From 31fa8cd73bab8480d38dae3a89ca578f337d6bbd Mon Sep 17 00:00:00 2001 From: Norbert Preining Date: Wed, 27 Mar 2024 03:02:20 +0000 Subject: CTAN sync 202403270302 --- .../doc/latex/TKZdoc-elements-classes-line.tex | 261 ++++++++++++--------- 1 file changed, 156 insertions(+), 105 deletions(-) (limited to 'macros/latex/contrib/tkz/tkz-elements/doc/latex/TKZdoc-elements-classes-line.tex') diff --git a/macros/latex/contrib/tkz/tkz-elements/doc/latex/TKZdoc-elements-classes-line.tex b/macros/latex/contrib/tkz/tkz-elements/doc/latex/TKZdoc-elements-classes-line.tex index 2aa2ce6537..741938eb66 100644 --- a/macros/latex/contrib/tkz/tkz-elements/doc/latex/TKZdoc-elements-classes-line.tex +++ b/macros/latex/contrib/tkz/tkz-elements/doc/latex/TKZdoc-elements-classes-line.tex @@ -5,7 +5,7 @@ \subsection{Attributes of a line} % (fold) \label{sub:attributes_of_a_line} -Writing |L.AB = line: new (z.A,z.B)| creates an object of the class \tkzname{line} (the notation is arbitrary for the moment). Geometrically it is, as much ,the line passing through the points $A$ and $B$ as the segment $[AB]$. Thus we can use the midpoint of |L.AB| which is, of course, the midpoint of the segment $[AB]$. This medium is obtained with |L.AB.mid|. Note that |L.AB.pa = z.A| and |L.AB.pb = z.B|. Finally, if a line $L$ is the result of a method, you can obtain the points with |z.A,z.B = get_points (L)| or with the previous remark. +Writing |L.AB = line: new (z.A,z.B)| creates an object of the class \tkzname{line} (the notation is arbitrary for the moment). Geometrically, it represents both the line passing through the points $A$ and $B$ as the segment $[AB]$. Thus, we can use the midpoint of |L.AB|, which is, of course, the midpoint of the segment $[AB]$. This medium is obtained with |L.AB.mid|. Note that |L.AB.pa = z.A| and |L.AB.pb = z.B|. Finally, if a line $L$ is the result of a method, you can obtain the points with |z.A,z.B = get_points (L)| or with the previous remark. \begin{mybox} Creation |L.AB = line : new ( z.A , z.B ) | @@ -26,15 +26,15 @@ The attributes are : \Iattr{line}{pb} & Second point of the segment & \\ \Iattr{line}{type} & Type is 'line' & |L.AB.type = 'line'| \\ \Iattr{line}{mid} & Middle of the segment& |z.M = L.AB.mid|\\ -\Iattr{line}{slope} & Slope of the line & see (\ref{ssub:example_class_line})\\ -\Iattr{line}{length} &|l = L.AB.length|&see (\ref{sub:transfer_from_lua_to_tex} ; \ref{ssub:example_class_line})\\ -\Iattr{line}{north\_pa} & &See (\ref{ssub:example_class_line}) \\ +\Iattr{line}{slope} & Slope of the line & Refer to (\ref{ssub:example_class_line})\\ +\Iattr{line}{length} &|l = L.AB.length|&Refer to (\ref{sub:transfer_from_lua_to_tex} ; \ref{ssub:example_class_line})\\ +\Iattr{line}{north\_pa} & &Refer to (\ref{ssub:example_class_line}) \\ \Iattr{line}{north\_pb} & &\\ \Iattr{line}{south\_pa} & &\\ -\Iattr{line}{south\_pb} & &See (\ref{ssub:example_class_line}) \\ +\Iattr{line}{south\_pb} & &Refer to (\ref{ssub:example_class_line}) \\ \Iattr{line}{east} & &\\ \Iattr{line}{west} & &\\ -\Iattr{line}{vec} & |V.AB = L.AB.vec|& defines $\overrightarrow{AB}$ See (\ref{sec:class_vector})\\ +\Iattr{line}{vec} & |V.AB = L.AB.vec|& defines $\overrightarrow{AB}$ Refer to (\ref{sec:class_vector})\\ \bottomrule \end{tabular} \egroup @@ -42,7 +42,7 @@ The attributes are : \subsubsection{Example: attributes of class line} % (fold) \label{ssub:example_class_line} \begin{minipage}{.5\textwidth} -\begin{verbatim} +\begin{Verbatim} \begin{tkzelements} scale = .5 z.a = point: new (1, 1) @@ -66,7 +66,7 @@ The attributes are : \tkzLabelSegment[sloped](a,b){ab = \tkzUseLua{len}} \tkzLabelSegment[above=12pt,sloped](a,b){slope of (ab) = \tkzUseLua{sl}} \end{tikzpicture} -\end{verbatim} +\end{Verbatim} \end{minipage} \begin{minipage}{.5\textwidth}\begin{tkzelements} scale = .5 @@ -101,14 +101,14 @@ len = L.ab.length \subsubsection{Method \Imeth{line}{new} and line attributes} \label{ssub:example_line_attributes} -Notation |L| or |L.AB| or |L.euler|. The notation is actually free. +The notation can be |L| or |L.AB| or |L.euler|. The notation is actually free. |L.AB| can also represent the segment. With | L.AB = line : new (z.A,z.B)|, a line is defined. \begin{minipage}{0.5\textwidth} -\begin{tkzexample}[latex=0cm,small,code only] +\begin{Verbatim} \begin{tkzelements} z.A = point : new (1,1) z.B = point : new (3,2) @@ -124,7 +124,7 @@ With | L.AB = line : new (z.A,z.B)|, a line is defined. \tkzMarkRightAngle(B,A,C) \tkzMarkSegments(A,C A,B A,D) \end{tikzpicture} -\end{tkzexample} +\end{Verbatim} \end{minipage} \begin{minipage}{0.5\textwidth} \begin{tkzelements} @@ -150,7 +150,7 @@ With | L.AB = line : new (z.A,z.B)|, a line is defined. \newpage \subsection{Methods of the class line} % (fold) \label{sub:methods_from_class_line} -Here's the list of methods for the \tkzNameObj{line} object. The results are either reals, points, lines, circles or triangles. The triangles obtained are similar to the triangles defined below. +Here's the list of methods for the \tkzNameObj{line} object. The results can be real numbers, points, lines, circles or triangles. The triangles obtained are similar to the triangles defined below. \begin{minipage}{\textwidth} \bgroup @@ -161,46 +161,56 @@ Here's the list of methods for the \tkzNameObj{line} object. The results are eit \toprule \textbf{Methods} & \textbf{Comments} & \\ \midrule -\Imeth{line}{new(pt, pt)} & |L.AB = line : new(z.A,z.B)| line $(AB)$& see (\ref{ssub:altshiller})\\ +\Igfct{line}{new(pt, pt)} & |L.AB = line : new(z.A,z.B)| & Create line $(AB)$ ; Refer to (\ref{ssub:altshiller})\\ \midrule \textbf{Points} &&\\ \midrule -\Imeth{line}{gold\_ratio ()} & |z.C = L.AB : gold_ratio()| & see (\ref{sub:gold_ratio_with_segment} ; \ref{sub:the_figure_pappus_circle} ; \ref{sub:bankoff_circle}) \\ -\Imeth{line}{normalize ()} & |z.C = L.AB : normalize()| & AC =1 and $C\in (AB)$ see (\ref{ssub:normalize}) \\ -\Imeth{line}{normalize\_inv ()} & |z.C = L.AB : normalize_inv()| & CB =1 and $C\in (AB)$ \\ - \Imeth{line}{barycenter (r,r)} & |z.C = L.AB : barycenter (1,2)| & see (\ref{ssub:barycenter_with_a_line})\\ - \Imeth{line}{point (r)} & |z.C = L.AB : point (2)| & $\overrightarrow{AC} = 2\overrightarrow{AB}$ See (\ref{sub:ellipse} ; \ref{ssub:method_point})\\ -\Imeth{line}{midpoint ()} & |z.M = L.AB : midpoint ()| & better is |z.M = L.AB.mid| \\ -\Imeth{line}{harmonic\_int (pt)} & |z.D = L.AB : harmonic_int (z.C)| & See (\ref{sub:bankoff_circle})\\ -\Imeth{line}{harmonic\_ext (pt)} & |z.D = L.AB : harmonic_ext (z.C)| & See (\ref{sub:bankoff_circle})\\ -\Imeth{line}{harmonic\_both (r)} & |z.C,z.D = L.AB : harmonic_both|($\varphi$) & \ref{sub:harmonic_division_with_tkzphi}\\ -\Imeth{line}{square ()} & |S.AB = L.AB : square () | & create a square |S.AB|.\footnote{ |_,_,z.C,z.D = get_points(S.AB)|}\\ +\Imeth{line}{gold\_ratio ()} & |z.C=L.AB : gold_ratio()| & Refer to (\ref{sub:gold_ratio_with_segment} ; \ref{sub:the_figure_pappus_circle} ; \ref{sub:bankoff_circle}) \\ +\Imeth{line}{normalize ()} & |z.C=L.AB : normalize()| & AC =1 and $C\in (AB)$ Refer to (\ref{ssub:normalize}) \\ +\Imeth{line}{normalize\_inv ()} & |z.C=L.AB : normalize_inv()| & CB=1 and $C\in (AB)$ \\ +\Imeth{line}{barycenter (r,r)} & |z.C=L.AB : barycenter (1,2)| & Refer to (\ref{ssub:barycenter_with_a_line})\\ +\Imeth{line}{point (r)} & |z.C=L.AB : point (2)| & $\overrightarrow{AC} = 2\overrightarrow{AB}$ Refer to (\ref{sub:ellipse} ; \ref{ssub:method_point})\\ +\Imeth{line}{midpoint ()} & |z.M=L.AB : midpoint ()| & better is |z.M = L.AB.mid| \\ +\Imeth{line}{harmonic\_int (pt)} & |z.D=L.AB : harmonic_int (z.C)| & Refer to (\ref{sub:bankoff_circle})\\ +\Imeth{line}{harmonic\_ext (pt)} & |z.D=L.AB : harmonic_ext (z.C)| & Refer to (\ref{sub:bankoff_circle})\\ +\Imeth{line}{harmonic\_both (r)} & |z.C,z.D=L.AB : harmonic_both|($\varphi$) & \ref{sub:harmonic_division_with_tkzphi}\\ +\Imeth{line}{\_east(d)} & |z.M=L.AB : _east(2)| & |BM = 2| $A,B,M$ aligned \\ +\Imeth{line}{\_west(d)} & |z.M=L.AB : _east(2)| & |BM = 2| $A,B,M$ aligned \\ +\Imeth{line}{\_north\_pa(d)} &|z.M=L.AB: _north_pa(2)| &|AM=2| $AM\perp AB$ ; $\overrightarrow{AB},\overrightarrow{AM}$ counterclockwise \\ +\Imeth{line}{\_south\_pa(d)} &|z.M=L.AB:_south_pa(2)| &|AM=2|; $AM\perp AB$ ; $\overrightarrow{AB},\overrightarrow{AM}$ clockwise \\ +\Imeth{line}{\_north\_pb(d)} &|z.M=L.AB:_north_pb(2)| &|BM=2|; $BM\perp BA$ ; $\overrightarrow{BA},\overrightarrow{BM}$ clockwise \\ +\Imeth{line}{\_south\_pb(d)} &|z.M=L.AB:_south_pb(2)| &|BM=2|; $BM\perp BA$ ; $\overrightarrow{AB},\overrightarrow{AM}$ counterclockwise \\ +\Imeth{line}{report(d,pt)} &|z.M=L.AB:report(2,z.N)| &|MN=2|; $AB\parallel MN$ ; Refer to ex. (\ref{ssub:method_report})\\ \midrule \textbf{Lines} &&\\ \midrule -\Imeth{line}{ll\_from ( pt )} & |L.CD = L.AB : ll_from (z.C)| & $(CD) \parallel (AB)$ \\ -\Imeth{line}{ortho\_from ( pt )} & |L.CD = L.AB : ortho_from (z.C)|& $(CD) \perp (AB)$\\ -\Imeth{line}{mediator ()}&|L.uv = L.AB : mediator ()| & $(u,v)$ mediator of $(A,B)$\\ +\Imeth{line}{ll\_from ( pt )} &|L.CD=L.AB: ll_from(z.C)| &$(CD) \parallel (AB)$ \\ +\Imeth{line}{ortho\_from ( pt )} &|L.CD=L.AB: ortho_from(z.C)|&$(CD) \perp (AB)$\\ +\Imeth{line}{mediator ()}&|L.uv=L.AB: mediator()| & $(u,v)$mediator of $(A,B)$\\ \midrule \textbf{Triangles}&&\\ \midrule -\Imeth{line}{equilateral ()} & |T.ABC = L.AB : equilateral ()| & $(\overrightarrow{AB},\overrightarrow{AC})>0$ or $<0$ with swap \footnote{Triangles are defined in the direct sense of rotation, unless the "swap" option is present.} \\ -\Imeth{line}{isosceles (an<,swap>)} & |T.ABC = L.AB : isosceles (math.pi/6)|& \\ -\Imeth{line}{two\_angles (an,an)} & |T.ABC = L.AB : two_angles (an,an)|¬e \footnote{The given side is between the two angles} see ( ) \\ -\Imeth{line}{school ()} & Angle measurements are 30°,60° and 90°. & \\ +\Imeth{line}{equilateral ()} & |T.ABC=L.AB:equilateral()| & $(\overrightarrow{AB},\overrightarrow{AC})>0$ or $<0$ with swap \footnote{Triangles are defined in the direct sense of rotation, unless the "swap" option is present.} \\ +\Imeth{line}{isosceles (an<,swap>)}&|T.ABC=L.AB:isosceles(math.pi/6)|&\\ +\Imeth{line}{two\_angles (an,an)} &|T.ABC=L.AB:two_angles(an,an)|¬e \footnote{The given side is between the two angles} Refer to (\ref{ssub:triangle_with_two__angles}) \\ +\Imeth{line}{school ()} & 30°,60°, 90° & \\ \Imeth{line}{sss (r,r)} & $AC=r$ $BC=r$ & \\ \Imeth{line}{as (r,an)} & $AC =r$ $\widehat{BAC} = an$& \\ \Imeth{line}{sa (r,an)} & $AC =r$ $\widehat{ABC} = an$& \\ \midrule \textbf{Sacred triangles}&&\\ \midrule -\Imeth{line}{gold ()} & |T.ABC = L.AB : gold ()| & right in $B$ and $AC = \varphi \times AB $ \\ -\Imeth{line}{euclide ()} & |T.ABC = L.AB : euclide ()| & $AB=AC$ and $(\overrightarrow{AB},\overrightarrow{AC}) = \pi/5$ \\ -\Imeth{line}{golden ()} & |T.ABC = L.AB : golden ()| & +\Imeth{line}{gold ()} &|T.ABC=L.AB:gold()| & right in $B$ and $AC = \varphi \times AB $ \\ +\Imeth{line}{euclide ()} &|T.ABC=L.AB:euclide()| &$AB=AC$ ; $(\overrightarrow{AB},\overrightarrow{AC}) = \pi/5$ \\ +\Imeth{line}{golden ()} &|T.ABC=L.AB:golden()| & $(\overrightarrow{AB},\overrightarrow{AC}) = 2\times \pi/5$ \\ \Imeth{line}{divine ()} & & \\ -\Imeth{line}{egyptian ()} & & \\ +\Imeth{line}{egyptian ()} & & \\ \Imeth{line}{cheops ()} & & \\ +\midrule +\textbf{Squares}&&\\ +\midrule +\Imeth{line}{square ()} &|S.AB=L.AB : square () | & create a square |S.AB|.