\section{Miscellaneous tools} \subsection{Duplicate a segment} This involves constructing a segment on a given half-line of the same length as a given segment. \begin{NewMacroBox}{tkzDuplicateSegment}{\parg{pt1,pt2}\parg{pt3,pt4}\marg{pt5}}% This involves creating a segment on a given half-line of the same length as a given segment . It is in fact the definition of a point. \medskip \begin{tabular}{lll}% \toprule arguments & example & explanation \\ \midrule \TAline{(pt1,pt2)(pt3,pt4)\{pt5\}} {\tkzcname{tkzDuplicateLen}(A,B)(E,F)\{C\}}{AC=EF et $C \in [AB)$} \\ \bottomrule \end{tabular} \medskip The macro \tkzcname{tkzDuplicateLength} is identical to this one. \end{NewMacroBox} \begin{tkzexample}[latex=6cm,small] \begin{tikzpicture} \tkzDefPoint(0,0){A} \tkzDefPoint(2,-3){B} \tkzDefPoint(2,5){C} \tkzDrawSegments[red](A,B A,C) \tkzDuplicateSegment(A,B)(A,C) \tkzGetPoint{D} \tkzDrawSegment[green](A,D) \tkzDrawPoints[color=red](A,B,C,D) \tkzLabelPoints[above right=3pt](A,B,C,D) \end{tikzpicture} \end{tkzexample} \subsubsection{Gold proportion with \tkzcname{tkzDuplicateSegment}} \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture}[rotate=-90,scale=.75] \tkzInit[xmax=10,ymax=10] \tkzClip[space=1] \tkzDefPoint(0,0){A} \tkzDefPoint(10,0){B} \tkzDefMidPoint(A,B) \tkzGetPoint{I} \tkzDefPointWith[orthogonal,K=-.75](B,A) \tkzGetPoint{C} \tkzInterLC(B,C)(B,I) \tkzGetSecondPoint{D} \tkzDuplicateSegment(B,D)(D,A) \tkzGetPoint{E} \tkzInterLC(A,B)(A,E) \tkzGetPoints{N}{M} \tkzDrawArc[delta=10](D,E)(B) \tkzDrawArc[delta=10](A,M)(E) \tkzDrawLines(A,B B,C A,D) \tkzDrawArc[delta=10](B,D)(I) \tkzDrawPoints(A,B,D,C,M,I,N) \tkzLabelPoints(A,B,D,C,M,I,N) \end{tikzpicture} \end{tkzexample} % <--------------------------------------------------------------------> % About Angles % <--------------------------------------------------------------------> \subsection{Determining a slope} It is a question of determining whether it exists, the slope of a straight line defined by two points. No verification of the existence is made. \begin{NewMacroBox}{tkzFindSlope}{\parg{pt1,pt2}\marg{name of macro}}% The result is stored in a macro. \medskip \begin{tabular}{lll} \toprule arguments & example & explication \\ \midrule \TAline{(pt1,pt2){pt3}} {\tkzcname{tkzFindSlope}(A,B)\{slope\}}{\tkzcname{slope} will give the result of $\frac{y_B-y_A}{x_B-x_A}$} \\ \bottomrule \end{tabular} \medskip \tkzHandBomb\ Careful not to have $x_B=x_A$ \end{NewMacroBox} \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture}[scale=1.5] \tkzInit[xmax=4,ymax=5]\tkzGrid[sub] \tkzDefPoint(1,2){A} \tkzDefPoint(3,4){B} \tkzDefPoint(3,2){C} \tkzDefPoint(3,1){D} \tkzDrawSegments(A,B A,C A,D) \tkzDrawPoints[color=red](A,B,C,D) \tkzLabelPoints(A,B,C,D) \tkzFindSlope(A,B){SAB} \tkzFindSlope(A,C){SAC} \tkzFindSlope(A,D){SAD} \pgfkeys{/pgf/number format/.cd,fixed,precision=2} \tkzText[fill=Gold!50,draw=brown](1,4)% {La pente de (AB) est: $\pgfmathprintnumber{\SAB}$} \tkzText[fill=Gold!50,draw=brown](1,3.5)% {La pente de (AC) est: $\pgfmathprintnumber{\SAC}$} \tkzText[fill=Gold!50,draw=brown](1,3)% {La pente de (AD) est: $\pgfmathprintnumber{\SAD}$} \end{tikzpicture} \end{tkzexample} \subsection{Angle formed by a straight line with the horizontal axis} Much more interesting than the last one. The result is between -180 degrees and +180 degrees. \begin{NewMacroBox}{tkzFindSlopeAngle}{\parg{pt1,pt2}}% The result is stored in a macro \tkzcname{tkzAngleResult}. \medskip \begin{tabular}{lll} \toprule arguments & example & explication \\ \midrule \TAline{(pt1,pt2)} {\tkzcname{tkzFindSlopeAngle}(A,B)}{\tkzcname{tkzGetAngle} can retrieve the result} \bottomrule \end{tabular} \medskip {If recovery is not necessary, it is possible to use \tkzcname{tkzAngleResult}} \end{NewMacroBox} \subsubsection{Example