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diff --git a/Master/texmf-dist/doc/latex/tkz-euclide/latex/TKZdoc-euclide-triangles.tex b/Master/texmf-dist/doc/latex/tkz-euclide/latex/TKZdoc-euclide-triangles.tex
index 13cf231e458..038e93892cb 100644
--- a/Master/texmf-dist/doc/latex/tkz-euclide/latex/TKZdoc-euclide-triangles.tex
+++ b/Master/texmf-dist/doc/latex/tkz-euclide/latex/TKZdoc-euclide-triangles.tex
@@ -1,46 +1,46 @@
-\section{Les triangles}
+\section{Triangles}
-\subsection{Définition des triangles \tkzcname{tkzDefTriangle}}
-Les macros suivantes vont permettre de définir ou de construire un triangle à partir \tkzname{au moins} de deux points.
+\subsection{Definition of triangles \tkzcname{tkzDefTriangle}}
+The following macros will allow you to define or construct a triangle from \tkzname{at least} two points.
- Pour le moment, il est possible de définir les triangles suivants :
+ At the moment, it is possible to define the following triangles:
\begin{itemize}
-\item \tkzname{two angles} détermine un triangle connaissant deux angles,
-\item \tkzname{equilateral} détermine un triangle équilatéral,
-\item \tkzname{half} détermine un triangle rectangle tel que le rapport des mesures des deux côtés adjacents à l'angle droit soit égal à $2$,
-\item \tkzname{pythagore} détermine un triangle rectangle dont les mesures des côtés sont proportionnelles à 3, 4 et 5,
-\item \tkzname{school} détermine un triangle rectangle dont les angles sont 30, 60 et 90 degrés,
-\item \tkzname{golden} détermine un triangle rectangle tel que le rapport des mesures des deux côtés adjacents à l'angle droit soit égal $\Phi=1,618034$, J'ai choisi comme dénomination « triangle doré » car il rpovient du rectangle d'or et j'ai conservé la dénomination « triangle d'or » ou encore « triangle d'Euclide » pour le triangle isocèle dont les angles à la base sont de 72 degrés,
+\item \tkzname{two angles} determines a triangle with two angles;
+\item \tkzname{equilateral} determines an equilateral triangle;
+\item \tkzname{half} determines a right-angled triangle such that the ratio of the measurements of the two adjacent sides to the right angle is equal to $2$;
+\item \tkzname{pythagore} determines a right-angled triangle whose side measurements are proportional to 3, 4 and 5;
+\item \tkzname{school} determines a right-angled triangle whose angles are 30, 60 and 90 degrees;
+\item \tkzname{golden} determines a right-angled triangle such that the ratio of the measurements on the two adjacent sides to the right angle is equal to $\Phi=1.618034$, I chose "golden triangle" as the denomination because it comes from the golden rectangle and I kept the denomination "gold triangle" or "Euclid's triangle" for the isosceles triangle whose angles at the base are 72 degrees;
-\item \tkzname{gold} ou \tkzname{euclide} pour le triangle d'or,
+\item \tkzname{euclide} or \tkzname{gold} for the gold triangle;
-\item \tkzname{cheops} détermine un troisième point tel que le triangle soit isocèle dont les mesures des côtés sont proportionnelles à $2$, $\Phi$ et $\Phi$.
+\item \tkzname{cheops} determines a third point such that the triangle is isosceles with side measurements proportional to $2$, $\Phi$ and $\Phi$.
\end{itemize}
-\begin{NewMacroBox}{tkzDefTriangle}{\oarg{local options}\parg{A,B}}
-les points sont ordonnés car le triangle est construit en suivant le sens direct du cercle trigonométrique. Cette macro est soit utilisée en partenariat avec \tkzcname{tkzGetPoint} soit en utilisant \tkzname{tkzPointResult} s'il n'est pas nécessaire de conserver le nom.
