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diff --git a/Master/texmf-dist/doc/latex/tkz-euclide/latex/TKZdoc-euclide-exemples.tex b/Master/texmf-dist/doc/latex/tkz-euclide/latex/TKZdoc-euclide-exemples.tex
index 4a380e355f2..fb591ba5e01 100644
--- a/Master/texmf-dist/doc/latex/tkz-euclide/latex/TKZdoc-euclide-exemples.tex
+++ b/Master/texmf-dist/doc/latex/tkz-euclide/latex/TKZdoc-euclide-exemples.tex
@@ -1,33 +1,39 @@
-\section{Des exemples}
-\subsection{Quelques exemples intéressants}
+\section{Some examples}
+\subsection{Some interesting examples}
-\subsubsection{Triangles isocèles semblables}
+\subsubsection{Similar isosceles triangles}
-Ce qui suit provient de l'excellent site \textbf{Descartes et les Mathématiques}. Je n'ai pas modifié le texte et je ne suis l'auteur que de la programmation des figures.
+The following is from the excellent site \textbf{Descartes et les Mathématiques}. I did not modify the text and I am only the author of the programming of the figures.
\url{http://debart.pagesperso-orange.fr/seconde/triangle.html}
-Bibliographie : Géométrie au Bac - Tangente, hors série no 8 - Exercice 11, page 11
+Bibliography:
-Élisabeth Busser et Gilles Cohen : 200 nouveaux problèmes du Monde - POLE 2007
+\begin{itemize}
-Affaire de logique n° 364 - Le Monde 17 février 2004
+\item Géométrie au Bac - Tangente, special issue no. 8 - Exercise 11, page 11
-Deux énoncés ont été proposés, l'un par la revue \emph{Tangente}, et l'autre par le journal \emph{Le Monde}.
+\item Elisabeth Busser and Gilles Cohen: 200 nouveaux problèmes du "Monde" - POLE 2007 (200 new problems of "Le Monde")
+
+
+\item Affaire de logique n° 364 - Le Monde February 17, 2004
+\end{itemize}
+
+
+Two statements were proposed, one by the magazine \textit{Tangente} and the other by \textit{Le Monde}.
\vspace*{2cm}
-\emph{Rédaction de la revue Tangente} : \textcolor{orange}{On construit deux triangles isocèles semblables AXB et BYC de sommets principaux X et Y, tels que A, B et C soient alignés et que ces triangles soient « indirect ». Soit $\alpha$ l'angle au sommet $\widehat{AXB}$ = $\widehat{BYC}$. On construit ensuite un troisième triangle isocèle XZY semblable aux deux premiers, de sommet principal Z et « indirect ».\\
-On demande de démontrer que le point Z appartient à la droite (AC).}
+\emph{Editor of the magazine "Tangente"}: \textcolor{orange}{Two similar isosceles triangles $AXB$ and $BYC$ are constructed with main vertices $X$ and $Y$, such that $A$, $B$ and $C$ are aligned and that these triangles are "indirect". Let $\alpha$ be the angle at vertex $\widehat{AXB}$ = $\widehat{BYC}$. We then construct a third isosceles triangle $XZY$ similar to the first two, with main vertex $Z$ and "indirect".
+We ask to demonstrate that point $Z$ belongs to the straight line $(AC)$.}
\vspace*{2cm}
-\emph{Rédaction du Monde} : \textcolor{orange}{On construit deux triangles isocèles semblables AXB et BYC de sommets principaux X et Y, tels que A, B et C soient alignés et que ces triangles soient « indirect ». Soit $\alpha$ l'angle au sommet $\widehat{AXB}$ = $\widehat{BYC}$. Le point Z du segment [AC] est équidistant des deux sommets X et Y.\\
-Sous quel angle voit-il ces deux sommets ?}
+\emph{Editor of "Le Monde"}: \textcolor{orange}{We construct two similar isosceles triangles $AXB$ and $BYC$ with principal vertices $X$ and $Y$, such that $A$, $B$ and $C$ are aligned and that these triangles are "indirect". Let $\alpha$ be the angle at vertex $\widehat{AXB}$ = $\widehat{BYC}$. The point Z of the line segment $[AC]$ is equidistant from the two vertices $X$ and $Y$.\\
+At what angle does he see these two vertices?}
-\vspace*{2cm} Les constructions et leurs codes associés sont sur les deux pages suivantes, mais vous pouvez chercher avant de regarder. La programmation respecte (il me semble ...), mon raisonnement dans les deux cas.
-\newpage
+\vspace*{2cm} The constructions and their associated codes are on the next two pages, but you can search before looking. The programming respects (it seems to me ...) my reasoning in both cases.
