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-rw-r--r--Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-presentation.tex72
1 files changed, 36 insertions, 36 deletions
diff --git a/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-presentation.tex b/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-presentation.tex
index 0073d788032..027cb7ea0dd 100644
--- a/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-presentation.tex
+++ b/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-presentation.tex
@@ -51,27 +51,27 @@ Here are some comparisons between \tkzname{\TIKZ } and \tkzname{\tkznameofpack}
Explanation :
-The fourth tutorial of the \emph{PgfManual} is about geometric constructions. \emph{T. Tantau} proposes to get the drawing with its beautiful tool Ti\emph{k}Z. Here I propose the same construction with \emph{tkz-elements}. The color of the Ti\emph{k}Z code is orange and that of \emph{tkz-elements} is red.
+The fourth tutorial of the \emph{PgfManual} is about geometric constructions. \emph{T. Tantau} proposes to get the drawing with its beautiful tool Ti\emph{k}Z. Here I propose the same construction with \emph{tkz-elements}. The color of the Ti\emph{k}Z code is green!50!black and that of \emph{tkz-elements} is red.
\medskip
-\hspace*{1cm}\vbox{\color{orange} |\usepackage{tikz}|\\
+\vbox{\color{green!50!black} |\usepackage{tikz}|\\
|\usetikzlibrary{calc,intersections,through,backgrounds}|}
\medskip
-\hspace*{1cm}\vbox{\color{red} |\usepackage{tkz-euclide}|}
+\vbox{\color{red} |\usepackage{tkz-euclide}|}
\medskip
How to get the line AB ? To get this line, we use two fixed points.\\
\medskip
-\hspace*{1cm}\vbox{\color{orange}
+\vbox{\color{green!50!black}
|\coordinate [label=left:$A$] (A) at (0,0);|\\
|\coordinate [label=right:$B$] (B) at (1.25,0.25);|\\
|\draw (A) -- (B);|}
\medskip
-\hspace*{1cm}\vbox{\color{red}
+\vbox{\color{red}
|\tkzDefPoint(0,0){A}|\\
|\tkzDefPoint(1.25,0.25){B}|\\
|\tkzDrawSegment(A,B)|\\
@@ -81,14 +81,14 @@ How to get the line AB ? To get this line, we use two fixed points.\\
We want to draw a circle around the points $A$ and $B$ whose radius is given by the length of the line AB.
\medskip
-\hspace*{1cm}\vbox{\color{orange}
+\vbox{\color{green!50!black}
|\draw let \p1 = ($ (B) - (A) $),|\\
|\n2 = {veclen(\x1,\y1)} in|\\
| (A) circle (\n2)|\\
| (B) circle (\n2);|}
\medskip
-\hspace*{1cm}\vbox{\color{red}
+\vbox{\color{red}
|\tkzDrawCircles(A,B B,A)|
}
@@ -96,27 +96,27 @@ The intersection of the circles $\mathcal{D}$ and $\mathcal{E}$
\medskip
-\hspace*{1cm}\vbox{\color{orange}
+\vbox{\color{green!50!black}
|draw [name path=A--B] (A) -- (B);|\\
|node (D) [name path=D,draw,circle through=(B),label=left:$D$] at (A) {}; |\\
|node (E) [name path=E,draw,circle through=(A),label=right:$E$] at (B) {};|\\
-|path [name intersections={of=D and E, by={[label=above:$C$]C, [label=below:$C'$]C'}}]; |\\
+|path [name intersections={of=D and E, by={[label=above:$C$]C,[label=below:$C'$]C'}}]; |\\
|draw [name path=C--C',red] (C) -- (C');|\\
|path [name intersections={of=A--B and C--C',by=F}];|\\
|node [fill=red,inner sep=1pt,label=-45:$F$] at (F) {};|\\}
\medskip
-\hspace*{1cm}\vbox{\color{red} |\tkzInterCC(A,B)(B,A) \tkzGetPoints{C}{X}|\\}
+\vbox{\color{red} |\tkzInterCC(A,B)(B,A) \tkzGetPoints{C}{X}|\\}
How to draw points :
\medskip
-\hspace*{1cm}\vbox{\color{orange} |\foreach \point in {A,B,C}|\\
+\vbox{\color{green!50!black} |\foreach \point in {A,B,C}|\\
|\fill [black,opacity=.5] (\point) circle (2pt);|\\}
\medskip
-\hspace*{1cm}\vbox{\color{red}| \tkzDrawPoints[fill=gray,opacity=.5](A,B,C)|\\}
+\vbox{\color{red}| \tkzDrawPoints[fill=gray,opacity=.5](A,B,C)|\\}
\subsubsection{Complete code with \pkg{tkz-euclide}}
@@ -124,12 +124,12 @@ We need to define colors
|\colorlet{input}{red!80!black} |\\