\footnote{ |_,_,z.C,z.D = get_points(S.AB)|}\\ \bottomrule \end{tabular} \egroup @@ -229,7 +239,7 @@ Here's the list of methods for the \tkzNameObj{line} object. The results are eit \midrule \textbf{Miscellaneous} &&\\ \midrule -\Imeth{line}{distance (pt)} & |d = L.Ab : distance (z.C)| & see \ref{ssub:example_distance_and_projection}\\ +\Imeth{line}{distance (pt)} & |d = L.Ab : distance (z.C)| & Refer to \ref{ssub:example_distance_and_projection}\\ \Imeth{line}{in\_out (pt)} & |b = L.AB: in_out(z.C)| & b=true if $C\in (AB)$ \\ \Imeth{line}{slope ()} & |a = L.AB : slope()| & better is L.AB.slope \\ \Imeth{line}{in\_out\_segment (pt)} & |b = L.AB : in_out_segment(z.C)| & b=true if $C\in [AB$] \\ @@ -238,8 +248,46 @@ Here's the list of methods for the \tkzNameObj{line} object. The results are eit \egroup \end{minipage} -\vspace{1 em} -Here are a few examples. +\subsubsection{Method report} % (fold) +\label{ssub:method_report} + +|report (d,pt)| If the point is absent, the transfer is made from the first point that defines the line. + +\begin{minipage}{.5\textwidth} + \begin{Verbatim} + \begin{tkzelements} + z.A = point : new (1,-1) + z.B = point : new (5,0) + L.AB = line : new ( z.A , z.B ) + z.M = point : new (2,3) + z.N = L.AB : report (3,z.M) + z.O = L.AB : report (3) + \end{tkzelements} + \begin{tikzpicture} + \tkzGetNodes + \tkzDrawSegments(A,B M,N) + \tkzDrawPoints(A,B,M,N,O) + \tkzLabelPoints(A,B,M,N,O) + \end{tikzpicture} + \end{Verbatim} +\end{minipage} +\begin{minipage}{.5\textwidth} + \begin{tkzelements} +z.A = point : new (1,-1) +z.B = point : new (5,0) +L.AB = line : new ( z.A , z.B ) +z.M = point : new (2,3) +z.N = L.AB : report (3,z.M) +z.O = L.AB : report (3) +\end{tkzelements} +\begin{tikzpicture} +\tkzGetNodes +\tkzDrawSegments(A,B M,N) +\tkzDrawPoints(A,B,M,N,O) +\tkzLabelPoints(A,B,M,N,O) +\end{tikzpicture} +\end{minipage} +% subsubsection method_report (end) \subsubsection{Triangle with two\_angles} % (fold) \label{ssub:triangle_with_two__angles} @@ -247,7 +295,7 @@ Here are a few examples. The angles are on either side of the given segment \begin{minipage}{.4\textwidth} -\begin{verbatim} +\begin{Verbatim} \begin{tkzelements} z.A = point : new ( 0 , 0 ) z.B = point : new ( 4 , 0 ) @@ -262,7 +310,7 @@ The angles are on either side of the given segment \tkzLabelPoints(A,B) \tkzLabelPoints[above](C) \end{tikzpicture} -\end{verbatim} +\end{Verbatim} \end{minipage} \begin{minipage}{.6\textwidth} \begin{tkzelements} @@ -290,9 +338,8 @@ The angles are on either side of the given segment In the following example, a small difficulty arises. The given lengths are not affected by scaling, so it's necessary to use the \Igfct{math}{value (r) } function, which will modify the lengths according to the scale. \begin{minipage}{.4\textwidth} -\begin{verbatim} +\begin{Verbatim} \begin{tkzelements} - scale =1.25 z.A = point : new ( 0 , 0 ) z.B = point : new ( 5 , 0 ) L.AB = line : new ( z.A , z.B ) @@ -306,18 +353,18 @@ In the following example, a small difficulty arises. The given lengths are not a \tkzLabelPoints(A,B) \tkzLabelPoints[above](C) \end{tikzpicture} -\end{verbatim} +\end{Verbatim} \end{minipage} \begin{minipage}{.6\textwidth} \begin{tkzelements} - scale =1.25 z.A = point : new ( 0 , 0 ) z.B = point : new ( 5 , 0 ) L.AB = line : new ( z.A , z.B ) T.ABC = L.AB : sss (value(3),value(4)) z.C = T.ABC.pc \end{tkzelements} -\hspace{\fill} \begin{tikzpicture}[gridded] +\hspace{\fill} + \begin{tikzpicture}[gridded] \tkzGetNodes \tkzDrawPolygons(A,B,C) \tkzDrawPoints(A,B,C) @@ -333,9 +380,9 @@ In the following example, a small difficulty arises. The given lengths are not a In some cases, two solutions are possible. \begin{minipage}{.4\textwidth} -\begin{verbatim} +\begin{Verbatim} \begin{tkzelements} - scale =1.2 + scale =1 z.A = point : new ( 0 , 0 ) z.B = point : new ( 5 , 0 ) L.AB = line : new ( z.A , z.B ) @@ -349,15 +396,15 @@ In some cases, two solutions are possible. \tkzDrawPoints(A,B,C,D) \tkzLabelPoints(A,B) \tkzLabelPoints[above](C,D) - \tkzLabelAngle(C,B,A){$\pi/3$} + \tkzLabelAngle[teal](C,B,A){$\pi/6$} \tkzLabelSegment[below left](A,C){$7$} \tkzLabelSegment[below left](A,D){$7$} \end{tikzpicture} -\end{verbatim} +\end{Verbatim} \end{minipage} \begin{minipage}{.6\textwidth} \begin{tkzelements} - scale =1.2 + scale =1 z.A = point : new ( 0 , 0 ) z.B = point : new ( 5 , 0 ) L.AB = line : new ( z.A , z.B ) @@ -371,7 +418,7 @@ In some cases, two solutions are possible. \tkzDrawPoints(A,B,C,D) \tkzLabelPoints(A,B) \tkzLabelPoints[above](C,D) - \tkzLabelAngle(C,B,A){$\pi/3$} + \tkzLabelAngle[teal](C,B,A){$\pi/6$} \tkzLabelSegment[below left](A,C){$7$} \tkzLabelSegment[below left](A,D){$7$} \end{tikzpicture} @@ -398,7 +445,7 @@ The side lengths are proportional to the lengths given in the table. They depend \end{tabular} \begin{minipage}{.4\textwidth} -\begin{verbatim} +\begin{Verbatim} \begin{tkzelements} z.A = point : new ( 0 , 0 ) z.B = point : new ( 4 , 0 ) @@ -422,7 +469,7 @@ The side lengths are proportional to the lengths given in the table. They depend \tkzDrawPoints(A,...,H) \tkzLabelPoints(A,...,H) \end{tikzpicture} -\end{verbatim} +\end{Verbatim} \end{minipage} \begin{minipage}{.6\textwidth} \begin{tkzelements} @@ -462,7 +509,7 @@ This method exists for all objects except quadrilaterals. \begin{minipage}{.4\textwidth} -\begin{verbatim} +\begin{Verbatim} \begin{tkzelements} z.A = point : new (-1,-1) z.B = point : new (1,1) @@ -477,7 +524,7 @@ This method exists for all objects except quadrilaterals. \tkzDrawPoints(A,B,I,J,K) \tkzLabelPoints(A,B,I,J,K) \end{tikzpicture} -\end{verbatim} +\end{Verbatim} \end{minipage} \begin{minipage}{.6\textwidth} \begin{tkzelements} @@ -502,22 +549,22 @@ This method exists for all objects except quadrilaterals. \begin{minipage}{.4\textwidth} - \begin{verbatim} - \begin{tkzelements} - z.a = point: new (1, 1) - z.b = point: new (5, 4) - L.ab = line : new (z.a,z.b) - z.c = L.ab : normalize () - \end{tkzelements} +\begin{Verbatim} +\begin{tkzelements} + z.a = point: new (1, 1) + z.b = point: new (5, 4) + L.ab = line : new (z.a,z.b) + z.c = L.ab : normalize () +\end{tkzelements} - \begin{tikzpicture}[gridded] - \tkzGetNodes - \tkzDrawSegments(a,b) - \tkzDrawCircle(a,c) - \tkzDrawPoints(a,b,c) - \tkzLabelPoints(a,b,c) - \end{tikzpicture} - \end{verbatim} +\begin{tikzpicture}[gridded] +\tkzGetNodes +\tkzDrawSegments(a,b) +\tkzDrawCircle(a,c) +\tkzDrawPoints(a,b,c) +\tkzLabelPoints(a,b,c) +\end{tikzpicture} +\end{Verbatim} \end{minipage} \begin{minipage}{.6\textwidth} \begin{tkzelements} @@ -543,7 +590,7 @@ This method exists for all objects except quadrilaterals. \label{ssub:barycenter_with_a_line} \begin{minipage}{.4\textwidth} -\begin{verbatim} +\begin{Verbatim} \begin{tkzelements} z.A = point : new ( 0 , -1 ) z.B = point : new ( 4 , 2 ) @@ -556,7 +603,7 @@ This method exists for all objects except quadrilaterals. \tkzDrawPoints(A,B,G) \tkzLabelPoints(A,B,G) \end{tikzpicture} -\end{verbatim} +\end{Verbatim} \end{minipage} \begin{minipage}{.6\textwidth} \begin{tkzelements} @@ -577,7 +624,7 @@ This method exists for all objects except quadrilaterals. \subsubsection{Example: new line from a defined line} % (fold) \label{ssub:new_line_from_a_defined_line} \begin{minipage}{0.5\textwidth} -\begin{tkzexample}[latex=0cm,small,code only] +\begin{Verbatim} \begin{tkzelements} scale = 1.25 z.A = point : new (1,1) @@ -597,7 +644,7 @@ This method exists for all objects except quadrilaterals. \tkzMarkRightAngle(B,A,C) \tkzMarkSegments(A,C A,B A,D) \end{tikzpicture} -\end{tkzexample} +\end{Verbatim} \end{minipage} \begin{minipage}{0.5\textwidth} \begin{tkzelements} @@ -629,7 +676,7 @@ _,z.E = get_points ( L.CD: ll_from (z.B)) \subsubsection{Example: projection of several points} % (fold) \label{ssub:example_projection_of_several_points} \begin{minipage}{0.5\textwidth} -\begin{tkzexample}[latex=0cm,small,code only] +\begin{Verbatim} \begin{tkzelements} scale = .8 z.a = point: new (0, 0) @@ -645,7 +692,7 @@ _,z.E = get_points ( L.CD: ll_from (z.B)) \tkzDrawPoints(a,...,d,c',d') \tkzLabelPoints(a,...,d,c',d') \end{tikzpicture} - \end{tkzexample} +\end{Verbatim} \end{minipage} \begin{minipage}{0.5\textwidth} \begin{tkzelements} @@ -674,7 +721,7 @@ z.cp,z.dp = L.ab : projection(z.c,z.d) \label{ssub:example_combination_of_methods} \begin{minipage}{0.6\textwidth} -\begin{tkzexample}[small,code only] +\begin{Verbatim} \begin{tkzelements} z.A = point: new (0 , 0) z.B = point: new (6 , 0) @@ -699,9 +746,11 @@ z.cp,z.dp = L.ab : projection(z.c,z.d) \tkzFillAngles[teal!30,opacity=.4](A,C,B b,A,B A,O,H) \tkzMarkAngles[mark=|](A,C,B b,A,B A,O,H H,O,B) \tkzDrawPoints(A,B,C,H,O) - \tkzLabelPoints(A,B,C,H,O) + \tkzLabelPoints(B,H) + \tkzLabelPoints[above](O,C) + \tkzLabelPoints[left](A) \end{tikzpicture} -\end{tkzexample} +\end{Verbatim} \end{minipage} \begin{minipage}{0.4\textwidth} \begin{tkzelements} @@ -730,7 +779,9 @@ z.a,z.b = L.ab.pa,L.ab.pb \tkzFillAngles[teal!30,opacity=.4,,size=.5](A,C,B b,A,B A,O,H) \tkzMarkAngles[mark=|,size=.5](A,C,B