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authorNorbert Preining <norbert@preining.info>2022-01-05 03:00:57 +0000
committerNorbert Preining <norbert@preining.info>2022-01-05 03:00:57 +0000
commitc03c0d515f1bfbd111d40e78805db4696e080fb8 (patch)
tree4d72d1df66079b3937776da41a78d71189719dad /macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-pointsSpc.tex
parent1903b598f8dda613f5c584bb18ccfd2fd8c278bb (diff)
CTAN sync 202201050300
Diffstat (limited to 'macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-pointsSpc.tex')
-rw-r--r--macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-pointsSpc.tex746
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diff --git a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-pointsSpc.tex b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-pointsSpc.tex
index f2bdf788ce..2949a6a262 100644
--- a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-pointsSpc.tex
+++ b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-pointsSpc.tex
@@ -1,33 +1,323 @@
-\section{Special points relating to a triangle}
+Now that the fixed points are defined, we can with their references using macros from the package or macros that you will create get new points. The calculations may not be apparent but they are usually done by the package.
+Vous aurez peut-être besoin d'utiliser certains constantes mathématiques, voici la liste des constantes définies par le package.
-\subsection{Triangle center: \tkzcname{tkzDefTriangleCenter}}
+\section{Auxiliary tools}
+\subsection{Constants}
-This macro allows you to define the center of a triangle.
+\tkzname{\tkznameofpack} knows some constants, here is the list:
+\begin{tkzltxexample}[]
+ \def\tkzPhi{1.618034}
+ \def\tkzInvPhi{0.618034}
+ \def\tkzSqrtPhi{1.27202}
+ \def\tkzSqrTwo{1.414213}
+ \def\tkzSqrThree{1.7320508}
+ \def\tkzSqrFive{2.2360679}
+ \def\tkzSqrTwobyTwo{0.7071065}
+ \def\tkzPi{3.1415926}
+ \def\tkzEuler{2.71828182}
+\end{tkzltxexample}
+\subsection{New point by calculation }
-\begin{NewMacroBox}{tkzDefTriangleCenter}{\oarg{local options}\parg{A,B,C}}%
-\tkzHandBomb\ Be careful, the arguments are lists of three points. This macro is used in conjunction with \tkzcname{tkzGetPoint} to get the center you are looking for. You can use \tkzname{tkzPointResult} if it is not necessary to keep the results.
+When a macro of \tkzname{tkznameofpack} creates a new point, it is stored internally with the reference \tkzname{tkzPointResult}. You can assign your own reference to it. This is done with the macro \tkzcname{tkzGetPoint}. A new reference is created, your choice of reference must be placed between braces.
-\medskip
+\begin{NewMacroBox}{tkzGetPoint}{\marg{ref}}%
+If the result is in \tkzname{tkzPointResult}, you can access it with \tkzcname{tkzGetPoint}.
+
+ \medskip
+\begin{tabular}{lll}%
+\toprule
+arguments & default & example \\
+\midrule
+\TAline{ref}{no default}{ \tkzcname{tkzGetPoint\{M\} } see the next example}
+\end{tabular}
+\end{NewMacroBox}
+
+Sometimes you need to get two points. It's possible with
+
+\begin{NewMacroBox}{tkzGetPoints}{\marg{ref1}\marg{ref2}}%
+The result is in \tkzname{tkzPointFirstResult} and \tkzname{tkzPointSecondResult}.
+
+ \medskip
+\begin{tabular}{lll}%
+\toprule
+arguments & default & example \\
+\midrule
+\TAline{\{ref1,ref2\}}{no default}{ \tkzcname{tkzGetPoints\{M,N\} } It's the case with \tkzcname{tkzInterCC}}
+\end{tabular}
+\end{NewMacroBox}
+
+If you need only the first or the second point you can also use :
+
+\begin{NewMacroBox}{tkzGetFirstPoint}{\marg{ref1}}%
+
+ \medskip
\begin{tabular}{lll}%
\toprule
+arguments & default & example \\
+\midrule
+\TAline{ref1}{no default}{ \tkzcname{tkzGetFirstPoint\{M\} }}
+\end{tabular}
+\end{NewMacroBox}
+
+\begin{NewMacroBox}{tkzGetSecondPoint}{\marg{ref2}}%
+
+ \medskip
+\begin{tabular}{lll}%
+\toprule
+arguments & default & example \\
+\midrule
+\TAline{ref2}{no default}{ \tkzcname{tkzGetSecondPoint\{M\} }}
+\end{tabular}
+\end{NewMacroBox}
+
+Parfois les résultats consistent en un point et une dimension. Vous obtenez le point avec \tkzcname{tkzGetPoint} et la dimension avec \tkzcname{tkzGetLength}.
+
+\begin{NewMacroBox}{tkzGetLength}{\marg{name of a macro}}%
+
+ \medskip
+\begin{tabular}{lll}%
+\toprule
+arguments & default & example \\
+\midrule
+\TAline{name of a macro}{no default}{ \tkzcname{tkzGetLength\{rAB\} \tkzcname{rAB} gives the length in cm}}
+\end{tabular}
+\end{NewMacroBox}
+
+%\tkzcname{tkzCalcLength}(A,B) After \tkzcname{tkzGetLength\{dAB\}} \tkzcname{dAB} gives $AB$ in cm}
+
+
+\section{Special points}
+Here are some special points.
+%<--------------------------------------------------------------------------->
+\subsection{Middle of a segment \tkzcname{tkzDefMidPoint}}
+It is a question of determining the middle of a segment.
+
+\begin{NewMacroBox}{tkzDefMidPoint}{\parg{pt1,pt2}}%
+The result is in \tkzname{tkzPointResult}. We can access it with \tkzcname{tkzGetPoint}.
+
+ \medskip
+\begin{tabular}{lll}%
+\toprule
+arguments & default & definition \\
+\midrule
+\TAline{(pt1,pt2)}{no default}{pt1 and pt2 are two points}
+\end{tabular}
+\end{NewMacroBox}
+
+\subsubsection{Use of \tkzcname{tkzDefMidPoint}}
+Review the use of \tkzcname{tkzDefPoint}.
