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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 index 038e93892cb..5becdf0e2f1 100644 --- a/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-triangles.tex +++ b/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-triangles.tex @@ -7,12 +7,13 @@ The following macros will allow you to define or construct a triangle from \tkzn \begin{itemize} \item \tkzname{two angles} determines a triangle with two angles; \item \tkzname{equilateral} determines an equilateral triangle; +\item \tkzname{isosceles right} determines an isoxsceles right 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{euclid} or \tkzname{gold} for the gold triangle; in the previous version the option was "euclide" with an "e". \item \tkzname{cheops} determines a third point such that the triangle is isosceles with side measurements proportional to $2$, $\Phi$ and $\Phi$. \end{itemize} @@ -26,13 +27,14 @@ The points are ordered because the triangle is constructed following the direct 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 $\Phi$.} +\TOline{equilateral} {equilateral}{equilateral triangle } +\TOline{isosceles right} {equilateral}{isosceles right triangle } +\TOline{pythagore}{equilateral}{proportional to the pythagorean triangle 3-4-5} +\TOline{school} {equilateral}{angles of 30, 60 and 90 degrees } +\TOline{gold}{equilateral}{angles of 72, 72 and 36 degrees, $A$ is the apex} +\TOline{euclid} {equilateral}{same as above but $[AB]$ is the base} +\TOline{golden} {equilateral}{B rectangle and $AB/AC = \Phi$} +\TOline{cheops} {equilateral}{AC=BC, AC and BC are proportional to $2$ and $\Phi$.} \bottomrule \end{tabular} @@ -40,130 +42,134 @@ options & default & definition \\ \tkzcname{tkzGetPoint} allows you to store the point otherwise \tkzname{tkzPointResult} allows for immediate use. \end{NewMacroBox} -\subsubsection{Option \tkzname{golden}} +\subsubsection{Option \tkzname{two angles}} \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) +\tkzDefPoint(0,0){A} +\tkzDefPoint(5,0){B} +\tkzDefTriangle[two angles = 50 and 70](A,B) +\tkzGetPoint{C} +\tkzDrawSegment(A,B) +\tkzDrawPoints(A,B) +\tkzLabelPoints(A,B) +\tkzDrawSegments[new](A,C B,C) +\tkzDrawPoints[new](C) +\tkzLabelPoints[above,new](C) +\tkzLabelAngle[pos=1.4](B,A,C){$50^\circ$} +\tkzLabelAngle[pos=0.8](C,B,A){$70^\circ$} \end{tikzpicture} \end{tkzexample} -\subsubsection{Option \tkzname{gold} or \tkzname{euclide} } -\begin{tkzexample}[latex=7 cm,small] +\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} - \tkzDefTriangle[euclide](A,B)\tkzGetPoint{C} - \tkzDrawPolygon(A,B,C) - \tkzDrawPoints(A,B,C) - \tkzLabelPoints(A,B) - \tkzLabelPoints[above](C) - \tkzDrawBisector(A,C,B) + \tkzDefPoints{0/0/A,4/0/B} + \tkzDefTriangle[school](A,B) + \tkzGetPoint{C} + \tkzMarkRightAngles(C,B,A) + \tkzLabelAngle[pos=0.8](B,A,C){$30^\circ$} + \tkzLabelAngle[pos=0.8](C,B,A){$90^\circ$} + \tkzLabelAngle[pos=0.8](A,C,B){$60^\circ$} + \tkzDrawSegments(A,B) + \tkzDrawSegments[new](A,C B,C) \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) +\begin{tikzpicture} + \tkzDefPoints{0/0/A,4/0/B} + \tkzDefTriangle[pythagore](A,B) + \tkzGetPoint{C} + \tkzDrawSegments(A,B) + \tkzDrawSegments[new](A,C B,C) + \tkzMarkRightAngles(A,B,C) + \tkzLabelPoint[above,new](C){$C$} + \tkzDrawPoints[new](C) + \tkzDrawPoints(A,B) + \tkzLabelPoints(A,B) \end{tikzpicture} \end{tkzexample} -\subsubsection{Option \tkzname{school}} -The angles are 30, 60 and 90 degrees. - +\subsubsection{Option \tkzname{golden}} \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) +\begin{tikzpicture}[scale=.8] +\tkzDefPoint(0,0){A} \tkzDefPoint(4,0){B} +\tkzDefTriangle[golden](A,B)\tkzGetPoint{C} +\tkzDefSpcTriangle[in,name=M](A,B,C){a,b,c} +\tkzDrawPolygon(A,B,C) +\tkzDrawPoints(A,B) +\tkzDrawSegment(C,Mc) +\tkzDrawPoints[new](C) +\tkzLabelPoints(A,B) +\tkzLabelPoints[above,new](C) \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) +\subsubsection{Option \tkzname{equilateral} and \tkzname{isosceles right}} +\begin{tkzexample}[latex=7 cm,small] +\begin{tikzpicture} + \tkzDefPoint(0,0){A} + \tkzDefPoint(4,0){B} + \tkzDefTriangle[equilateral](A,B) + \tkzGetPoint{C} + \tkzDefTriangle[isosceles right](A,B) + \tkzGetPoint{E} + \tkzDrawPolygons(A,B,C A,B,E) + \tkzDefTriangle[equilateral](B,A) + \tkzGetPoint{D} + \tkzDrawPolygon(B,A,D) + \tkzMarkRightAngles(B,E,A) + \tkzDrawPoints(A,B,C,D,E) + \tkzLabelPoints(A,B,C,D,E) \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) +\subsubsection{Option \tkzname{gold} } +\begin{tkzexample}[latex=7 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){$72^\circ$} \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} +\subsubsection{Option \tkzname{euclid}} +\tkzimp{Euclid} and \tkzimp{gold} are identical but the segment AB is a base in one and a side in the other. + +\begin{tkzexample}[latex=7 cm,small] +\begin{tikzpicture}[scale=.75] + \tkzDefPoint(0,0){A} \tkzDefPoint(4,0){B} + \tkzDefTriangle[euclid](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){$72^\circ$} + \tkzLabelAngle[pos=0.8](C,B,A){$72^\circ$} + \tkzLabelAngle[pos=0.8](A,C,B){$36^\circ$} +\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! +\begin{NewMacroBox}{tkzDefSpcTriangle}{\oarg{local options}\parg{p1,p2,p3}\marg{r1,r2,r3}} +The order of the points is important! p1p2p3 defines a triangle then the result is a triangle whose vertices have as reference a combination with \tkzname{name} and r1,r2, r3. If \tkzname{name} is empty then the references are r1,r2 and r3. \medskip @@ -171,25 +177,36 @@ The order of the points is important! \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 $\Phi$.} +\TOline{orthic} {centroid}{determined by endpoints of the altitudes ...} +\TOline{centroid or medial}{centroid}{intersection of the triangle's three triangle medians} +\TOline{in or incentral}{centroid}{determined with the angle bisectors} +\TOline{ex or excentral} {centroid}{determined with the excenters} +\TOline{extouch}{centroid}{formed by the points of tangency with the excircles} +\TOline{intouch or contact} {centroid}{formed by the points of tangency of the incircle} +\TOline{} {}{each of the vertices} +\TOline{euler} {centroid}{formed by Euler points on the nine-point circle} +\TOline{symmedial} {centroid}{intersection points of the symmedians} +\TOline{tangential}{centroid}{formed by the lines tangent to the circumcircle} +\TOline{feuerbach} {centroid}{formed by the points of tangency of the nine-point ...} +\TOline{} {} {circle with the excircles} +\TOline{name} {empty}{used to name the vertices} \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.\\ +\subsection{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 $M_C$. + +With \tkzcname{tkzDefSpcTriangle[medial](A,B,C)\{a,b,c\}} you get three vertices named and labeled $a$, $b$ and $c$. + +Possible \tkzcname{tkzDefSpcTriangle[medial,name=M\_](A,B,C)\{A,B,C\}} you get three vertices named $M_A$, $M_B$ and $M_C$. + +\subsection{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. @@ -198,24 +215,25 @@ In the following example, we obtain the Euler circle which passes through the pr \begin{tikzpicture}[rotate=90,scale=.75] \tkzDefPoints{0/0/A,6/0/B,0.8/4/C} \tkzDefTriangleCenter[centroid](A,B,C) - \tkzGetPoint{M} + \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]% + \tkzDrawPolygon(A,B,C) + \tkzDrawSegments[dashed,new](A,M_A B,M_B C,M_C) + \tkzDrawPolygon[new](M_A,M_B,M_C) + \tkzDrawPoints(A,B,C) + \tkzDrawPoints[new](M,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} } +\subsection{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.