-
+\begin{NewMacroBox}{tkzDefTriangle}{\oarg{local options}\parg{A,B}}%
+The points are ordered because the triangle is constructed following the direct direction of the trigonometric circle. This macro is either used in partnership with \tkzcname{tkzGetPoint} or by using \tkzname{tkzPointResult} if it is not necessary to keep the name.
\medskip
-\begin{tabular}{lll}
+\begin{tabular}{lll}%
\toprule
options & default & definition \\
\midrule
-\TOline{two angles= \#1 and \#2}{no defaut}{triangle connaissant deux angles}
-\TOline{equilateral} {no defaut}{triangle équilatéral }
-\TOline{pythagore}{no defaut}{proportionnel au triangle de pythagore 3-4-5}
-\TOline{school} {no defaut}{ angles de 30, 60 et 90 degrés }
-\TOline{gold}{no defaut}{ angles de 72, 72 et 36 degrés, $A$ est le sommet }
-\TOline{euclide} {no defaut}{identique au précédent mais $[AB]$ est la base}
-\TOline{golden} {no defaut}{rectangle en B et $AB/AC = \Phi$}
-\TOline{cheops} {no defaut}{AC=BC, AC et BC sont proportionnels à $2$ et $\Phi$.}
+\TOline{two angles= \#1 and \#2}{no defaut}{triangle knowing two angles}
+\TOline{equilateral} {no defaut}{equilateral triangle }
+\TOline{pythagore}{no defaut}{proportional to the pythagorean triangle 3-4-5}
+\TOline{school} {no defaut}{angles of 30, 60 and 90 degrees }
+\TOline{gold}{no defaut}{angles of 72, 72 and 36 degrees, $A$ is the apex}
+\TOline{euclide} {no defaut}{same as above but $[AB]$ is the base}
+\TOline{golden} {no defaut}{B rectangle and $AB/AC = \Phi$}
+\TOline{cheops} {no defaut}{AC=BC, AC and BC are proportional to $2$ and $\Phi$.}
+\bottomrule
\end{tabular}
\medskip
-\tkzcname{tkzGetPoint} permet de stocker le point sinon \tkzname{tkzPointResult} permet une utilisation immédiate.
+\tkzcname{tkzGetPoint} allows you to store the point otherwise \tkzname{tkzPointResult} allows for immediate use.
\end{NewMacroBox}
-\subsubsection{triangle doré (golden)}
+\subsubsection{Option \tkzname{golden}}
\begin{tkzexample}[latex=6 cm,small]
\begin{tikzpicture}[scale=.8]
\tkzInit[xmax=5,ymax=3] \tkzClip[space=.5]
@@ -52,7 +52,7 @@ options & default & definition \\
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{triangle équilatéral}\label{def_equilateral}
+\subsubsection{Option \tkzname{equilateral}}
\begin{tkzexample}[latex=7 cm,small]
\begin{tikzpicture}
\tkzDefPoint(0,0){A}
@@ -68,7 +68,7 @@ options & default & definition \\
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{triangle d'or (euclide)}
+\subsubsection{Option \tkzname{gold} or \tkzname{euclide} }
\begin{tkzexample}[latex=7 cm,small]
\begin{tikzpicture}
\tkzDefPoint(0,0){A} \tkzDefPoint(4,0){B}
@@ -82,33 +82,32 @@ options & default & definition \\
\end{tkzexample}
\newpage
-\subsection{Tracé des triangles}
- \begin{NewMacroBox}{tkzDrawTriangle}{\oarg{local options}\parg{A,B}}
-Macro semblable à la macro précédente mais les côtés sont tracés.
+\subsection{Drawing of triangles}
+ \begin{NewMacroBox}{tkzDrawTriangle}{\oarg{local options}\parg{A,B}}%
+Macro similar to the previous macro but the sides are drawn.