- \subsubsection{version revue "Tangente"}
+ \subsubsection{Revised version of "Tangente"}
\begin{tkzexample}[]
\begin{tikzpicture}[scale=.8,rotate=60]
\tkzDefPoint(6,0){X} \tkzDefPoint(3,3){Y}
@@ -47,8 +53,8 @@ Sous quel angle voit-il ces deux sommets ?}
\tkzLabelPoints(A,B,C,Z) \tkzLabelPoints[above right](X,Y,O)
\end{tikzpicture}
\end{tkzexample}
-\newpage
-\subsubsection{version "Le Monde"}
+
+\subsubsection{"Le Monde" version}
\begin{tkzexample}[]
\begin{tikzpicture}[scale=1.25]
@@ -74,13 +80,13 @@ Sous quel angle voit-il ces deux sommets ?}
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{Hauteurs d'un triangle}
+\subsubsection{Triangle altitudes}
-Ce qui suit provient encore de l'excellent site \textbf{Descartes et les Mathématiques}.
+The following is again from the excellent site \textbf{Descartes et les Mathématiques} (Descartes and the Mathematics).
\url{http://debart.pagesperso-orange.fr/geoplan/geometrie_triangle.html}
-Les trois hauteurs d'un triangle sont concourantes au même point H.
+The three altitudes of a triangle intersect at the same H-point.
\begin{tkzexample}[latex=7cm]
\begin{tikzpicture}[scale=.8]
@@ -109,7 +115,7 @@ Les trois hauteurs d'un triangle sont concourantes au même point H.
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{Hauteurs - autre construction}
+\subsubsection{Altitudes - other construction}
\begin{tkzexample}[latex=7cm]
\begin{tikzpicture}[scale=.75]
@@ -140,10 +146,9 @@ Les trois hauteurs d'un triangle sont concourantes au même point H.
\end{tikzpicture}
\end{tkzexample}
-\newpage
\subsection{Different authors}
-\subsubsection{ Square root of the integers }
+\subsubsection{ Square root of the integers}
How to get $1$, $\sqrt{2}$, $\sqrt{3}$ with a rule and a compass.
\begin{tkzexample}[latex=7cm,small]
@@ -159,38 +164,12 @@ How to get $1$, $\sqrt{2}$, $\sqrt{3}$ with a rule and a compass.
\end{tkzexample}
-\subsubsection{Circle and tangent}
-We have a point A $(8,2)$, a circle with center A and radius=3cm and a line
- $\delta$ $y=4$. The line intercepts the circle at B. We want to draw the tangent at the circle in B.
-
-
-\begin{tkzexample}[]
-\begin{tikzpicture}
- \edef\alphaR{\fpeval{asin(2/3)}}
- \edef\xB{8-3*cos(\alphaR)}
- \tkzDrawX[noticks,label=$(d)$]
- \tkzDefPoint["$A$" above right](8,2){A}
- \tkzDefPoint[color=red,"$O$" above right](0,0){O}
- \tkzDefPoint["$B$" above left](\xB,4){B}
- \tkzDefLine[orthogonal=through B](A,B) \tkzGetPoint{b}
- \tkzDefPoint(1,0){i}
- \tkzInterLL(B,b)(O,i) \tkzGetPoint{B'}
- \tkzDrawSegment[line width=1pt](A,B)
- \tkzHLine[color=red,style=dashed]{4}
- \tkzText[above](12,4){$\delta$}
- \tkzDrawCircle[R,color=blue,line width=.8pt](A,3 cm)
- \tkzDrawPoint(B')
- \tkzDrawLine(B,B')
- \end{tikzpicture}
-\end{tkzexample}
-
-
\subsubsection{About right triangle}
-We have a segment $[AB]$ and we want to determine a point $C$ such as $AC=8 cm$ and $ABC$ is a right triangle in $B$.
+We have a segment $[AB]$ and we want to determine a point $C$ such that $AC=8$~cm and $ABC$ is a right triangle in $B$.
\begin{tkzexample}[latex=7cm]
-\begin{tikzpicture}
+\begin{tikzpicture}[scale=.5]
\tkzDefPoint["$A$" left](2,1){A}
\tkzDefPoint(6,4){B}
\tkzDrawSegment(A,B)
@@ -211,7 +190,7 @@ We have a segment $[AB]$ and we want to determine a point $C$ such as $AC=8 cm$
\subsubsection{Archimedes}
This is an ancient problem proved by the great Greek mathematician Archimedes .
