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diff --git a/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-triangles.tex b/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-triangles.tex new file mode 100644 index 00000000000..038e93892cb --- /dev/null +++ b/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-triangles.tex @@ -0,0 +1,403 @@ +\section{Triangles} + +\subsection{Definition of triangles \tkzcname{tkzDefTriangle}} +The following macros will allow you to define or construct a triangle from \tkzname{at least} two points. + + At the moment, it is possible to define the following triangles: + \begin{itemize} +\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{euclide} or \tkzname{gold} for the gold triangle; + +\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}}% +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}% +\toprule +options & default & definition \\ +\midrule +\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} allows you to store the point otherwise \tkzname{tkzPointResult} allows for immediate use. +\end{NewMacroBox} + +\subsubsection{Option \tkzname{golden}} +\begin{tkzexample}[latex=6 cm,small] +\begin{tikzpicture}[scale=.8] +\tkzInit[xmax=5,ymax=3] \tkzClip[space=.5] + \tkzDefPoint(0,0){A} \tkzDefPoint(4,0){B} + \tkzDefTriangle[golden](A,B)\tkzGetPoint{C} + \tkzDrawPolygon(A,B,C) \tkzDrawPoints(A,B,C) + \tkzLabelPoints(A,B) \tkzDrawBisector(A,C,B) + \tkzLabelPoints[above](C) +\end{tikzpicture} +\end{tkzexample} + +\subsubsection{Option \tkzname{equilateral}} +\begin{tkzexample}[latex=7 cm,small] +\begin{tikzpicture} + \tkzDefPoint(0,0){A} + \tkzDefPoint(4,0){B} + \tkzDefTriangle[equilateral](A,B) + \tkzGetPoint{C} + \tkzDrawPolygon(A,B,C) + \tkzDefTriangle[equilateral](B,A) + \tkzGetPoint{D} + \tkzDrawPolygon(B,A,D) + \tkzDrawPoints(A,B,C,D) + \tkzLabelPoints(A,B,C,D) +\end{tikzpicture} +\end{tkzexample} + +\subsubsection{Option \tkzname{gold} or \tkzname{euclide} } +\begin{tkzexample}[latex=7 cm,small] +\begin{tikzpicture} + \tkzDefPoint(0,0){A} \tkzDefPoint(4,0){B} + \tkzDefTriangle[euclide](A,B)\tkzGetPoint{C} + \tkzDrawPolygon(A,B,C) + \tkzDrawPoints(A,B,C) + \tkzLabelPoints(A,B) + \tkzLabelPoints[above](C) + \tkzDrawBisector(A,C,B) +\end{tikzpicture} +\end{tkzexample} + +\newpage +\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}% +\toprule +options & default & definition \\ +\midrule +\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 +In all its definitions, the dimensions of the triangle depend on the two starting points. +\end{NewMacroBox} + +\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} + \tkzDefPoint(0,0){A} + \tkzDefPoint(4,0){B} + \tkzDrawTriangle[pythagore,fill=blue!30](A,B) + \tkzMarkRightAngles(A,B,tkzPointResult) +\end{tikzpicture} +\end{tkzexample} + + +\subsubsection{Option \tkzname{school}} +The angles are 30, 60 and 90 degrees. + +\begin{tkzexample}[latex=6 cm,small] +\begin{tikzpicture} + \tkzDefPoint(0,0){A} \tkzDefPoint(4,0){B} + \tkzDrawTriangle[school,fill=red!30](A,B) + \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} + +\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} + +\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}% +\toprule +options & default & definition \\ +\midrule +\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} allows you to store the point otherwise \tkzname{tkzPointResult} allows for immediate use. +\end{NewMacroBox} + +\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.} + +In the following example, we obtain the Euler circle which passes through the previously defined points. + +\begin{tkzexample}[latex=7cm,small] +\begin{tikzpicture}[rotate=90,scale=.75] + \tkzDefPoints{0/0/A,6/0/B,0.8/4/C} + \tkzDefTriangleCenter[centroid](A,B,C) + \tkzGetPoint{M} + \tkzDefSpcTriangle[medial,name=M](A,B,C){_A,_B,_C} + \tkzDrawPolygon[color=blue](A,B,C) + \tkzDrawSegments[dashed,red](A,M_A B,M_B C,M_C) + \tkzDrawPolygon[color=red](M_A,M_B,M_C) + \tkzDrawPoints(A,B,C,M) + \tkzDrawPoints[red](M_A,M_B,M_C) +\tkzAutoLabelPoints[center=M,font=\scriptsize]% +(A,B,C,M_A,M_B,M_C) + \tkzLabelPoints[font=\scriptsize](M) +\end{tikzpicture} +\end{tkzexample} + +\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.