summaryrefslogtreecommitdiff
path: root/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-triangles.tex
diff options
context:
space:
mode:
Diffstat (limited to 'macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-triangles.tex')
-rw-r--r--macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-triangles.tex96
1 files changed, 71 insertions, 25 deletions
diff --git a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-triangles.tex b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-triangles.tex
index 990862afff..7eb2875ec7 100644
--- a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-triangles.tex
+++ b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-triangles.tex
@@ -163,18 +163,17 @@ This triangle has sides whose lengths are proportional to 3, 4 and 5.
\end{tkzexample}
\subsubsection{Option \tkzname{gold} }
-\begin{tkzexample}[latex=7 cm,small]
+\begin{tkzexample}[latex=6 cm,small]
\begin{tikzpicture}
- \tkzDefPoints{0/0/A,4/0/B}
- \tkzDefTriangle[gold](A,B)
- \tkzGetPoint{C}
- \tkzDrawPolygon(A,B,C)
- \tkzDrawPoints(A,B,C)
- \tkzLabelPoints(A,B)
- \tkzLabelPoints[above](C)
- \tkzLabelAngle[pos=0.8](B,A,C){$36^\circ$}
- \tkzLabelAngle[pos=0.8](C,B,A){$72^\circ$}
- \tkzLabelAngle[pos=0.8](A,C,B){$72^\circ$}
+ \tkzDefPoints{0/0/A,4/0/B}
+ \tkzDefTriangle[gold](A,B)
+ \tkzGetPoint{C}
+ \tkzDrawPolygon(A,B,C)
+ \tkzDrawPoints(A,B,C)
+ \tkzLabelPoints(B) \tkzLabelPoints[below](A,C)
+ \tkzLabelAngle[pos=0.8](C,A,B){$36^\circ$}
+ \tkzLabelAngle[pos=0.8](A,B,C){$72^\circ$}
+ \tkzLabelAngle[pos=0.8](B,C,A){$72^\circ$}
\end{tikzpicture}
\end{tkzexample}
@@ -196,7 +195,7 @@ This triangle has sides whose lengths are proportional to 3, 4 and 5.
\end{tikzpicture}
\end{tkzexample}
-\section{Specific triangles with \tkzcname{tkzDefSpcTriangle}}
+\subsection{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.
@@ -227,7 +226,7 @@ options & default & definition \\
\end{NewMacroBox}
-\subsection{How to name the vertices}
+\subsubsection{How to name the vertices}
With \tkzcname{tkzDefSpcTriangle[medial,name=M](A,B,C)\{\_A,\_B,\_C\}} you get three vertices named $M_A$, $M_B$ and $M_C$.
@@ -260,7 +259,7 @@ In the following example, we obtain the Euler circle which passes through the pr
\end{tikzpicture}
\end{tkzexample}
-\subsection{Option \tkzname{in} or \tkzname{incentral} }
+\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
@@ -286,7 +285,7 @@ respective opposite sides.
\end{tikzpicture}
\end{tkzexample}
-\subsection{Option \tkzname{ex} or \tkzname{excentral} }
+\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 $ABC$.
@@ -308,7 +307,7 @@ The excentral triangle of a triangle $ABC$ is the triangle $J_aJ_bJ_c$ with vert
\end{tkzexample}
-\subsection{Option \tkzname{intouch} or \tkzname{contact}}
+\subsubsection{Option \tkzname{intouch} or \tkzname{contact}}
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.}
@@ -330,11 +329,11 @@ We obtain the intersections of the bisectors with the sides.
\end{tikzpicture}
\end{tkzexample}
-\subsection{Option \tkzname{extouch}}
+\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.
+We obtain the points of contact of the exinscribed circles as well as the triangle formed by the centers of the exinscribed circles.
\begin{tkzexample}[latex=8cm,small]
\begin{tikzpicture}[scale=.7]
@@ -365,7 +364,7 @@ We obtain the points of contact of the exinscribed circles as well as the triang
\end{tikzpicture}
\end{tkzexample}
-\subsection{Option \tkzname{orthic}}
+\subsubsection{Option \tkzname{orthic}}
Given a triangle $ABC$, the triangle $H_AH_BH_C$ whose vertices are endpoints of the altitudes from each of the vertices of ABC is called the orthic triangle, or sometimes the altitude triangle. The three lines $AH_A$, $BH_B$, and $CH_C$ are concurrent at the orthocenter H of ABC.
