzLabelPoints[above](A)
-\end{tikzpicture}
-\end{tkzexample}
-
-\section{Point on line or circle}
-\subsection{Point on a line}
-
-\begin{NewMacroBox}{tkzDefPointOnLine}{\oarg{local options}\parg{A,B}}
-\begin{tabular}{lll}
-arguments & default & definition \\
-\midrule
-\TAline{pt1,pt2} {no default} {Two points to define a line}
-\bottomrule
-\end{tabular}
-
-\medskip
-\begin{tabular}{lll}
-\toprule
-options & default & definition \\
-\midrule
-\TOline{pos=nb} {}{nb is a decimal }
-\bottomrule
-\end{tabular}
-
-\medskip
-
-\end{NewMacroBox}
-
-\subsubsection{Use of option \tkzname{pos} 1}
-\begin{tkzexample}[latex=9cm,small]
- \begin{tikzpicture}
- \tkzDefPoints{0/0/A,4/0/B}
- \tkzDrawLine[red](A,B)
- \tkzDefPointOnLine[pos=1.2](A,B)
- \tkzGetPoint{P}
- \tkzDefPointOnLine[pos=-0.2](A,B)
- \tkzGetPoint{R}
- \tkzDefPointOnLine[pos=0.5](A,B)
- \tkzGetPoint{S}
- \tkzDrawPoints(A,B,P)
- \tkzLabelPoints(A,B)
- \tkzLabelPoint[above](P){pos=$1.2$}
- \tkzLabelPoint[above](R){pos=$-.2$}
- \tkzLabelPoint[above](S){pos=$.5$}
- \tkzDrawPoints(A,B,P,R,S)
- \tkzLabelPoints(A,B)
- \end{tikzpicture}
-\end{tkzexample}
-
-\subsection{Point on a circle}
-
-\begin{NewMacroBox}{tkzDefPointOnCircle}{\oarg{local options}\parg{A,B}}
-\begin{tabular}{lll}
-arguments & default & definition \\
-\midrule
-\TAline{pt1,pt2} {no default} {Two points to define a line}
-\bottomrule
-\end{tabular}
-
-\medskip
-\begin{tabular}{lll}
-\toprule
-options & default & definition \\
-\midrule
-\TOline{angle} {0}{angle formed with the abscissa axis}
-\TOline{center} {tkzPointResult}{circle center}
-\TOline{radius} {|\tkzLengthResult pt|}{radius circle}
-\bottomrule
-\end{tabular}
-
-
-\end{NewMacroBox}
-
-\begin{tkzexample}[latex=7cm,small]
-\begin{tikzpicture}
- \tkzDefPoints{0/0/A,4/0/B,0.8/3/C}
- \tkzDefPointOnCircle[angle=90,center=B,
- radius=1 cm]
- \tkzGetPoint{I}
- \tkzDrawCircle[R,teal](B,1cm)
- \tkzDrawPoint[teal](I)
- \tkzDefCircle[circum](A,B,C)
- \tkzGetPoint{G} \tkzGetLength{rG}
- \tkzDefPointOnCircle[angle=30,center=G,
- radius=\rG pt]
- \tkzGetPoint{J}
- \tkzDrawPoints(A,B,C)
- \tkzDrawCircle(G,J)
- \tkzDrawPoint(G)
- \tkzDrawPoint[red](J)
-\end{tikzpicture}
-\end{tkzexample}
-
-
\endinput