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-rw-r--r--Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-triangles.tex428
1 files changed, 261 insertions, 167 deletions
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 \ No newline at end of file