@@ -395,7 +394,7 @@ Given a triangle $ABC$, the triangle $H_AH_BH_C$ whose vertices are endpoints of
\end{tikzpicture}
\end{tkzexample}
-\subsection{Option \tkzname{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.}
@@ -426,7 +425,7 @@ The Feuerbach triangle is the triangle formed by the three points of tangency of
\end{tikzpicture}
\end{tkzexample}
-\subsection{Option \tkzname{tangential}}
+\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. }
@@ -448,7 +447,7 @@ The tangential triangle is the triangle $T_aT_bT_c$ formed by the lines tangent
\end{tikzpicture}
\end{tkzexample}
-\subsection{Option \tkzname{euler}}
+\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.
\\
\href{https://mathworld.wolfram.com/EulerTriangle.html}{Weisstein, Eric W. "Euler Triangle." From MathWorld--A Wolfram Web Resource.}
@@ -482,7 +481,7 @@ The Euler triangle of a triangle $ABC$ is the triangle $E_AE_BE_C$ whose vertice
\end{tikzpicture}
\end{tkzexample}
-\subsection{Option \tkzname{euler} and Option \tkzname{orthic}}
+\subsubsection{Option \tkzname{euler} and Option \tkzname{orthic}}
\begin{tkzexample}[vbox,small]
\begin{tikzpicture}[scale=1.25]
\tkzDefPoints{0/0/A,6/0/B,0.8/4/C}
@@ -509,7 +508,7 @@ The Euler triangle of a triangle $ABC$ is the triangle $E_AE_BE_C$ whose vertice
\end{tkzexample}
-\subsection{Option \tkzname{symmedial}}
+\subsubsection{Option \tkzname{symmedial}}
The symmedial triangle$ K_AK_BK_C$ is the triangle whose vertices are the intersection points of the symmedians with the reference triangle $ABC$.
\begin{tkzexample}[latex=7cm,small]
@@ -525,5 +524,52 @@ The symmedial triangle$ K_AK_BK_C$ is the triangle whose vertices are the inters
\tkzLabelPoints[font=\scriptsize](A,B,C,K,K_A,K_B,K_C)
\end{tikzpicture}
\end{tkzexample}
-
+
+\subsection{Permutation of two points of a triangle}
+
+\begin{NewMacroBox}{tkzPermute}{\parg{$pt1$,$pt2$,$pt3$}}%
+\begin{tabular}{lll}%
+arguments & example & explanation \\
+\midrule
+\TAline{(pt1,pt2,pt3)} {\tkzcname{tkzPermute}(A,B,C)}{$A$, $\widehat{B,A,C}$ are unchanged, $B$, $C$ exchange their position}
+\midrule
+\end{tabular}
+
+\emph{The triangle is unchanged.}
+\end{NewMacroBox}
+
+\subsubsection{Modification of the \tkzname{school} triangle}
+This triangle is constructed from the segment $[AB]$ on $[A,x)$
+\begin{tkzexample}[latex=7cm,small]
+\begin{tikzpicture}
+ \tkzDefPoints{0/0/A,4/0/B,6/0/x}
+ \tkzDefTriangle[school](A,B)
+ \tkzGetPoint{C}
+ \tkzDrawSegments(A,B B,x)
+ \tkzDrawSegments(A,C B,C)
+ \tkzDrawPoints(A,B,C)
+ \tkzLabelPoints(A,B,C,x)
+ \tkzMarkRightAngles(C,B,A)
+\end{tikzpicture}
+\end{tkzexample}
+
+If we want the segment $[AC]$ to be on $[A,x)$, we just have to swap $B$ and $C$.
+
+\begin{tkzexample}[latex=7cm,small]
+\begin{tikzpicture}
+ \tkzDefPoints{0/0/A,4/0/B,6/0/x}
+ \tkzDefTriangle[school](A,B)
+ \tkzGetPoint{C}
+ \tkzPermute(A,B,C)
+ \tkzDrawSegments(A,B C,x)
+ \tkzDrawSegments(A,C B,C)
+ \tkzDrawPoints(A,B,C)
+ \tkzLabelPoints(A,C,x)
+ \tkzLabelPoints[above](B)
+ \tkzMarkRightAngles(C,B,A)
+\end{tikzpicture}
+\end{tkzexample}
+
+Remark: Only the first point is unchanged. The order of the last two parameters is not important.
+
\endinput \ No newline at end of file