+\begin{tkzexample}[latex=7cm,small]
+\begin{tikzpicture}[scale=1]
+ \tkzDefPoint(2,3){A}
+ \tkzDefPoint(6,2){B}
+ \tkzDefMidPoint(A,B)
+ \tkzGetPoint{M}
+ \tkzDrawSegment(A,B)
+ \tkzDrawPoints(A,B,M)
+ \tkzLabelPoints[below](A,B,M)
+\end{tikzpicture}
+\end{tkzexample}
+
+\subsection{Barycentric coordinates }
+
+$pt_1$, $pt_2$, \dots, $pt_n$ being $n$ points, they define $n$ vectors $\overrightarrow{v_1}$, $\overrightarrow{v_2}$, \dots, $\overrightarrow{v_n}$ with the origin of the referential as the common endpoint. $\alpha_1$, $\alpha_2$,
+\dots $\alpha_n$ are $n$ numbers, the vector obtained by:
+\begin{align*}
+ \frac{\alpha_1 \overrightarrow{v_1} + \alpha_2 \overrightarrow{v_2} + \cdots + \alpha_n \overrightarrow{v_n}}{\alpha_1
+ + \alpha_2 + \cdots + \alpha_n}
+\end{align*}
+defines a single point.
+
+\begin{NewMacroBox}{tkzDefBarycentricPoint}{\parg{pt1=$\alpha_1$,pt2=$\alpha_2$,\dots}}%
+\begin{tabular}{lll}%
arguments & default & definition \\
+\midrule
+\TAline{(pt1=$\alpha_1$,pt2=$\alpha_2$,\dots)}{no default}{Each point has a assigned weight}
+\bottomrule
+\end{tabular}
+
+\medskip
+You need at least two points. Result in \tkzname{tkzPointResult}.
+\end{NewMacroBox}
+
+
+\subsubsection{Using \tkzcname{tkzDefBarycentricPoint} with two points}
+In the following example, we obtain the barycentre of points $A$ and $B$ with coefficients $1$ and $2$, in other words:
+\[
+ \overrightarrow{AI}= \frac{2}{3}\overrightarrow{AB}
+\]
+
+\begin{tkzexample}[latex=7cm,small]
+\begin{tikzpicture}
+ \tkzDefPoint(2,3){A}
+ \tkzDefShiftPointCoord[2,3](30:4){B}
+ \tkzDefBarycentricPoint(A=1,B=2)
+ \tkzGetPoint{G}
+ \tkzDrawLine(A,B)
+ \tkzDrawPoints(A,B,G)
+ \tkzLabelPoints(A,B,G)
+\end{tikzpicture}
+\end{tkzexample}
+
+\subsubsection{Using \tkzcname{tkzDefBarycentricPoint} with three points}
+This time $M$ is simply the center of gravity of the triangle.
+
+ For reasons of simplification and homogeneity, there is also \tkzcname{tkzCentroid}.
+\begin{tkzexample}[latex=7cm,small]
+\begin{tikzpicture}[scale=.8]
+ \tkzDefPoints{2/1/A,5/3/B,0/6/C}
+ \tkzDefBarycentricPoint(A=1,B=1,C=1)
+ \tkzGetPoint{G}
+ \tkzDefMidPoint(A,B) \tkzGetPoint{C'}
+ \tkzDefMidPoint(A,C) \tkzGetPoint{B'}
+ \tkzDefMidPoint(C,B) \tkzGetPoint{A'}
+ \tkzDrawPolygon(A,B,C)
+ \tkzDrawLines[add=0 and 1,new](A,G B,G C,G)
+ \tkzLabelPoint(G){$G$}
+ \tkzDrawPoints[new](A',B',C',G)
+ \tkzDrawPoints(A,B,C)
+ \tkzAutoLabelPoints[center=G](A,B,C)
+ \tkzAutoLabelPoints[center=G,above right](A',B',C')
+\end{tikzpicture}
+\end{tkzexample}
+
+\subsection{Golden ration}
+From Wikipedia : In mathematics, two quantities are in the golden ratio if their ratio is the same as the ratio of their sum to the larger of the two quantities. Expressed algebraically, for quantities $a$, $b$ $a > b > 0$ $a+b$ is to $a$ as $a$ is to $b$.
+
+$ \frac{a+b}{a} = \frac{a}{b} = \phi = \frac{1 + \sqrt{5}}{2}$
+
+
+One of the two solutions to the equation $x^2 - x - 1 = 0$
+is the golden ratio $\phi$, $\phi = \frac{1 + \sqrt{5}}{2}$.
+
+\begin{NewMacroBox}{tkzDefGoldenRatio}{\parg{pt1,pt2}}%
+\begin{tabular}{lll}%
+arguments & default & example \\
+\midrule
+\TAline{(pt1,pt2)}{no default}{\tkzcname{tkzDefGoldenRatio(A,C)} \tkzcname{tkzGetPoint}\{B\}}
+\bottomrule
+\end{tabular}
+
+\medskip
+$AB=a$, $BC=b$ and $\frac{AC}{AB} = \frac{AB}{BC} =\phi$
+\end{NewMacroBox}
+
+\subsection{Use the golden ratio to divide a line segment}
+\begin{tkzexample}[latex=7cm,small]
+\begin{tikzpicture}
+ \tkzDefPoints{0/0/A,6/0/C}
+ \tkzDefMidPoint(A,C) \tkzGetPoint{I}
+ %\tkzDefPointWith[linear,K=\tkzInvPhi](A,C)
+ \tkzDefGoldenRatio(A,C) \tkzGetPoint{B}
+ \tkzDrawSegments(A,C)
+ \tkzDrawPoints(A,B,C)
+ \tkzLabelPoints(A,B,C)
+\end{tikzpicture}
+\end{tkzexample}
+
+
+\subsection{Internal Similitude Center}
+The centres of the two homotheties in which two circles correspond are called external and internal centres of similitude.