\medskip
-\begin{tabular}{lll}
+\begin{tabular}{lll}%
\toprule
options & default & definition \\
\midrule
-\TOline{two angles= \#1 and \#2}{no defaut}{triangle connaissant deux angles}
-\TOline{equilateral} {no defaut}{triangle équilatéral }
-\TOline{pythagore}{no defaut}{proportionnel au triangle de pythagore 3-4-5}
-\TOline{school} {no defaut}{les angles sont 30, 60 et 90 degrés }
-\TOline{gold}{no defaut}{les angles sont 72, 72 et 36 degrés, $A$ est le sommet }
-\TOline{euclide} {no defaut}{identique au précédent mais $[AB]$ est la base}
-\TOline{golden} {no defaut}{rectangle en B et $AB/AC = \Phi$}
-\TOline{cheops} {no defaut}{isocèle en C et $AC/AB = \frac{\Phi}{2}$}
+\TOline{two angles= \#1 and \#2}{equilateral}{triangle knowing two angles}
+\TOline{equilateral} {equilateral}{equilateral triangle }
+\TOline{pythagore}{equilateral}{proportional to the pythagorean triangle 3-4-5}
+\TOline{school} {equilateral}{the angles are 30, 60 and 90 degrees }
+\TOline{gold}{equilateral}{the angles are 72, 72 and 36 degrees, $A$ is the vertex }
+\TOline{euclide} {equilateral}{identical to the previous one but $[AB]$ is the base}
+\TOline{golden} {equilateral}{B rectangle and $AB/AC = \Phi$}
+\TOline{cheops} {equilateral}{isosceles in C and $AC/AB = \frac{\Phi}{2}$}
\bottomrule
\end{tabular}
\medskip
-Dans toutes ses définitions, les dimensions du triangle dépendent des deux points de départ.
+In all its definitions, the dimensions of the triangle depend on the two starting points.
\end{NewMacroBox}
-
-\subsubsection{triangle de Pythagore}
-Ce triangle a des côtés dont les longueurs sont proportionnelles à 3, 4 et 5.
+\subsubsection{Option \tkzname{pythagore}}
+This triangle has sides whose lengths are proportional to 3, 4 and 5.
\begin{tkzexample}[latex=6 cm,small]
\begin{tikzpicture}
@@ -120,53 +119,76 @@ Ce triangle a des côtés dont les longueurs sont proportionnelles à 3, 4 et 5.
\end{tkzexample}
- \subsubsection{triangle 30 60 90 (school)}
- Les angles font 30, 60 et 90 degrés.
+\subsubsection{Option \tkzname{school}}
+The angles are 30, 60 and 90 degrees.
\begin{tkzexample}[latex=6 cm,small]
\begin{tikzpicture}
-\tkzInit[ymin=-2.5,ymax=0,xmin=-5,xmax=0]
-\tkzClip[space=.5]
-\begin{scope}[rotate=-180]
\tkzDefPoint(0,0){A} \tkzDefPoint(4,0){B}
\tkzDrawTriangle[school,fill=red!30](A,B)
- \tkzMarkRightAngles(B,A,tkzPointResult)
-\end{scope}
+ \tkzMarkRightAngles(tkzPointResult,B,A)
\end{tikzpicture}
\end{tkzexample}
+\subsubsection{Option \tkzname{golden}}
+\begin{tkzexample}[latex=6 cm,small]
+\begin{tikzpicture}[scale=1]
+ \tkzDefPoint(0,-10){M}
+ \tkzDefPoint(3,-10){N}
+ \tkzDrawTriangle[golden,color=brown](M,N)
+\end{tikzpicture}
+\end{tkzexample}
-\section{Triangles spécifiques avec \tkzcname{tkzDefSpcTriangle}}
+\subsubsection{Option \tkzname{gold}}
+\begin{tkzexample}[latex=6 cm,small]
+\begin{tikzpicture}[scale=1]
+ \tkzDefPoint(5,-5){I}
+ \tkzDefPoint(8,-5){J}
+ \tkzDrawTriangle[gold,color=blue!50](I,J)
+\end{tikzpicture}
+\end{tkzexample}
-Les centres de certains triangles ont été définis dans la section "points", ici il s'agit de déterminer les trois sommets de triangles spécifiques.