b,A,B A,O,H H,O,B) \tkzDrawPoints(A,B,C,H,O) -\tkzLabelPoints(A,B,C,H,O) +\tkzLabelPoints(B,H) +\tkzLabelPoints[above](O,C) +\tkzLabelPoints[left](A) \end{tikzpicture} \hspace*{\fill} \end{minipage} @@ -741,7 +792,7 @@ z.a,z.b = L.ab.pa,L.ab.pb \label{ssub:example_translation} \begin{minipage}{0.6\textwidth} -\begin{tkzexample}[small,code only] +\begin{Verbatim} \begin{tkzelements} z.A = point: new (0,0) z.B = point: new (1,2) @@ -756,7 +807,7 @@ z.a,z.b = L.ab.pa,L.ab.pb \tkzLabelPoints(A,...,F) \tkzDrawSegments[->,red,> =latex](C,E D,F A,B) \end{tikzpicture} -\end{tkzexample} +\end{Verbatim} \end{minipage} \begin{minipage}{0.4\textwidth} \begin{tkzelements} @@ -783,7 +834,7 @@ z.a,z.b = L.ab.pa,L.ab.pb \label{ssub:example_distance_and_projection} \begin{minipage}{0.5\textwidth} -\begin{tkzexample}[latex=0cm,small,code only] +\begin{Verbatim} \begin{tkzelements} z.A = point : new (0 , 0) z.B = point : new (5 , -2) @@ -800,7 +851,7 @@ z.a,z.b = L.ab.pa,L.ab.pb \tkzLabelSegment[above left, draw](C,H){$CH = \tkzUseLua{d}$} \end{tikzpicture} -\end{tkzexample} +\end{Verbatim} \end{minipage} \begin{minipage}{0.5\textwidth} \begin{tkzelements} @@ -829,26 +880,26 @@ z.a,z.b = L.ab.pa,L.ab.pb \label{ssub:reflection_of_object} \begin{minipage}{.5\textwidth} - \begin{verbatim} - \begin{tkzelements} - z.A = point : new ( 0 , 0 ) - z.B = point : new ( 4 , 1 ) - z.E = point : new ( 0 , 2 ) - z.F = point : new ( 3 , 3 ) - z.G = point : new ( 4 , 2 ) - L.AB = line : new ( z.A , z.B ) - T.EFG = triangle : new (z.E,z.F,z.G) - T.new = L.AB : reflection (T.EFG) - z.Ep,z.Fp,z.Gp = get_points(T.new) - \end{tkzelements} - \begin{tikzpicture} - \tkzGetNodes - \tkzDrawLine(A,B) - \tkzDrawPolygon(E,F,G) - \tkzDrawPolygon[new](E',F',G') - \tkzDrawSegment[red,dashed](E,E') - \end{tikzpicture} - \end{verbatim} +\begin{Verbatim} +\begin{tkzelements} + z.A = point : new ( 0 , 0 ) + z.B = point : new ( 4 , 1 ) + z.E = point : new ( 0 , 2 ) + z.F = point : new ( 3 , 3 ) + z.G = point : new ( 4 , 2 ) + L.AB = line : new ( z.A , z.B ) + T.EFG = triangle : new (z.E,z.F,z.G) + T.new = L.AB : reflection (T.EFG) + z.Ep,z.Fp,z.Gp = get_points(T.new) +\end{tkzelements} +\begin{tikzpicture} + \tkzGetNodes + \tkzDrawLine(A,B) + \tkzDrawPolygon(E,F,G) + \tkzDrawPolygon[new](E',F',G') + \tkzDrawSegment[red,dashed](E,E') +\end{tikzpicture} +\end{Verbatim} \end{minipage} \begin{minipage}{.5\textwidth} \begin{tkzelements} @@ -876,7 +927,7 @@ z.a,z.b = L.ab.pa,L.ab.pb \subsection{Apollonius circle MA/MB = k} % (fold) \label{sub:apollonius_circle_ma_mb_k} -\begin{verbatim} +\begin{Verbatim} \begin{tkzelements} z.A = point : new ( 0 , 0 ) z.B = point : new ( 6 , 0 ) @@ -901,7 +952,7 @@ z.a,z.b = L.ab.pa,L.ab.pb \tkzMarkRightAngle[opacity=.3,fill=lightgray](O,P,C) \tkzMarkAngles[mark=||](A,P,D D,P,B) \end{tikzpicture} -\end{verbatim} +\end{Verbatim} \begin{tkzelements} z.A = point : new ( 0 , 0 ) -- cgit v1.2.3