+
+\begin{tkzexample}[vbox,small]
+\begin{tikzpicture}[rotate=30]
+ \tkzDefPoints{0/0/O,4/-5/A}
+ \tkzDefPoints{3/0/x,5/-5/y}
+ \pgfmathsetmacro\R{3}\pgfmathsetmacro\r{1}
+ \tkzDefIntSimilitudeCenter[R](O,\R)(A,\r) \tkzGetPoint{I}
+ \tkzDefExtSimilitudeCenter[R](O,\R)(A,\r) \tkzGetPoint{J}
+ \tkzDefTangent[from with R= I](O,3) \tkzGetPoints{D}{E}
+ \tkzDefTangent[from with R= I](A,1) \tkzGetPoints{D'}{E'}
+ \tkzDefTangent[from with R= J](O,3) \tkzGetPoints{F}{G}
+ \tkzDefTangent[from with R= J](A,1) \tkzGetPoints{F'}{G'}
+ \tkzDrawCircles(O,x A,y) \tkzDrawCircles[R](O,3 A,1)
+ \tkzDrawSegments[add = .5 and .5,new](D,D' E,E')
+ \tkzDrawSegments[add= 0 and 0.25,new](J,F J,G)
+ \tkzDrawPoints(O,A,I,J,D,E,F,G,D',E',F',G')
+ \tkzLabelPoints[font=\scriptsize](O,A,I,J,D,E,F,G,D',E',F',G')
+\end{tikzpicture}
+\end{tkzexample}
+
+You can \tkzcname{tkzDefBarycentricPoint} to find a homothetic center
+
+|\tkzDefBarycentricPoint(O=\r,A=\R) \tkzGetPoint{I}| \\
+|\tkzDefBarycentricPoint(O={-\r},A=\R) \tkzGetPoint{J}|
+
+%<---------------------------------------------------------------------->
+\subsection{ Equidistant points}
+%<---------------------------------------------------------------------->
+\subsubsection{\tkzcname{tkzDefEquiPoints}}
+\begin{NewMacroBox}{tkzDefEquiPoints}{\oarg{local options}\parg{pt1,pt2}}%
+\begin{tabular}{lll}%
+arguments & default & definition \\
+\midrule
+\TAline{(pt1,pt2)}{no default}{unordered list of two items}
+\end{tabular}
+
+\begin{tabular}{lll}%
+options & default & definition \\
+\midrule
+\TOline{dist} {2 (cm)} {half the distance between the two points}
+\TOline{from=pt} {no default} {reference point}
+\TOline{show} {false} {if true displays compass traces}
+\TOline{/compass/delta} {0} {compass trace size }
+\end{tabular}
+\end{NewMacroBox}
+
+This macro makes it possible to obtain two points on a straight line equidistant from a given point.
+\subsubsection{Using \tkzcname{tkzDefEquiPoints} with options}
+\begin{tkzexample}[latex=7cm,small]
+\begin{tikzpicture}
+ \tkzSetUpCompass[color=purple,line width=1pt]
+ \tkzDefPoints{0/1/A,5/2/B,3/4/C}
+ \tkzDefEquiPoints[from=C,dist=1,show,
+ /tkzcompass/delta=20](A,B)
+ \tkzGetPoints{E}{H}
+ \tkzDrawLines[color=blue](C,E C,H A,B)
+ \tkzDrawPoints[color=blue](A,B,C)
+ \tkzDrawPoints[color=red](E,H)
+ \tkzLabelPoints(E,H)
+ \tkzLabelPoints[color=blue](A,B,C)
+\end{tikzpicture}
+\end{tkzexample}
+
+\newpage
+\section{Special points relating to a triangle}
+
+\subsection{Triangle center: \tkzcname{tkzDefTriangleCenter}}
+
+\begin{NewMacroBox}{tkzDefTriangleCenter}{\oarg{local options}\parg{A,B,C}}%
+\tkzHandBomb\ This macro allows you to define the center of a triangle.. Be careful, the arguments are lists of three points. This macro is used in conjunction with \tkzcname{tkzGetPoint} to get the center you are looking for.
+
+ You can use \tkzname{tkzPointResult} if it is not necessary to keep the results.
+
+\medskip
+\begin{tabular}{lll}%
+\toprule
+arguments & default & example \\
\midrule
-\TAline{(pt1,pt2,pt3)}{no default}{three points}
+\TAline{(pt1,pt2,pt3)}{no default}{ \tkzcname{tkzDefTriangleCenter[ortho](B,C,A)}}
\midrule
options & default & definition \\
\midrule
-\TOline{ortho} {circum}{intersection of the altitudes of a triangle}
-\TOline{centroid} {circum}{centre of gravity. Intersection of the medians }
+\TOline{ortho} {circum}{intersection of the altitudes}
+\TOline{orthic} {circum}{\dots}
+\TOline{centroid} {circum}{intersection of the medians}
+\TOline{median} {circum}{ \dots }
\TOline{circum}{circum}{circle center circumscribed}
\TOline{in} {circum}{center of the circle inscribed in a triangle }
+\TOline{in} {circum}{intersection of the bisectors}
\TOline{ex} {circum}{center of a circle exinscribed to a triangle }
\TOline{euler}{circum}{center of Euler's circle }
-\TOline{symmedian} {circum}{Lemoine's point or symmedian centre or Grebe's point }
-\TOline{spieker} {circum}{Spieker Circle Center}
+\TOline{gergonne}{circum}{defined with the Contact triangle}
+\TOline{symmedian} {circum}{Lemoine's point or symmedian center or Grebe's point }
+\TOline{lemoine} {circum}{ \dots}
+\TOline{grebe} {circum}{ \dots}
+\TOline{spieker} {circum}{Spieker circle center}
\TOline{nagel}{circum}{Nagel Center}
-\TOline{mittenpunkt} {circum}{also called the middlespoint}
+\TOline{mittenpunkt} {circum}{Or middlespoint}
\TOline{feuerbach}{circum}{Feuerbach Point}
\end{tabular}
@@ -36,35 +326,33 @@ options & default & definition \\
\subsubsection{Option \tkzname{ortho} or \tkzname{orthic}}
The intersection $H$ of the three altitudes of a triangle is called the orthocenter.