+\subsubsection{Option \tkzname{euclide}}
+\begin{tkzexample}[latex=6 cm,small]
+ \begin{tikzpicture}[scale=1]
+ \tkzDefPoint(10,-5){K}
+ \tkzDefPoint(13,-5){L}
+ \tkzDrawTriangle[euclide,color=blue,fill=blue!10](K,L)
+ \end{tikzpicture}
+\end{tkzexample}
+
+
+\section{Specific triangles with \tkzcname{tkzDefSpcTriangle}}
+
+The centers of some triangles have been defined in the "points" section, here it is a question of determining the three vertices of specific triangles.
\begin{NewMacroBox}{tkzDefSpcTriangle}{\oarg{local options}\parg{A,B,C}}
The order of the points is important!
\medskip
-\begin{tabular}{lll}
+\begin{tabular}{lll}%
\toprule
options & default & definition \\
\midrule
-\TOline{in or incentral}{centroid}{triangle connaissant deux angles}
-\TOline{ex or excentral} {centroid}{triangle équilatéral }
-\TOline{extouch}{centroid}{proportionnel au triangle de pythagore 3-4-5}
-\TOline{intouch or contact} {centroid}{ angles de 30, 60 et 90 degrés }
-\TOline{centroid or medial}{centroid}{ angles de 72, 72 et 36 degrés, $A$ est le sommet }
-\TOline{orthic} {centroid}{identique au précédent mais $[AB]$ est la base}
-\TOline{feuerbach} {centroid}{rectangle en B et $AB/AC = \Phi$}
-\TOline{euler} {centroid}{AC=BC, AC et BC sont proportionnels à $2$ et $\Phi$.}
-\TOline{tangential} {centroid}{AC=BC, AC et BC sont proportionnels à $2$ et $\Phi$.}
-\TOline{name} {no defaut}{AC=BC, AC et BC sont proportionnels à $2$ et $\Phi$.}
+\TOline{in or incentral}{centroid}{two-angled triangle}
+\TOline{ex or excentral} {centroid}{equilateral triangle }
+\TOline{extouch}{centroid}{proportional to the pythagorean triangle 3-4-5}
+\TOline{intouch or contact} {centroid}{ 30, 60 and 90 degree angles }
+\TOline{centroid or medial}{centroid}{ angles of 72, 72 and 36 degrees, $A$ is the vertex }
+\TOline{orthic} {centroid}{same as above but $[AB]$ is the base}
+\TOline{feuerbach} {centroid}{B rectangle and $AB/AC = \Phi$}
+\TOline{euler} {centroid}{AC=BC, AC and BC are proportional to $2$ and $\Phi$.}
+\TOline{tangential} {centroid}{AC=BC, AC and BC are proportional to $2$ and $\Phi$.}
+\TOline{name} {no defaut}{AC=BC, AC and BC are proportional to $2$ and $\Phi$.}
\midrule
\end{tabular}
\medskip
-\tkzcname{tkzGetPoint} permet de stocker le point sinon \tkzname{tkzPointResult} permet une utilisation immédiate.
+\tkzcname{tkzGetPoint} allows you to store the point otherwise \tkzname{tkzPointResult} allows for immediate use.