-The figure below shows a semicircle, with diameter $AB$. A tangent line is drawn and touches the semicircle at $B$. An other tangent line at a point, $C$, on the semicircle is drawn. We project the point $C$ on the segment$[AB]$ on a point $D$ . The two tangent lines intersect at the point $T$.
+The figure below shows a semicircle, with diameter $AB$. A tangent line is drawn and touches the semicircle at $B$. An other tangent line at a point, $C$, on the semicircle is drawn. We project the point $C$ on the line segment $[AB]$ on a point $D$. The two tangent lines intersect at the point $T$.
Prove that the line $(AT)$ bisects $(CD)$
@@ -235,7 +214,7 @@ Prove that the line $(AT)$ bisects $(CD)$
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{Exemple : Dimitris Kapeta}
+\subsubsection{Example: Dimitris Kapeta}
You need in this example to use \tkzname{mkpos=.2} with \tkzcname{tkzMarkAngle} because the measure of $ \widehat{CAM}$ is too small.
Another possiblity is to use \tkzcname{tkzFillAngle}.
@@ -266,9 +245,9 @@ Another possiblity is to use \tkzcname{tkzFillAngle}.
\end{tkzexample}
-\subsubsection{Example : John Kitzmiller }
+\subsubsection{Example 1: John Kitzmiller }
-Prove $\bigtriangleup LKJ$ is equilateral
+Prove that $\bigtriangleup LKJ$ is equilateral.
\begin{tkzexample}[vbox,small]
@@ -285,23 +264,26 @@ Prove $\bigtriangleup LKJ$ is equilateral
\tkzInterLL(B,B')(A,A') \tkzGetPoint{L}
\tkzLabelPoint[above](C){C}
\tkzDrawPolygon(A,B,C) \tkzDrawSegments(A,J B,L C,K)
- \tkzMarkAngles[fill= orange,size=1cm,opacity=.3](J,A,C K,C,B L,B,A)
- \tkzLabelPoint[right](J){J}
- \tkzLabelPoint[below](K){K}
- \tkzLabelPoint[above left](L){L}
- \tkzMarkAngles[fill=orange, opacity=.3,thick,size=1,](A,C,J C,B,K B,A,L)
- \tkzMarkAngles[fill=green, size=1, opacity=.5](A,C,J C,B,K B,A,L)
+ \tkzMarkAngles[size=1 cm](J,A,C K,C,B L,B,A)
+ \tkzMarkAngles[thick,size=1 cm](A,C,J C,B,K B,A,L)
+ \tkzMarkAngles[opacity=.5](A,C,J C,B,K B,A,L)
+ \tkzFillAngles[fill= orange,size=1 cm,opacity=.3](J,A,C K,C,B L,B,A)
+ \tkzFillAngles[fill=orange, opacity=.3,thick,size=1,](A,C,J C,B,K B,A,L)
+ \tkzFillAngles[fill=green, size=1, opacity=.5](A,C,J C,B,K B,A,L)
\tkzFillPolygon[color=yellow, opacity=.2](J,A,C)
\tkzFillPolygon[color=yellow, opacity=.2](K,B,C)
\tkzFillPolygon[color=yellow, opacity=.2](L,A,B)
\tkzDrawSegments[line width=3pt,color=cyan,opacity=0.4](A,J C,K B,L)
\tkzDrawSegments[line width=3pt,color=red,opacity=0.4](A,L B,K C,J)
\tkzMarkSegments[mark=o](J,K K,L L,J)
+ \tkzLabelPoint[right](J){J}
+ \tkzLabelPoint[below](K){K}
+ \tkzLabelPoint[above left](L){L}
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{Exemple : John Kitzmiller }
-Prove $\dfrac{AC}{CE}=\dfrac{BD}{DF} \qquad$
+\subsubsection{Example 2: John Kitzmiller }
+Prove that $\dfrac{AC}{CE}=\dfrac{BD}{DF}$.
Another interesting example from John, you can see how to use some extra options like \tkzname{decoration} and \tkzname{postaction} from \TIKZ\ with \tkzname{tkz-euclide}.
@@ -327,8 +309,8 @@ Another interesting example from John, you can see how to use some extra options
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{Exemple : John Kitzmiller }
-Prove $\dfrac{BC}{CD}=\dfrac{AB}{AD} \qquad$ (Angle Bisector)
+\subsubsection{Example 3: John Kitzmiller }
+Prove that $\dfrac{BC}{CD}=\dfrac{AB}{AD} \qquad$ (Angle Bisector).