-\begin{tkzexample}[latex=5cm,small]
+\begin{tkzexample}[latex=6cm,small]
\begin{tikzpicture}
\tkzDefPoint(0,0){A}
\tkzDefPoint(5,1){B}
\tkzDefPoint(1,4){C}
- \tkzClipPolygon(A,B,C)
\tkzDefTriangleCenter[ortho](B,C,A)
- \tkzGetPoint{H}
+ \tkzGetPoint{H}
\tkzDefSpcTriangle[orthic,name=H](A,B,C){a,b,c}
- \tkzDrawPolygon[color=blue](A,B,C)
+ \tkzDrawPolygon(A,B,C)
+ \tkzDrawSegments[new](A,Ha B,Hb C,Hc)
\tkzDrawPoints(A,B,C,H)
- \tkzDrawLines[add=0 and 1](A,Ha B,Hb C,Hc)
\tkzLabelPoint(H){$H$}
\tkzAutoLabelPoints[center=H](A,B,C)
- \tkzMarkRightAngles(A,Ha,B B,Hb,C C,Hc,A)
+ \tkzMarkRightAngles(A,Ha,B B,Hb,C C,Hc,A)
\end{tikzpicture}
\end{tkzexample}
\subsubsection{Option \tkzname{centroid}}
-\begin{tkzexample}[latex=5cm,small]
+\begin{tkzexample}[latex=6cm,small]
\begin{tikzpicture}[scale=.75]
- \tkzDefPoints{-1/1/A,5/1/B}
- \tkzDefEquilateral(A,B)
- \tkzGetPoint{C}
+ \tkzDefPoints{0/0/A,5/0/B,1/4/C}
\tkzDefTriangleCenter[centroid](A,B,C)
- \tkzGetPoint{G}
- \tkzDrawPolygon[color=brown](A,B,C)
+ \tkzGetPoint{G}
+ \tkzDrawPolygon(A,B,C)
+ \tkzDrawLines[add = 0 and 2/3,new](A,G B,G C,G)
\tkzDrawPoints(A,B,C,G)
- \tkzDrawLines[add = 0 and 2/3](A,G B,G C,G)
+ \tkzLabelPoint(G){$G$}
\end{tikzpicture}
\end{tkzexample}
@@ -73,97 +361,92 @@ options & default & definition \\
\begin{tikzpicture}
\tkzDefPoints{0/1/A,3/2/B,1/4/C}
\tkzDefTriangleCenter[circum](A,B,C)
- \tkzGetPoint{G}
- \tkzDrawPolygon[color=brown](A,B,C)
- \tkzDrawCircle(G,A)
- \tkzDrawPoints(A,B,C,G)
- \end{tikzpicture}
+ \tkzGetPoint{O}
+ \tkzDrawPolygon(A,B,C)
+ \tkzDrawCircle(O,A)
+ \tkzDrawPoints(A,B,C,O)
+ \tkzLabelPoint(O){$O$}
+\end{tikzpicture}
\end{tkzexample}
\subsubsection{Option \tkzname{in}}
In geometry, the incircle or inscribed circle of a triangle is the largest circle contained in the triangle; it touches (is tangent to) the three sides. The center of the incircle is a triangle center called the triangle's incenter.
-The center of the incircle, called the incenter, can be found as the intersection of the three internal angle bisectors. The center of an excircle is the intersection of the internal bisector of one angle (at vertex $A$, for example) and the external bisectors of the other two. The center of this excircle is called the excenter relative to the vertex $A$, or the excenter of $A$. Because the internal bisector of an angle is perpendicular to its external bisector, it follows that the center of the incircle together with the three excircle centers form an orthocentric system.(\url{https://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle})
+The center of the incircle, called the incenter, can be found as the intersection of the three internal angle bisectors. The center of an excircle is the intersection of the internal bisector of one angle (at vertex $A$, for example) and the external bisectors of the other two. The center of this excircle is called the excenter relative to the vertex $A$, or the excenter of $A$. Because the internal bisector of an angle is perpendicular to its external bisector, it follows that the center of the incircle together with the three excircle centers form an orthocentric system.\\
+(\url{https://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle})
\medskip
We get the centre of the inscribed circle of the triangle. The result is of course in \tkzname{tkzPointResult}. We can retrieve it with \tkzcname{tkzGetPoint}.
-\begin{tkzexample}[latex=6cm,small]
+\begin{tkzexample}[latex=8cm,small]
\begin{tikzpicture}
- \tkzDefPoints{0/1/A,3/2/B,1/4/C}
- \tkzDefTriangleCenter[in](A,B,C)\tkzGetPoint{I}
- \tkzDefPointBy[projection=onto A--C](I)
- \tkzGetPoint{Ib}
- \tkzDrawPolygon[color=blue](A,B,C)
- \tkzDrawPoints(A,B,C,I)
- \tkzDrawLines[add = 0 and 2/3](A,I B,I C,I)
- \tkzDrawCircle(I,Ib)
+\tkzDefPoints{0/0/A,6/0/B,0.8/4/C}
+\tkzDefTriangleCenter[in](A,B,C)
+ \tkzGetPoint{I}
+\tkzDrawLines(A,B B,C C,A)
+\tkzDrawCircle[in](A,B,C)
+\tkzDrawPoint[red](I)
+\tkzDrawPoints(A,B,C)
+\tkzLabelPoint(I){$I$}
\end{tikzpicture}
\end{tkzexample}
\subsubsection{Option \tkzname{ex}}
-An excircle or escribed circle of the triangle is a circle lying outside the triangle, tangent to one of its sides and tangent to the extensions of the other two. Every triangle has three distinct excircles, each tangent to one of the triangle's sides.
+An excircle or escribed circle of the triangle is a circle lying outside the triangle, tangent to one of its sides and tangent to the extensions of the other two. Every triangle has three distinct excircles, each tangent to one of the triangle's sides.\\
(\url{https://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle})
We get the centre of an inscribed circle of the triangle. The result is of course in \tkzname{tkzPointResult}. We can retrieve it with \tkzcname{tkzGetPoint}.