\end{NewMacroBox}
-\subsubsection{\tkzcname{tkzDefSpcTriangle} option "medial" ou "centroid"}
+\subsubsection{Option \tkzname{medial} or \tkzname{centroid} }
The geometric centroid of the polygon vertices of a triangle is the point $G$ (sometimes also denoted $M$) which is also the intersection of the triangle's three triangle medians. The point is therefore sometimes called the median point. The centroid is always in the interior of the triangle.\\
\href{http://mathworld.wolfram.com/TriangleCentroid.html}{Weisstein, Eric W. "Centroid triangle" From MathWorld--A Wolfram Web Resource.}
@@ -189,8 +211,9 @@ In the following example, we obtain the Euler circle which passes through the pr
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{Option : "in" ou "incentral"}
-The Incentral triangle is the triangle whose vertices are determined by
+\subsubsection{Option \tkzname{in} or \tkzname{incentral} }
+
+The incentral triangle is the triangle whose vertices are determined by
the intersections of the reference triangle’s angle bisectors with the
respective opposite sides.\\
\href{http://mathworld.wolfram.com/ContactTriangle.html}{Weisstein, Eric W. "Incentral triangle" From MathWorld--A Wolfram Web Resource.}
@@ -199,66 +222,66 @@ respective opposite sides.\\
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=1]
\tkzDefPoints{ 0/0/A,5/0/B,1/3/C}
- \tkzDefSpcTriangle[in,name=I](A,B,C){a,b,c}
+ \tkzDefSpcTriangle[in,name=I](A,B,C){_a,_b,_c}
\tkzInCenter(A,B,C)\tkzGetPoint{I}
\tkzDrawPolygon[red](A,B,C)
- \tkzDrawPolygon[blue](Ia,Ib,Ic)
- \tkzDrawPoints(A,B,C,I,Ia,Ib,Ic)
+ \tkzDrawPolygon[blue](I_a,I_b,I_c)
+ \tkzDrawPoints(A,B,C,I,I_a,I_b,I_c)
\tkzDrawCircle[in](A,B,C)
- \tkzDrawSegments[dashed](A,Ia B,Ib C,Ic)
- \tkzAutoLabelPoints[center=I,blue,font=\scriptsize]%
-(Ia,Ib,Ic)
- \tkzAutoLabelPoints[center=I,red,font=\scriptsize]%
-(A,B,C)
-(A,B,C,Ia,Ib,Ic)
+ \tkzDrawSegments[dashed](A,I_a B,I_b C,I_c)
+ \tkzAutoLabelPoints[center=I,
+ blue,font=\scriptsize](I_a,I_b,I_c)
+ \tkzAutoLabelPoints[center=I,red,
+ font=\scriptsize](A,B,C,I_a,I_b,I_c)
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{Option : "ex" ou "Excentral"}
-The excentral triangle of a triangle $ABC$ is the triangle $JaJbJc$ with vertices corresponding to the excenters of $ABC$.
+\subsubsection{Option \tkzname{ex} or \tkzname{excentral} }
+
+The excentral triangle of a triangle $ABC$ is the triangle $J_aJ_bJ_c$ with vertices corresponding to the excenters of $ABC$.
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=.6]
\tkzDefPoints{0/0/A,6/0/B,0.8/4/C}
- \tkzDefSpcTriangle[excentral,name=J](A,B,C){a,b,c}
- \tkzDefSpcTriangle[extouch,name=T](A,B,C){a,b,c}
+ \tkzDefSpcTriangle[excentral,name=J](A,B,C){_a,_b,_c}
+ \tkzDefSpcTriangle[extouch,name=T](A,B,C){_a,_b,_c}
\tkzDrawPolygon[blue](A,B,C)
- \tkzDrawPolygon[red](Ja,Jb,Jc)
+ \tkzDrawPolygon[red](J_a,J_b,J_c)
\tkzDrawPoints(A,B,C)
- \tkzDrawPoints[red](Ja,Jb,Jc)
+ \tkzDrawPoints[red](J_a,J_b,J_c)
\tkzLabelPoints(A,B,C)
- \tkzLabelPoints[red](Jb,Jc)
- \tkzLabelPoints[red,above](Ja)
+ \tkzLabelPoints[red](J_b,J_c)
+ \tkzLabelPoints[red,above](J_a)
\tkzClipBB \tkzShowBB
- \tkzDrawCircles[gray](Ja,Ta Jb,Tb Jc,Tc)
+ \tkzDrawCircles[gray](J_a,T_a J_b,T_b J_c,T_c)
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{Option : "intouch"}
-The contact triangle of a triangle ABC, also called the intouch triangle, is the triangle formed by the points of tangency of the incircle of $ABC$ with $ABC$.\\
+\subsubsection{Option \tkzname{intouch}}
+The contact triangle of a triangle $ABC$, also called the intouch triangle, is the triangle formed by the points of tangency of the incircle of $ABC$ with $ABC$.\\
\href{http://mathworld.wolfram.com/ContactTriangle.html}{Weisstein, Eric W. "Contact triangle" From MathWorld--A Wolfram Web Resource.}
We obtain the intersections of the bisectors with the sides.
