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Diffstat (limited to 'macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-triangles.tex')
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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
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