-\begin{tkzexample}[latex=8cm,small]
- \begin{tikzpicture}[scale=.5]
- \tkzDefPoints{0/1/A,3/2/B,1/4/C}
- \tkzDefTriangleCenter[ex](B,C,A)
- \tkzGetPoint{J_c}
- \tkzDefPointBy[projection=onto A--B](J_c)
- \tkzGetPoint{Tc}
- %or
- % \tkzDefCircle[ex](B,C,A)
- % \tkzGetFirstPoint{J_c}
- % \tkzGetSecondPoint{Tc}
- \tkzDrawPolygon[color=blue](A,B,C)
- \tkzDrawPoints(A,B,C,J_c)
- \tkzDrawCircle[red](J_c,Tc)
- \tkzDrawLines[add=1.5 and 0](A,C B,C)
- \tkzLabelPoints(J_c)
- \end{tikzpicture}
+\begin{tkzexample}[latex=7cm,small]
+\begin{tikzpicture}[scale=.5]
+ \tkzDefPoints{0/1/A,3/2/B,1/4/C}
+ \tkzDefTriangleCenter[ex](B,C,A)
+ \tkzGetPoint{J_c}
+ \tkzDefPointBy[projection=onto A--B](J_c)
+ \tkzGetPoint{Tc}
+ \tkzDrawPolygon(A,B,C)
+ \tkzDrawCircle[new](J_c,Tc)
+ \tkzDrawLines[add=1.5 and 0](A,C B,C)
+ \tkzDrawPoints(A,B,C,J_c)
+ \tkzLabelPoints(J_c)
+\end{tikzpicture}
\end{tkzexample}
\subsubsection{Option \tkzname{euler}}
-This macro allows to obtain the center of the circle of the nine points or euler's circle or Feuerbach's circle.
-The nine-point circle, also called Euler's circle or the Feuerbach circle, is the circle that passes through the perpendicular feet $H_A$, $H_B$, and $H_C$ dropped from the vertices of any reference triangle $ABC$ on the sides opposite them. Euler showed in 1765 that it also passes through the midpoints $M_A$, $M_B$, $M_C$ of the sides of $ABC$. By Feuerbach's theorem, the nine-point circle also passes through the midpoints $E_A$, $E_B$, and $E_C$ of the segments that join the vertices and the orthocenter $H$. These points are commonly referred to as the Euler points. (\url{http://mathworld.wolfram.com/Nine-PointCircle.html})
+This macro allows to obtain the center of the circle of the nine points or euler's circle or Feuerbach's circle. The nine-point circle, also called Euler's circle or the Feuerbach circle, is the circle that passes through the perpendicular feet $H_A$, $H_B$, and $H_C$ dropped from the vertices of any reference triangle $ABC$ on the sides opposite them. Euler showed in 1765 that it also passes through the midpoints $M_A$, $M_B$, $M_C$ of the sides of $ABC$. By Feuerbach's theorem, the nine-point circle also passes through the midpoints $E_A$, $E_B$, and $E_C$ of the segments that join the vertices and the orthocenter $H$. These points are commonly referred to as the Euler points.\\ (\url{https://mathworld.wolfram.com/Nine-PointCircle.html})
-\begin{tkzexample}[latex=7cm,small]
-\begin{tikzpicture}[scale=1]
+\begin{tkzexample}[latex=5cm,small]
+\begin{tikzpicture}[scale=1,rotate=90]
\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}
+ \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)
+ \tkzDrawPolygon(A,B,C)
+ \tkzDrawCircle[new](N,E_A)
+ \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)
+ \tkzDrawPoints[new](M_A,M_B,M_C)
+ \tkzDrawPoints( 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)
+ 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)
+ color=blue,line width=1pt](B,E_B E_B,H)
\end{tikzpicture}
\end{tkzexample}
\subsubsection{Option \tkzname{symmedian}}
+The point of concurrence $K$ of the symmedians, sometimes also called the Lemoine point (in England and France) or the Grebe point (in Germany).\\
+\href{https://mathworld.wolfram.com/SymmedianPoint.html}{Weisstein, Eric W. "Symmedian Point." From MathWorld--A Wolfram Web Resource.}
+
\begin{tkzexample}[latex=6cm,small]
\begin{tikzpicture}
\tkzDefPoint(0,0){A}
@@ -174,39 +457,83 @@ The nine-point circle, also called Euler's circle or the Feuerbach circle, is th
\tkzDefTriangleCenter[in](A,B,C)\tkzGetPoint{I}
\tkzDefSpcTriangle[centroid,name=M](A,B,C){a,b,c}
\tkzDefSpcTriangle[incentral,name=I](A,B,C){a,b,c}
- \tkzDrawPolygon[color=blue](A,B,C)
- \tkzDrawLines[add = 0 and 2/3,blue](A,K B,K C,K)
- \tkzDrawSegments[red,dashed](A,Ma B,Mb C,Mc)
- \tkzDrawSegments[orange,dashed](A,Ia B,Ib C,Ic)
+ \tkzDrawPolygon(A,B,C)
+ \tkzDrawLines[add = 0 and 2/3,new](A,K B,K C,K)
+ \tkzDrawSegments[color=cyan](A,Ma B,Mb C,Mc)
+ \tkzDrawSegments[color=green](A,Ia B,Ib C,Ic)
\tkzDrawLine[add=2 and 2](G,I)
\tkzDrawPoints(A,B,C,K,G,I)
+ \tkzLabelPoints[font=\scriptsize](A,B,K,G,I)
+ \tkzLabelPoints[above,font=\scriptsize](C)
\end{tikzpicture}
\end{tkzexample}
+\subsubsection{Option \tkzname{spieker}}
+The Spieker center is the center $Sp$ of the Spieker circle, i.e., the incenter of the medial triangle of a reference triangle.\\
+\href{https://mathworld.wolfram.com/SpiekerCenter.html}{Weisstein, Eric W. "Spieker Center." From MathWorld--A Wolfram Web Resource. }
-\subsubsection{Option \tkzname{nagel}}
-Let $Ta$ be the point at which the excircle with center $Ja$ meets the side $BC$ of a triangle $ABC$, and define $Tb$ and $Tc$ similarly. Then the lines $ATa$, $BTb$, and $CTc$ concur in the Nagel point $Na$.
-\href{http://mathworld.wolfram.com/NagelPoint.html}{Weisstein, Eric W. "Nagel point." From MathWorld--A Wolfram Web Resource. }
+\begin{tkzexample}[latex=8cm,small]
+\begin{tikzpicture}
+ \tkzDefPoints{0/0/A,6/0/B,5/5/C}
+ \tkzDefSpcTriangle[medial](A,B,C){Ma,Mb,Mc}
+ \tkzDefTriangleCenter[centroid](A,B,C)
+ \tkzGetPoint{G}
+ \tkzDefTriangleCenter[spieker](A,B,C)
+ \tkzGetPoint{Sp}
+ \tkzDrawPolygon[](A,B,C)
+ \tkzDrawPolygon[new](Ma,Mb,Mc)
+ \tkzDrawCircle[in](Ma,Mb,Mc)
+ \tkzDrawPoints(B,C,A,Sp,Ma,Mb,Mc)
+ \tkzAutoLabelPoints[center=G,dist=.3](Ma,Mb,Mc)
+ \tkzLabelPoints[right](Sp)
+ \tkzAutoLabelPoints[center=G](A,B,C)
+\end{tikzpicture}
+\end{tkzexample}
+
+\subsubsection{Option \tkzname{gergonne}}
+The Gergonne Point is the point of concurrency which results from connecting the vertices of a triangle to the opposite points of tangency of the triangle's incircle.