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=.75]
\tkzDefPoints{0/0/A,6/0/B,0.8/4/C}
- \tkzDefSpcTriangle[intouch,name=x](A,B,C){a,b,c}
+ \tkzDefSpcTriangle[intouch,name=X](A,B,C){_a,_b,_c}
\tkzInCenter(A,B,C)\tkzGetPoint{I}
\tkzDrawPolygon[red](A,B,C)
- \tkzDrawPolygon[blue](xa,xb,xc)
+ \tkzDrawPolygon[blue](X_a,X_b,X_c)
\tkzDrawPoints[red](A,B,C)
- \tkzDrawPoints[blue](xa,xb,xc)
+ \tkzDrawPoints[blue](X_a,X_b,X_c)
\tkzDrawCircle[in](A,B,C)
\tkzAutoLabelPoints[center=I,blue,font=\scriptsize]%
-(xa,xb,xc)
+(X_a,X_b,X_c)
\tkzAutoLabelPoints[center=I,red,font=\scriptsize]%
(A,B,C)
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{Option : "extouch"}
-The extouch triangle $TaTbTc$ is the triangle formed by the points of tangency of a triangle $ABC$ with its excircles $Ja$, $Jb$, and $Jc$. The points $Ta$, $Tb$, and $Tc$ can also be constructed as the points which bisect the perimeter of $A_1A_2A_3$ starting at $A$, $B$, and $C$.\\
+\subsubsection{Option \tkzname{extouch}}
+The extouch triangle $T_aT_bT_c$ is the triangle formed by the points of tangency of a triangle $ABC$ with its excircles $J_a$, $J_b$, and $J_c$. The points $T_a$, $T_b$, and $T_c$ can also be constructed as the points which bisect the perimeter of $A_1A_2A_3$ starting at $A$, $B$, and $C$.\\
\href{http://mathworld.wolfram.com/ExtouchTriangle.html}{Weisstein, Eric W. "Extouch triangle" From MathWorld--A Wolfram Web Resource.}
We obtain the points of contact of the exinscribed circles as well as the triangle formed by the centres of the exinscribed circles.
@@ -267,32 +290,32 @@ We obtain the points of contact of the exinscribed circles as well as the triang
\begin{tikzpicture}[scale=.7]
\tkzDefPoints{0/0/A,6/0/B,0.8/4/C}
\tkzDefSpcTriangle[excentral,
- name=J](A,B,C){a,b,c}
+ name=J](A,B,C){_a,_b,_c}
\tkzDefSpcTriangle[extouch,
- name=T](A,B,C){a,b,c}
+ name=T](A,B,C){_a,_b,_c}
\tkzDefTriangleCenter[nagel](A,B,C)
-\tkzGetPoint{Na}
+\tkzGetPoint{N_a}
\tkzDefTriangleCenter[centroid](A,B,C)
\tkzGetPoint{G}
-\tkzDrawPoints[blue](Ja,Jb,Jc)
+\tkzDrawPoints[blue](J_a,J_b,J_c)
\tkzClipBB \tkzShowBB
-\tkzDrawCircles[gray](Ja,Ta Jb,Tb Jc,Tc)
+\tkzDrawCircles[gray](J_a,T_a J_b,T_b J_c,T_c)
\tkzDrawLines[add=1 and 1](A,B B,C C,A)
-\tkzDrawSegments[gray](A,Ta B,Tb C,Tc)
-\tkzDrawSegments[gray](Ja,Ta Jb,Tb Jc,Tc)
+\tkzDrawSegments[gray](A,T_a B,T_b C,T_c)
+\tkzDrawSegments[gray](J_a,T_a J_b,T_b J_c,T_c)
\tkzDrawPolygon[blue](A,B,C)
-\tkzDrawPolygon[red](Ta,Tb,Tc)
-\tkzDrawPoints(A,B,C,Na)
-\tkzLabelPoints(Na)
+\tkzDrawPolygon[red](T_a,T_b,T_c)
+\tkzDrawPoints(A,B,C,N_a)
+\tkzLabelPoints(N_a)
\tkzAutoLabelPoints[center=Na,blue](A,B,C)
\tkzAutoLabelPoints[center=G,red,
- dist=.4](Ta,Tb,Tc)