\\ +respective opposite sides. +\\ \href{http://mathworld.wolfram.com/ContactTriangle.html}{Weisstein, Eric W. "Incentral triangle" From MathWorld--A Wolfram Web Resource.} @@ -224,19 +242,19 @@ respective opposite sides.\\ \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) + \tkzDrawPolygon(A,B,C) + \tkzDrawPolygon[new](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) + \tkzDrawSegments[dashed,new](A,I_a B,I_b C,I_c) + \tkzAutoLabelPoints[center=I,% + new,font=\scriptsize](I_a,I_b,I_c) + \tkzAutoLabelPoints[center=I, + font=\scriptsize](A,B,C) \end{tikzpicture} \end{tkzexample} -\subsubsection{Option \tkzname{ex} or \tkzname{excentral} } +\subsection{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 $ABC$. @@ -245,33 +263,33 @@ The excentral triangle of a triangle $ABC$ is the triangle $J_aJ_bJ_c$ with vert \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) + \tkzDrawPolygon(A,B,C) + \tkzDrawPolygon[new](J_a,J_b,J_c) + \tkzClipBB \tkzDrawPoints(A,B,C) - \tkzDrawPoints[red](J_a,J_b,J_c) + \tkzDrawPoints[new](J_a,J_b,J_c) \tkzLabelPoints(A,B,C) - \tkzLabelPoints[red](J_b,J_c) - \tkzLabelPoints[red,above](J_a) - \tkzClipBB \tkzShowBB + \tkzLabelPoints[new](J_b,J_c) + \tkzLabelPoints[new,above](J_a) \tkzDrawCircles[gray](J_a,T_a J_b,T_b J_c,T_c) \end{tikzpicture} \end{tkzexample} -\subsubsection{Option \tkzname{intouch}} +\subsection{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.} 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} + \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) + \tkzDrawPolygon(A,B,C) + \tkzDrawPolygon[new](X_a,X_b,X_c) + \tkzDrawPoints(A,B,C) + \tkzDrawPoints[new](X_a,X_b,X_c) \tkzDrawCircle[in](A,B,C) \tkzAutoLabelPoints[center=I,blue,font=\scriptsize]% (X_a,X_b,X_c) @@ -280,7 +298,7 @@ We obtain the intersections of the bisectors with the sides. \end{tikzpicture} \end{tkzexample} -\subsubsection{Option \tkzname{extouch}} +\subsection{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.} @@ -297,33 +315,62 @@ We obtain the points of contact of the exinscribed circles as well as the triang \tkzGetPoint{N_a} \tkzDefTriangleCenter[centroid](A,B,C) \tkzGetPoint{G} -\tkzDrawPoints[blue](J_a,J_b,J_c) +\tkzDrawPoints[new](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) +\tkzDrawSegments[new](A,T_a B,T_b C,T_c) +\tkzDrawSegments[new](J_a,T_a J_b,T_b J_c,T_c) +\tkzDrawPolygon(A,B,C) +\tkzDrawPolygon[new](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, +\tkzAutoLabelPoints[center=N_a](A,B,C) +\tkzAutoLabelPoints[center=G,new, 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}} +\subsection{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. + +\begin{tkzexample}[latex=7cm,small] +\begin{tikzpicture}[scale=.75] +\tkzDefPoints{1/5/A,0/0/B,7/0/C} + \tkzDefSpcTriangle[orthic](A,B,C){H_A,H_B,H_C} + \tkzDefTriangleCenter[ortho](B,C,A) + \tkzGetPoint{H} + \tkzDefPointWith[orthogonal,normed](H_A,B) + \tkzGetPoint{a} + \tkzDrawSegments[new](A,H_A B,H_B C,H_C) + \tkzMarkRightAngles[fill=gray!20, + opacity=.5](A,H_A,C B,H_B,A