+(Joseph Gergonne French mathematician )
\begin{tkzexample}[latex=8cm,small]
+\begin{tikzpicture}
+\tkzDefPoints{0/0/B,3.6/0/C,2.8/4/A}
+\tkzDefTriangleCenter[gergonne](A,B,C)
+\tkzGetPoint{Ge}
+\tkzDefSpcTriangle[intouch](A,B,C){C_1,C_2,C_3}
+\tkzDrawCircle[in](A,B,C)
+\tkzDrawLines[add=.25 and .25,teal](A,B A,C B,C)
+\tkzDrawSegments[new](A,C_1 B,C_2 C,C_3)
+\tkzDrawPoints(A,...,C,C_1,C_2,C_3)
+\tkzDrawPoints[red](Ge)
+\tkzLabelPoints(A,...,C,C_1,C_2,C_3,Ge)
+\end{tikzpicture}
+\end{tkzexample}
+
+\subsubsection{Option \tkzname{nagel}}
+Let $Ta$ be the point at which the excircle with center $Ja$ meets the side $BC$ of a triangle $ABC$, and define $Tb$ and $Tc$ similarly. Then the lines $ATa$, $BTb$, and $CTc$ concur in the Nagel point $Na$.\\
+\href{https://mathworld.wolfram.com/NagelPoint.html}{Weisstein, Eric W. "Nagel point." From MathWorld--A Wolfram Web Resource. }
+
+
+\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=.5]
\tkzDefPoints{0/0/A,6/0/B,4/6/C}
\tkzDefSpcTriangle[ex](A,B,C){Ja,Jb,Jc}
\tkzDefSpcTriangle[extouch](A,B,C){Ta,Tb,Tc}
- \tkzDrawPoints(Ja,Jb,Jc,Ta,Tb,Tc)
- \tkzLabelPoints(Ja,Jb,Jc,Ta,Tb,Tc)
+ \tkzDefTriangleCenter[nagel](A,B,C)
+ \tkzGetPoint{Na}
\tkzDrawPolygon[blue](A,B,C)
- \tkzDefTriangleCenter[nagel](A,B,C) \tkzGetPoint{Na}
- \tkzDrawPoints[blue](B,C,A)
- \tkzDrawPoints[red](Na)
- \tkzLabelPoints[blue](B,C,A)
- \tkzLabelPoints[red](Na)
\tkzDrawLines[add=0 and 1](A,Ta B,Tb C,Tc)
- \tkzShowBB\tkzClipBB
+ \tkzDrawPoints[new](Ja,Jb,Jc,Ta,Tb,Tc)
+ \tkzClipBB
\tkzDrawLines[add=1 and 1,dashed](A,B B,C C,A)
- \tkzDrawCircles[ex,gray](A,B,C C,A,B B,C,A)
- \tkzDrawSegments[dashed](Ja,Ta Jb,Tb Jc,Tc)
+ \tkzDrawCircles[ex,new](A,B,C C,A,B B,C,A)
+ \tkzDrawSegments[new,dashed](Ja,Ta Jb,Tb Jc,Tc)
+ \tkzDrawPoints(B,C,A)
+ \tkzDrawPoints[new](Na)
+ \tkzLabelPoints(B,C,A)
+ \tkzLabelPoints[new](Na)
+ \tkzLabelPoints[new](Ja,Jb,Jc,Ta,Tb,Tc)
\tkzMarkRightAngles[fill=gray!20](Ja,Ta,C
Jb,Tb,A Jc,Tc,B)
\end{tikzpicture}
@@ -214,6 +541,11 @@ Let $Ta$ be the point at which the excircle with center $Ja$ meets the side $BC$
\subsubsection{Option \tkzname{mittenpunkt}}
+
+The mittenpunkt (also called the middlespoint) of a triangle $ABC$ is the symmedian point of the excentral triangle, i.e., the point of concurrence M of the lines from the excenters through the corresponding triangle side midpoints.\\
+\href{https://mathworld.wolfram.com/Mittenpunkt.html}{Weisstein, Eric W. "Mittenpunkt." From MathWorld--A Wolfram Web Resource.}
+
+
\begin{tkzexample}[latex=8cm,small]
\begin{tikzpicture}[scale=.5]
\tkzDefPoints{0/0/A,6/0/B,4/6/C}
@@ -222,123 +554,141 @@ Let $Ta$ be the point at which the excircle with center $Ja$ meets the side $BC$
\tkzDefSpcTriangle[extouch](A,B,C){Ta,Tb,Tc}
\tkzDefTriangleCenter[mittenpunkt](A,B,C)
\tkzGetPoint{Mi}
- \tkzDrawPoints(Ma,Mb,Mc,Ja,Jb,Jc)
+ \tkzDrawPoints[new](Ma,Mb,Mc,Ja,Jb,Jc)
\tkzClipBB
\tkzDrawPolygon[blue](A,B,C)
\tkzDrawLines[add=0 and 1](Ja,Ma
Jb,Mb Jc,Mc)
\tkzDrawLines[add=1 and 1](A,B A,C B,C)
- \tkzDrawCircles[gray](Ja,Ta Jb,Tb Jc,Tc)
- \tkzDrawPoints[blue](B,C,A)
- \tkzDrawPoints[red](Mi)
- \tkzLabelPoints[red](Mi)
+ \tkzDrawCircles[new](Ja,Ta Jb,Tb Jc,Tc)
+ \tkzDrawPoints(B,C,A)
+ \tkzDrawPoints[new](Mi)
+ \tkzLabelPoints(Mi)
\tkzLabelPoints[left](Mb)
- \tkzLabelPoints(Ma,Mc,Jb,Jc)
+ \tkzLabelPoints[new](Ma,Mc,Jb,Jc)
\tkzLabelPoints[above left](Ja,Jc)
- \tkzShowBB
\end{tikzpicture}
\end{tkzexample}
-%<---------------------------------------------------------------------->
-%<---------------------------------------------------------------------->
-\section{Draw a point}
-\subsubsection{Drawing points \tkzcname{tkzDrawPoint}} \hypertarget{tdrp}{}
-
-\begin{NewMacroBox}{tkzDrawPoint}{\oarg{local options}\parg{name}}%
-\begin{tabular}{lll}%
-arguments & default & definition \\
-\midrule
-\TAline{name of point} {no default} {Only one point name is accepted}
-\bottomrule
-\end{tabular}
-\medskip
-The argument is required. The disc takes the color of the circle, but lighter. It is possible to change everything. The point is a node and therefore it is invariant if the drawing is modified by scaling.