|\colorlet{output}{red!70!black}|\\
-|\colorlet{triangle}{orange!40} |
+|\colorlet{triangle}{green!50!black!40} |
\begin{tkzexample}[vbox,small]
\colorlet{input}{red!80!black}
\colorlet{output}{red!70!black}
- \colorlet{triangle}{orange!40}
+ \colorlet{triangle}{green!50!black!40}
\begin{tikzpicture}[scale=1.25,thick,help lines/.style={thin,draw=black!50}]
\tkzDefPoint(0,0){A}
\tkzDefPoint(1.25+rand(),0.25+rand()){B}
@@ -164,7 +164,7 @@ Explanation
In the first part, we need to find the midpoint of the straight line $AB$. With \TIKZ\ we can use the calc library
\medskip
-\hspace*{1cm}\vbox{\color{orange} |\coordinate [label=left:$A$] (A) at (0,0);|\\
+\vbox{\color{green!50!black} |\coordinate [label=left:$A$] (A) at (0,0);|\\
|\coordinate [label=right:$B$] (B) at (1.25,0.25);|\\
|\draw (A) -- (B);|\\
|\node [fill=red,inner sep=1pt,label=below:$X$] (X) at ($ (A)!.5!(B) $) {};|\\}
@@ -173,40 +173,40 @@ With \pkg{tkz-euclide} we have a macro \tkzcname{tkzDefMidPoint}, we get the poi
\medskip
-\hspace*{1cm}\vbox{\red |\tkzDefPoints{0/0/A,0.75/0.25/B,1/1.5/C}|\\
-|\tkzDefMidPoint(A,B) \tkzGetPoint{X}|}\\
+\vbox{\red |\tkzDefPoints{0/0/A,0.75/0.25/B,1/1.5/C}|\\
+|\tkzDefMidPoint(A,B) \tkzGetPoint{X}|}
\medskip
Then we need to construct a triangle equilateral. It's easy with \pkg{tkz-euclide} . With TikZ you need some effort because you need to use the midpoint $X$ to get the point $D$ with trigonometry calculation.
\medskip
-\hspace*{1cm}\vbox{\color{orange}
+\vbox{\color{green!50!black}
|\node [fill=red,inner sep=1pt,label=below:$X$] (X) at ($ (A)!.5!(B) $) {}; | \\
|\node [fill=red,inner sep=1pt,label=above:$D$] (D) at | \\
|($ (X) ! {sin(60)*2} ! 90:(B) $) {}; | \\
|\draw (A) -- (D) -- (B); | \\
-} \\
+}
\medskip
-\hspace*{1cm}\vbox{\color{red} |\tkzDefTriangle[equilateral](A,B) \tkzGetPoint{D}|}\\
+\vbox{\color{red} |\tkzDefTriangle[equilateral](A,B) \tkzGetPoint{D}|}
We can draw the triangle at the end of the picture with
\medskip
-\hspace*{1cm}\vbox{\color{red} |\tkzDrawPolygon{A,B,C}|}
+\vbox{\color{red} |\tkzDrawPolygon{A,B,C}|}
\medskip
We know how to draw the circle $\mathcal{H}$ around $B$ through $C$ and how to place the points $E$ and $F$
\medskip
-\hspace*{1cm}\vbox{\color{orange}
+\vbox{\color{green!50!black}
|\node (H) [label=135:$H$,draw,circle through=(C)] at (B) {};| \\
|\draw (D) -- ($ (D) ! 3.5 ! (B) $) coordinate [label=below:$F$] (F);| \\
-|\draw (D) -- ($ (D) ! 2.5 ! (A) $) coordinate [label=below:$E$] (E);|} \\
+|\draw (D) -- ($ (D) ! 2.5 ! (A) $) coordinate [label=below:$E$] (E);|}
\medskip
-\hspace*{1cm}\vbox{\color{red} |\tkzDrawCircle(B,C)|\\
+\vbox{\color{red} |\tkzDrawCircle(B,C)|\\
|\tkzDrawLines[add=0 and 2](D,A D,B)|}
\medskip
@@ -216,32 +216,32 @@ Intersecting a Line and a Circle : here we search the intersection of the circle
The infinite straight line $DB$ intercepts the circle but with \TIKZ\ we need to extend the lines $DB$ and that can be done using partway calculations. We get the point $F$ and $BF$ or $DF$ intercepts the circle
\medskip
-\hspace*{1cm}\vbox{\color{orange}| \node (H) [label=135:$H$,draw,circle through=(C)] at (B) {}; | \\
+\vbox{\color{green!50!black}| \node (H) [label=135:$H$,draw,circle through=(C)] at (B) {}; | \\
|\path let \p1 = ($ (B) - (C) $) in| \\
| coordinate [label=left:$G$] (G) at ($ (B) ! veclen(\x1,\y1) ! (F) $); | \\
-|\fill[red,opacity=.5] (G) circle (2pt);|} \\
+|\fill[red,opacity=.5] (G) circle (2pt);|}
\medskip
Like the intersection of two circles, it's easy to find the intersection of a line and a circle with \pkg{tkz-euclide}. We don't need $F$
\medskip
-\hspace*{1cm}\vbox{\color{red} | \tkzInterLC(B,D)(B,C)\tkzGetFirstPoint{G}|}
+\vbox{\color{red} | \tkzInterLC(B,D)(B,C)\tkzGetFirstPoint{G}|}
\medskip
There are no more difficulties. Here the final code with some simplications.