C,H_C,A) + \tkzDrawPolygon[fill=teal!20,opacity=.3](A,B,C) + \tkzDrawPoints(A,B,C) + \tkzDrawPoints[new](H_A,H_B,H_C) + \tkzDrawPolygon[new,fill=orange!20, + opacity=.3](H_A,H_B,H_C) + \tkzDrawPoint(a) + \tkzLabelPoints(C) + \tkzLabelPoints[left](B) + \tkzLabelPoints[above](A) + \tkzLabelPoints[new](H_A) + \tkzLabelPoints[new,above left](H_C) + \tkzLabelPoints[new,above right](H_B,H) +\end{tikzpicture} +\end{tkzexample} + +\subsection{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] +\begin{tikzpicture}[scale=1.25] \tkzDefPoint(0,0){A} \tkzDefPoint(3,0){B} \tkzDefPoint(0.5,2.5){C} @@ -334,18 +381,20 @@ The Feuerbach triangle is the triangle formed by the three points of tangency of 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) + \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) + \tkzDrawPolygon[new](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) + 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}} +\subsection{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. } @@ -355,20 +404,22 @@ The tangential triangle is the triangle $T_aT_bT_c$ formed by the lines tangent \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) + \tkzDrawPolygon(A,B,C) + \tkzDrawPolygon[new](T_a,T_b,T_c) + \tkzDrawPoints(A,B,C) + \tkzDrawPoints[new](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) + \tkzLabelPoints(A,B,C) + \tkzLabelPoints[new](T_a,T_b,T_c) \end{tikzpicture} \end{tkzexample} -\subsubsection{Option \tkzname{euler}} +\subsection{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.} \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture}[rotate=90,scale=1.25] @@ -383,9 +434,9 @@ The Euler triangle of a triangle $ABC$ is the triangle $E_AE_BE_C$ whose vertice \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) + \tkzDrawPolygon(A,B,C) \tkzDrawCircle(N,E_A) - \tkzDrawSegments[blue](A,H_A B,H_B C,H_C) + \tkzDrawSegments[new](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) @@ -395,9 +446,52 @@ The Euler triangle of a triangle $ABC$ is the triangle $E_AE_BE_C$ whose vertice \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) + \tkzDrawPolygon[color=cyan](M_A,M_B,M_C) \end{tikzpicture} \end{tkzexample} +\subsection{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} + \tkzDefSpcTriangle[euler,name=E](A,B,C){a,b,c} + \tkzDefSpcTriangle[orthic,name=H](A,B,C){a,b,c} + \tkzDefExCircle(A,B,C) \tkzGetPoint{I} \tkzGetLength{rI} + \tkzDefExCircle(C,A,B) \tkzGetPoint{J} \tkzGetLength{rJ} + \tkzDefExCircle(B,C,A) \tkzGetPoint{K} \tkzGetLength{rK} + \tkzDrawPoints[orange](I,J,K) + \tkzLabelPoints[font=\scriptsize](A,B,C,I,J,K) + \tkzClipBB + \tkzInterLC[R](I,C)(I,\rI) \tkzGetSecondPoint{Fc} + \tkzInterLC[R](J,B)(J,\rJ) \tkzGetSecondPoint{Fb} + \tkzInterLC[R](K,A)(K,\rK) \tkzGetSecondPoint{Fa} + \tkzDrawLines[add=1.5 and 1.5](A,B A,C B,C) + \tkzDrawCircle[euler,orange](A,B,C) \tkzGetPoint{E} + \tkzDrawSegments[orange](E,I E,J E,K) + \tkzDrawSegments[dashed](A,Ha B,Hb C,Hc) + \tkzDrawCircles[R](J,{\rJ} I,{\rI} K,{\rK}) + \tkzDrawPoints(A,B,C) + \tkzDrawPoints[orange](E,I,J,K,Ha,Hb,Hc,Ea,Eb,Ec,Fa,Fb,Fc) + \tkzLabelPoints[font=\scriptsize](E,Ea,Eb,Ec,Ha,Hb,Hc,Fa,Fb,Fc) + \end{tikzpicture} +\end{tkzexample} + +\subsection{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] +\begin{tikzpicture} +\tkzDefPoint(0,0){A} +\tkzDefPoint(5,0){B} +\tkzDefPoint(.75,4){C} +\tkzDefTriangleCenter[symmedian](A,B,C)\tkzGetPoint{K} +\tkzDefSpcTriangle[symmedial,name=K_](A,B,C){A,B,C} +\tkzDrawPolygon(A,B,C) +\tkzDrawSegments[new](A,K_A B,K_B C,K_C) +\tkzDrawPoints(A,B,C,K,K_A,K_B,K_C) +\tkzLabelPoints[font=\scriptsize](A,B,C,K,K_A,K_B,K_C) +\end{tikzpicture} +\end{tkzexample} + \endinput
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