\begin{tkzexample}[vbox,small]
\begin{tikzpicture}[scale=2]
@@ -347,10 +329,13 @@ Prove $\dfrac{BC}{CD}=\dfrac{AB}{AD} \qquad$ (Angle Bisector)
\tkzDrawSegments[color=magenta, line width=3pt, opacity=0.4](B,C P,A)
\tkzDrawSegments[color=teal, line width=3pt, opacity=0.4](C,D A,D)
\tkzDrawSegments[color=magenta, line width=3pt, opacity=0.4](A,B)
- \tkzMarkAngles[size=0.7](B,A,C C,A,D)
- \tkzMarkAngles[size=0.7, fill=green, opacity=0.5](B,A,C A,B,P)
- \tkzMarkAngles[size=0.7, fill=yellow, opacity=0.3](B,P,A C,A,D)
- \tkzMarkAngles[size=0.7, fill=green, opacity=0.6](B,A,C A,B,P B,P,A C,A,D)
+ \tkzMarkAngles[size=3mm](B,A,C C,A,D)
+ \tkzMarkAngles[size=3mm](B,A,C A,B,P)
+ \tkzMarkAngles[size=3mm](B,P,A C,A,D)
+ \tkzMarkAngles[size=3mm](B,A,C A,B,P B,P,A C,A,D)
+ \tkzFillAngles[fill=green, opacity=0.5](B,A,C A,B,P)
+ \tkzFillAngles[fill=yellow, opacity=0.3](B,P,A C,A,D)
+ \tkzFillAngles[fill=green, opacity=0.6](B,A,C A,B,P B,P,A C,A,D)
\tkzLabelAngle[pos=1](B,A,C){1} \tkzLabelAngle[pos=1](C,A,D){2}
\tkzLabelAngle[pos=1](A,B,P){3} \tkzLabelAngle[pos=1](B,P,A){4}
\tkzMarkSegments[mark=|](A,B A,P)
@@ -358,8 +343,8 @@ Prove $\dfrac{BC}{CD}=\dfrac{AB}{AD} \qquad$ (Angle Bisector)
\end{tkzexample}
-\subsubsection{Exemple : author John Kitzmiller }
-Prove $\overline{AG}\cong\overline{EF} \qquad$ (Detour)
+\subsubsection{Example 4: author John Kitzmiller }
+Prove that $\overline{AG}\cong\overline{EF} \qquad$ (Detour).
\begin{tkzexample}[vbox,small]
\begin{tikzpicture}[scale=2]
@@ -370,10 +355,14 @@ Prove $\overline{AG}\cong\overline{EF} \qquad$ (Detour)
\tkzFillPolygon[yellow, opacity=0.4](D,F,C)
\tkzFillPolygon[blue, opacity=0.3](A,B,G)
\tkzFillPolygon[blue, opacity=0.3](E,D,F)
- \tkzMarkAngles[size=0.6,fill=green](B,G,A D,F,E)
- \tkzMarkAngles[size=0.6,fill=orange](B,C,G D,C,F)
- \tkzMarkAngles[size=0.6,fill=yellow](G,B,C F,D,C)
- \tkzMarkAngles[size=0.6,fill=red](A,B,G E,D,F)
+ \tkzMarkAngles[size=0.5 cm](B,G,A D,F,E)
+ \tkzMarkAngles[size=0.5 cm](B,C,G D,C,F)
+ \tkzMarkAngles[size=0.5 cm](G,B,C F,D,C)
+ \tkzMarkAngles[size=0.5 cm](A,B,G E,D,F)
+ \tkzFillAngles[size=0.5 cm,fill=green](B,G,A D,F,E)
+ \tkzFillAngles[size=0.5 cm,fill=orange](B,C,G D,C,F)
+ \tkzFillAngles[size=0.5 cm,fill=yellow](G,B,C F,D,C)
+ \tkzFillAngles[size=0.5 cm,fill=red](A,B,G E,D,F)
\tkzMarkSegments[mark=|](B,C D,C) \tkzMarkSegments[mark=s||](G,C F,C)
\tkzMarkSegments[mark=o](A,G E,F) \tkzMarkSegments[mark=s](B,G D,F)
\tkzDrawSegment[color=red](A,E)
@@ -383,7 +372,7 @@ Prove $\overline{AG}\cong\overline{EF} \qquad$ (Detour)
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{Example from Indonesia}
+\subsubsection{Example 1: from Indonesia}
\begin{tkzexample}[vbox,small]
\begin{tikzpicture}[scale=3]
@@ -404,7 +393,7 @@ Prove $\overline{AG}\cong\overline{EF} \qquad$ (Detour)
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{Another example from Indonesia}
+\subsubsection{Example 2: from Indonesia}
\begin{tkzexample}[vbox,small]
\begin{tikzpicture}[pol/.style={fill=brown!40,opacity=.5},
seg/.style={tkzdotted,color=gray},