We draw the circle $\mathcal{K}$ with center $D$ and passing through $G$. It intersects the line $AD$ at point $L$. $AL = BC$.
-\hspace*{1cm}\vbox{\color{red} | \tkzDrawCircle(D,G)|}
-\hspace*{1cm}\vbox{\color{red} | \tkzInterLC(D,A)(D,G)\tkzGetSecondPoint{L}|}
+\vbox{\color{red} | \tkzDrawCircle(D,G)|}
+\vbox{\color{red} | \tkzInterLC(D,A)(D,G)\tkzGetSecondPoint{L}|}
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=1.5]
\tkzDefPoint(0,0){A}
\tkzDefPoint(0.75,0.25){B}
\tkzDefPoint(1,1.5){C}
-\tkzDefTriangle[equilateral](A,B) \tkzGetPoint{D}
-\tkzInterLC[near](D,B)(B,C) \tkzGetSecondPoint{G}
-\tkzInterLC[near](A,D)(D,G) \tkzGetFirstPoint{L}
+\tkzDefTriangle[equilateral](A,B)\tkzGetPoint{D}
+\tkzInterLC[near](D,B)(B,C) \tkzGetSecondPoint{G}
+\tkzInterLC[near](A,D)(D,G) \tkzGetFirstPoint{L}
\tkzDrawCircles(B,C D,G)
\tkzDrawLines[add=0 and 2](D,A D,B)
\tkzDrawSegment(A,B)
@@ -399,7 +399,7 @@ Let's analyze the figure
\[\widehat{BCA}=90^\circ -\alpha/2 \]
\item Finally \[\widehat{CBD}=\alpha=36^\circ \]
- the triangle $CBD$ is a "golden" triangle.
+ the triangle $CBD$ is a \code{golden} triangle.
\end{enumerate}
\vspace*{24pt}
@@ -448,7 +448,7 @@ After building the golden triangle $BCD$, we build the point $A$ by noticing tha
\subsubsection{Part II: two others methods with golden and euclid triangle}
-\tkzname{\tkznameofpack} knows how to define a "golden" or "euclide" triangle. We can define $BCD$ and $BCA$ like gold triangles.
+\tkzname{\tkznameofpack} knows how to define a \code{golden} or \code{euclide} triangle. We can define $BCD$ and $BCA$ like gold triangles.
\begin{center}
@@ -529,7 +529,7 @@ $IB=a$, $AI=1$
\tkzDefMidPoint(A,B) \tkzGetPoint{M}
\tkzDefPointWith[orthogonal](I,M) \tkzGetPoint{H}
\tkzInterLC(I,H)(M,B) \tkzGetFirstPoint{C}
- \tkzDrawSegment[style=orange](I,C)
+ \tkzDrawSegment[style=purple](I,C)
\tkzDrawArc(M,B)(A)
\tkzDrawSegment[dim={$1$,-16pt,}](A,I)
\tkzDrawSegment[dim={$(a-1)/2$,-10pt,}](I,M)
@@ -587,7 +587,7 @@ The following code consists of several parts:
\item The third one includes the different drawings;
\begin{tkzltxexample}[]
- \tkzDrawSegment[style=orange](I,H)
+ \tkzDrawSegment[style=purple](I,H)
\tkzDrawPoints(O,I,A,B,M)
\tkzDrawArc(M,A)(O)
\tkzDrawSegment[dim={$1$,-16pt,}](A,I)