} + + +\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} + \tkzInCenter(A,B,C)\tkzGetPoint{I} + \tkzDrawPolygon[red](A,B,C) + \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,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 \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} + \tkzDrawPolygon[blue](A,B,C) + \tkzDrawPolygon[red](J_a,J_b,J_c) + \tkzDrawPoints(A,B,C) + \tkzDrawPoints[red](J_a,J_b,J_c) + \tkzLabelPoints(A,B,C) + \tkzLabelPoints[red](J_b,J_c) + \tkzLabelPoints[red,above](J_a) + \tkzClipBB \tkzShowBB + \tkzDrawCircles[gray](J_a,T_a J_b,T_b J_c,T_c) +\end{tikzpicture} +\end{tkzexample} + + +\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} + \tkzInCenter(A,B,C)\tkzGetPoint{I} + \tkzDrawPolygon[red](A,B,C) + \tkzDrawPolygon[blue](X_a,X_b,X_c) + \tkzDrawPoints[red](A,B,C) + \tkzDrawPoints[blue](X_a,X_b,X_c) + \tkzDrawCircle[in](A,B,C) + \tkzAutoLabelPoints[center=I,blue,font=\scriptsize]% +(X_a,X_b,X_c) + \tkzAutoLabelPoints[center=I,red,font=\scriptsize]% +(A,B,C) +\end{tikzpicture} +\end{tkzexample} + +\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. + +\begin{tkzexample}[latex=8cm,small] +\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} +\tkzDefSpcTriangle[extouch, + name=T](A,B,C){_a,_b,_c} +\tkzDefTriangleCenter[nagel](A,B,C) +\tkzGetPoint{N_a} +\tkzDefTriangleCenter[centroid](A,B,C) +\tkzGetPoint{G} +\tkzDrawPoints[blue](J_a,J_b,J_c) +\tkzClipBB \tkzShowBB +\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,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](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](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 \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.} + + The points of tangency define the Feuerbach triangle. + + +\begin{tkzexample}[latex=8cm,small] +\begin{tikzpicture}[scale=1] + \tkzDefPoint(0,0){A} + \tkzDefPoint(3,0){B} + \tkzDefPoint(0.5,2.5){C} + \tkzDefCircle[euler](A,B,C) \tkzGetPoint{N} + \tkzDefSpcTriangle[feuerbach, + name=F](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} + \tkzDrawPoints[blue](J_a,J_b,J_c,F_a,F_b,F_c,A,B,C) + \tkzClipBB \tkzShowBB + \tkzDrawCircle[purple](N,F_a) + \tkzDrawPolygon(A,B,C) + \tkzDrawPolygon[blue](F_a,F_b,F_c) + \tkzDrawCircles[gray](J_a,F_a J_b,F_b J_c,F_c) + \tkzAutoLabelPoints[center=N,dist=.3, + font=\scriptsize](A,B,C,F_a,F_b,F_c,J_a,J_b,J_c) +\end{tikzpicture} +\end{tkzexample} + +\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. } + + +\begin{tkzexample}[latex=8cm,small] +\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} + \tkzDrawPolygon[red](A,B,C) + \tkzDrawPolygon[blue](T_a,T_b,T_c) + \tkzDrawPoints[red](A,B,C) + \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](T_a,T_b,T_c) +\end{tikzpicture} +\end{tkzexample} + +\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] + \tkzDefPoints{0/0/A,6/0/B,0.8/4/C} + \tkzDefSpcTriangle[medial, + name=M](A,B,C){_A,_B,_C} + \tkzDefTriangleCenter[euler](A,B,C) + \tkzGetPoint{N} % I= N nine points + \tkzDefTriangleCenter[ortho](A,B,C) + \tkzGetPoint{H} + \tkzDefMidPoint(A,H) \tkzGetPoint{E_A} + \tkzDefMidPoint(C,H) \tkzGetPoint{E_C} + \tkzDefMidPoint(B,H) \tkzGetPoint{E_B} + \tkzDefSpcTriangle[ortho,name=H](A,B,C){_A,_B,_C} + \tkzDrawPolygon[color=blue](A,B,C) + \tkzDrawCircle(N,E_A) + \tkzDrawSegments[blue](A,H_A B,H_B C,H_C) + \tkzDrawPoints(A,B,C,N,H) + \tkzDrawPoints[red](M_A,M_B,M_C) + \tkzDrawPoints[blue]( H_A,H_B,H_C) + \tkzDrawPoints[green](E_A,E_B,E_C) + \tkzAutoLabelPoints[center=N,font=\scriptsize]% +(A,B,C,M_A,M_B,M_C,H_A,H_B,H_C,E_A,E_B,E_C) +\tkzLabelPoints[font=\scriptsize](H,N) +\tkzMarkSegments[mark=s|,size=3pt, + color=blue,line width=1pt](B,E_B E_B,H) + \tkzDrawPolygon[color=red](M_A,M_B,M_C) +\end{tikzpicture} +\end{tkzexample} + + +\endinput
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