-\tkzMarkRightAngles[fill=gray!15](Ja,Ta,B
- Jb,Tb,C Jc,Tc,A)
+ dist=.4](T_a,T_b,T_c)
+\tkzMarkRightAngles[fill=gray!15](J_a,T_a,B
+ J_b,T_b,C J_c,T_c,A)
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{Option : "feuerbach"}
+\subsubsection{Option \tkzname{feuerbach}}
The Feuerbach triangle is the triangle formed by the three points of tangency of the nine-point circle with the excircles.\\
\href{http://mathworld.wolfram.com/FeuerbachTriangle.html}{Weisstein, Eric W. "Feuerbach triangle" From MathWorld--A Wolfram Web Resource.}
@@ -322,8 +345,8 @@ The Feuerbach triangle is the triangle formed by the three points of tangency of
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{Option Triangle "tangential"}
-The tangential triangle is the triangle $T_AT_BT_C $formed by the lines tangent to the circumcircle of a given triangle ABC at its vertices. It is therefore antipedal triangle of ABC with respect to the circumcenter O.\\
+\subsubsection{Option \tkzname{tangential}}
+The tangential triangle is the triangle $T_aT_bT_c$ formed by the lines tangent to the circumcircle of a given triangle $ABC$ at its vertices. It is therefore antipedal triangle of $ABC$ with respect to the circumcenter $O$.\\
\href{http://mathworld.wolfram.com/TangentialTriangle.html}{Weisstein, Eric W. "Tangential Triangle." From MathWorld--A Wolfram Web Resource. }
@@ -331,21 +354,21 @@ The tangential triangle is the triangle $T_AT_BT_C $formed by the lines tangent
\begin{tikzpicture}[scale=.5,rotate=80]
\tkzDefPoints{0/0/A,6/0/B,1.8/4/C}
\tkzDefSpcTriangle[tangential,
- name=T](A,B,C){a,b,c}
+ name=T](A,B,C){_a,_b,_c}
\tkzDrawPolygon[red](A,B,C)
- \tkzDrawPolygon[blue](Ta,Tb,Tc)
+ \tkzDrawPolygon[blue](T_a,T_b,T_c)
\tkzDrawPoints[red](A,B,C)
- \tkzDrawPoints[blue](Ta,Tb,Tc)
+ \tkzDrawPoints[blue](T_a,T_b,T_c)
\tkzDefCircle[circum](A,B,C)
\tkzGetPoint{O}
\tkzDrawCircle(O,A)
\tkzLabelPoints[red](A,B,C)
- \tkzLabelPoints[blue](Ta,Tb,Tc)
+ \tkzLabelPoints[blue](T_a,T_b,T_c)
\end{tikzpicture}
\end{tkzexample}
- \subsubsection{Option Triangle "euler"}
-The Euler triangle of a triangle ABC is the triangle $E_AE_BE_C$ whose vertices are the midpoints of the segments joining the orthocenter H with the respective vertices. The vertices of the triangle are known as the Euler points, and lie on the nine-point circle.
+\subsubsection{Option \tkzname{euler}}
+The Euler triangle of a triangle $ABC$ is the triangle $E_AE_BE_C$ whose vertices are the midpoints of the segments joining the orthocenter $H$ with the respective vertices. The vertices of the triangle are known as the Euler points, and lie on the nine-point circle.
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[rotate=90,scale=1.25]