+\subsubsection{Example : relation between \tkzname{gergonne}, \tkzname{centroid} and \tkzname{mittenpunkt}}
-\medskip
-\begin{tabular}{lll}%
-\toprule
-options & default & definition \\
-\midrule
-\TOline{shape} {circle}{Possible \tkzname{cross} or \tkzname{cross out}}
-\TOline{size} {6}{$6 \times$ \tkzcname{pgflinewidth}}
-\TOline{color} {black}{the default color can be changed }
-\bottomrule
-\end{tabular}
+The Gergonne point $Ge$, triangle centroid $G$, and mittenpunkt $M$ are collinear, with GeG/GM=2.
-\medskip
-{We can create other forms such as \tkzname{cross}}
-\end{NewMacroBox}
-
-\subsubsection{Example of point drawings}
-Note that \tkzname{scale} does not affect the shape of the dots. Which is normal. Most of the time, we are satisfied with a single point shape that we can define from the beginning, either with a macro or by modifying a configuration file.
-
-
-\begin{tkzexample}[latex=5cm,small]
- \begin{tikzpicture}[scale=.5]
- \tkzDefPoint(1,3){A}
- \tkzDefPoint(4,1){B}
- \tkzDefPoint(0,0){O}
- \tkzDrawPoint[color=red](A)
- \tkzDrawPoint[fill=blue!20,draw=blue](B)
- \tkzDrawPoint[color=green](O)
- \end{tikzpicture}
+\begin{tkzexample}[vbox,small]
+\begin{tikzpicture}
+\tkzDefPoints{0/0/A,2/2/B,8/0/C}
+\tkzDefTriangleCenter[gergonne](A,B,C) \tkzGetPoint{Ge}
+\tkzDefTriangleCenter[centroid](A,B,C)
+\tkzGetPoint{G}
+\tkzDefTriangleCenter[mittenpunkt](A,B,C)
+\tkzGetPoint{M}
+\tkzDrawLines[add=.25 and .25,teal](A,B A,C B,C)
+\tkzDrawLines[add=.25 and .25,new](Ge,M)
+\tkzDrawPoints(A,...,C)
+\tkzDrawPoints[red,size=2](G,M,Ge)
+\tkzLabelPoints(A,...,C,M,G,Ge)
+\tkzMarkSegment[mark=s||](Ge,G)
+\tkzMarkSegment[mark=s|](G,M)
+\end{tikzpicture}
\end{tkzexample}
-It is possible to draw several points at once but this macro is a little slower than the previous one. Moreover, we have to make do with the same options for all the points.
+\newpage
+\section{Projection of excenters}
+
+\begin{NewMacroBox}{tkzDefProjExcenter}{\oarg{local options}\parg{A,B,C}\parg{a,b,c}\marg{X,Y,Z}}%
+Each excenter has three projections on the sides of the triangle ABC. We can do this with one macro\\ \tkzcname{tkzDefProjExcenter[name=J](A,B,C)(a,b,c)\{Y,Z,X\}}.
-\hypertarget{tdrps}{}
-\begin{NewMacroBox}{tkzDrawPoints}{\oarg{local options}\parg{liste}}%
+\medskip
\begin{tabular}{lll}%
-arguments & default & definition \\
+\toprule
+options & default & definition \\
\midrule
-\TAline{points list}{no default}{example \tkzcname{tkzDrawPoints(A,B,C)}}
+\TOline{name} {no defaut}{used to name the vertices}
\bottomrule
\end{tabular}
-\medskip
\begin{tabular}{lll}%
-options & default & definition \\
+arguments & default & definition \\
\midrule
-\TOline{shape} {circle}{Possible \tkzname{cross} or \tkzname{cross out}}
-\TOline{size} {6}{$6 \times$ \tkzcname{pgflinewidth}}
-\TOline{color} {black}{the default color can be changed }
+\TAline{(pt1=$\alpha_1$,pt2=$\alpha_2$,\dots)}{no default}{Each point has a assigned weight}
\bottomrule
\end{tabular}
\medskip
-\tkzHandBomb\ Beware of the final "s", an oversight leads to cascading errors if you try to draw multiple points. The options are the same as for the previous macro.
\end{NewMacroBox}
-\subsubsection{First example}
+\subsubsection{Excircles}
-\begin{tkzexample}[latex=7cm,small]
-\begin{tikzpicture}
- \tkzDefPoint(1,3){A}
- \tkzDefPoint(4,1){B}
- \tkzDefPoint(0,0){C}
- \tkzDrawPoints[size=6,color=red,
- fill=red!50](A,B,C)
-\end{tikzpicture}
-\end{tkzexample}
-
-\subsubsection{Second example}
-
-\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=.5]
- \tkzDefPoint(2,3){A} \tkzDefPoint(5,-1){B}
- \tkzDefPoint[label=below:$\mathcal{C}$,
- shift={(2,3)}](-30:5.5){E}
- \begin{scope}[shift=(A)]
- \tkzDefPoint(30:5){C}
- \end{scope}
- \tkzCalcLength[cm](A,B)\tkzGetLength{rAB}
- \tkzDrawCircle[R](A,\rAB cm)
- \tkzDrawSegment(A,B)
- \tkzDrawPoints(A,B,C)
- \tkzLabelPoints(B,C)
- \tkzLabelPoints[above](A)
-\end{tikzpicture}
+\tkzDefPoints{0/0/A,5/0/B,0.8/4/C}
+\tkzDefSpcTriangle[excentral,name=J](A,B,C){a,b,c}
+\tkzDefSpcTriangle[intouch,name=I](A,B,C){a,b,c}
+\tkzDefProjExcenter[name=J](A,B,C)(a,b,c){X,Y,Z}
+
+\tkzDefCircle[in](A,B,C) \tkzGetPoint{I} \tkzGetSecondPoint{T}
+\tkzDrawCircles[red](Ja,Xa Jb,Yb Jc,Zc)
+\tkzDrawCircle(I,T)
+\tkzDrawPolygon[dashed,color=blue](Ja,Jb,Jc)
+\tkzDrawLines[add=2 and 2,line width=1pt](A,C A,B B,C)
+\tkzDrawSegments(Ja,Xa Ja,Ya Ja,Za
+ Jb,Xb Jb,Yb Jb,Zb
+ Jc,Xc Jc,Yc Jc,Zc
+ I,Ia I,Ib I,Ic)
+\tkzMarkRightAngles[size=.2,fill=gray!15](%
+ Ja,Za,B
+ Ja,Xa,B
+ Ja,Ya,C
+ Jb,Yb,C
+ Jb,Zb,B
+ Jb,Xb,C
+ Jc,Yc,A
+ Jc,Zc,B
+ Jc,Xc,C
+ I,Ia,B
+ I,Ib,C
+ I,Ic,A)
+\tkzDrawSegments[blue](Jc,C Ja,A Jb,B)
+\tkzLabelPoints(Xb,Yc,A,B,C,Xa,Xc,Ya,Yb,Ja,Jb,Jc,I)
+\tkzLabelPoints[above right](Za,Zb,Zc)
+\tkzLabelPoints[below](Ia,Ib,Ic)
+\end{tikzpicture}
+
+\begin{tkzexample}[code only,small]
+ \begin{tikzpicture}[scale=.5]
+ \tkzDefPoints{0/0/A,5/0/B,0.8/4/C}
+ \tkzDefSpcTriangle[excentral,name=J](A,B,C){a,b,c}
+ \tkzDefSpcTriangle[intouch,name=I](A,B,C){a,b,c}
+ \tkzDefProjExcenter[name=J](A,B,C)(a,b,c){X,Y,Z}
+
+ \tkzDefCircle[in](A,B,C) \tkzGetPoint{I} \tkzGetSecondPoint{T}
+ \tkzDrawCircles[red](Ja,Xa Jb,Yb Jc,Zc)
+ \tkzDrawCircle(I,T)
+ \tkzDrawPolygon[dashed,color=blue](Ja,Jb,Jc)
+ \tkzDrawLines[add=2 and 2,line width=1pt](A,C A,B B,C)
+ \tkzDrawSegments(Ja,Xa Ja,Ya Ja,Za
+ Jb,Xb Jb,Yb Jb,Zb
+ Jc,Xc Jc,Yc Jc,Zc
+ I,Ia I,Ib I,Ic)
+ \tkzMarkRightAngles[size=.2,fill=gray!15](%
+ Ja,Za,B
+ Ja,Xa,B
+ Ja,Ya,C
+ Jb,Yb,C
+ Jb,Zb,B
+ Jb,Xb,C
+ Jc,Yc,A
+ Jc,Zc,B
+ Jc,Xc,C
+ I,Ia,B
+ I,Ib,C
+ I,Ic,A)
+ \tkzDrawSegments[blue](Jc,C Ja,A Jb,B)
+ \tkzLabelPoints(Xb,Yc,A,B,C,Xa,Xc,Ya,Yb,Ja,Jb,Jc,I)
+ \tkzLabelPoints[above right](Za,Zb,Zc)
+ \tkzLabelPoints[below](Ia,Ib,Ic)
+ \end{tikzpicture}
\end{tkzexample}
+
\section{Point on line or circle}
\subsection{Point on a line}
@@ -363,13 +713,13 @@ options & default & definition \\
\begin{tkzexample}[latex=9cm,small]
\begin{tikzpicture}
\tkzDefPoints{0/0/A,4/0/B}
- \tkzDrawLine[red](A,B)
- \tkzDefPointOnLine[pos=1.2](A,B)
+ \tkzDefPointOnLine[pos=1.2](A,B)
\tkzGetPoint{P}
- \tkzDefPointOnLine[pos=-0.2](A,B)
+ \tkzDefPointOnLine[pos=-0.2](A,B)
\tkzGetPoint{R}
\tkzDefPointOnLine[pos=0.5](A,B)
\tkzGetPoint{S}
+ \tkzDrawLine[new](A,B)
\tkzDrawPoints(A,B,P)
\tkzLabelPoints(A,B)
\tkzLabelPoint[above](P){pos=$1.2$}
@@ -394,20 +744,20 @@ options & default & definition \\
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}
- \tkzDefPoints{0/0/A,4/0/B,0.8/3/C}
- \tkzDefPointOnCircle[angle=90,center=B,radius=1 cm]
- \tkzGetPoint{I}
- \tkzDefCircle[circum](A,B,C)
- \tkzGetPoint{G} \tkzGetLength{rG}
- \tkzDefPointOnCircle[angle=30,center=G,radius=\rG pt]
- \tkzGetPoint{J}
- \tkzDrawCircle[R,teal](B,1cm)
- \tkzDrawPoint[teal](I)
- \tkzDrawPoints(A,B,C)
- \tkzDrawCircle(G,J)
- \tkzDrawPoints(G,J)
- \tkzDrawPoint[red](J)
- \tkzLabelPoints(G,J)
+\tkzDefPoints{0/0/A,4/0/B,0.8/3/C}
+\tkzDefPointOnCircle[angle=90,center=B,radius=1]
+\tkzGetPoint{I}
+\tkzDefCircle[circum](A,B,C)
+\tkzGetPoint{G} \tkzGetLength{rG}
+\tkzDefPointOnCircle[angle=30,center=G,radius=\rG]
+\tkzGetPoint{J}
+\tkzDrawCircle[R,teal](B,1)
+\tkzDrawPoint[teal](I)
+\tkzDrawPoints(A,B,C)
+\tkzDrawCircle(G,J)
+\tkzDrawPoints(G,J)
+\tkzDrawPoint[red](J)
+\tkzLabelPoints(G,J)
\end{tikzpicture}
\end{tkzexample}
\endinput \ No newline at end of file