of use of \tkzcname{tkzFindSlopeAngle}} Here is another version of the construction of a mediator \begin{tkzexample}[latex=6cm,small] \begin{tikzpicture} \tkzInit \tkzDefPoint(0,0){A} \tkzDefPoint(3,2){B} \tkzDefLine[mediator](A,B) \tkzGetPoints{I}{J} \tkzCalcLength[cm](A,B) \tkzGetLength{dAB} \tkzFindSlopeAngle(A,B) \tkzGetAngle{tkzangle} \begin{scope}[rotate=\tkzangle] \tikzset{arc/.style={color=gray,delta=10}} \tkzDrawArc[R,arc](B,3/4*\dAB)(120,240) \tkzDrawArc[R,arc](A,3/4*\dAB)(-45,60) \tkzDrawLine(I,J) \tkzDrawSegment(A,B) \end{scope} \tkzDrawPoints(A,B,I,J) \tkzLabelPoints(A,B) \tkzLabelPoints[right](I,J) \end{tikzpicture} \end{tkzexample} \subsection{from an angle} In the previous example, I used the macro \tkzcname{tkzGetAngle} to retrieve an angle. \begin{NewMacroBox}{tkzGetAngle}{\marg{name of macro}}% This macro retrieves \tkzcname{tkzAngleResult} and stores the result in a new macro. \medskip \begin{tabular}{lll}% \toprule arguments & example & explication \\ \midrule \TAline{name of macro} {\tkzcname{tkzGetAngle}\{ang\}}{\tkzcname{ang} contains the value of the angle. } \end{tabular} \end{NewMacroBox} \subsection{Example of the use of \tkzcname{tkzGetAngle}} The point here is that $(AB)$ is the bisector of $\widehat{CAD}$, such that the $AD$ slope is zero. We recover the slope of $(AB)$ and then rotate twice. \begin{tkzexample}[vbox,small] \begin{tikzpicture} \tkzInit \tkzDefPoint(1,5){A} \tkzDefPoint(5,2){B} \tkzDrawSegment(A,B) \tkzFindSlopeAngle(A,B)\tkzGetAngle{tkzang} \tkzDefPointBy[rotation= center A angle \tkzang ](B) \tkzGetPoint{C} \tkzDefPointBy[rotation= center A angle -\tkzang ](B) \tkzGetPoint{D} \tkzCompass[length=1,dashed,color=red](A,C) \tkzCompass[delta=10,brown](B,C) \tkzDrawPoints(A,B,C,D) \tkzLabelPoints(B,C,D) \tkzLabelPoints[above left](A) \tkzDrawSegments[style=dashed,color=orange!30](A,C A,D) \end{tikzpicture} \end{tkzexample} \subsection{Angle formed by three points} \begin{NewMacroBox}{tkzFindAngle}{\parg{pt1,pt2,pt3}}% The result is stored in a macro \tkzcname{tkzAngleResult}. \medskip \begin{tabular}{lll} \toprule arguments & example & explication \\ \midrule \TAline{(pt1,pt2,pt3)} {\tkzcname{tkzFindAngle}(A,B,C)}{\tkzcname{tkzAngleResult} gives the angle ($\overrightarrow{BA},\overrightarrow{BC}$)} \bottomrule \end{tabular} \medskip The result is between -180 degrees and +180 degrees. pt2 is the vertex and \tkzcname{tkzGetAngle} can retrieve the angle. \end{NewMacroBox} \subsection{Example of use of \tkzcname{tkzFindAngle} } \begin{tkzexample}[vbox,small] \begin{tikzpicture} \tkzInit[xmin=-1,ymin=-1,xmax=7,ymax=7] \tkzClip \tkzDefPoint (0,0){O} \tkzDefPoint (6,0){A} \tkzDefPoint (5,5){B} \tkzDefPoint (3,4){M} \tkzFindAngle (A,O,M) \tkzGetAngle{an} \tkzDefPointBy[rotation=center O angle \an](A) \tkzGetPoint{C} \tkzDrawSector[fill = blue!50,opacity=.5](O,A)(C) \tkzFindAngle(M,B,A) \tkzGetAngle{am} \tkzDefPointBy[rotation = center O angle \am](A) \tkzGetPoint{D} \tkzDrawSector[fill = red!50,opacity = .5](O,A)(D) \tkzDrawPoints(O,A,B,M,C,D) \tkzLabelPoints(O,A,B,M,C,D) \edef\an{\fpeval{round(\an,2)}}\edef\am{\fpeval{round(\am,2)}} \tkzDrawSegments(M,B B,A) \tkzText(4,2){$\widehat{AOC}=\widehat{AOM}=\an^{\circ}$} \tkzText(1,4){$\widehat{AOD}=\widehat{MBA}=\am^{\circ}$} \end{tikzpicture} \end{tkzexample} \subsection{\tkzcname{tkzCalcLength}} There's an option in \TIKZ\ name in \tkzname{veclen}. This option is used to calculate AB if A and B are two points. The only problem for me is that the version of \TIKZ\ is not accurate enough in some cases. My version uses the \tkzNamePack{xfp} package and is slower, but more accurate. \hypertarget{tpsc}{} \begin{NewMacroBox}{tkzCalcLength}{\oarg{local options}\parg{pt1,pt2}\marg{name of macro}}% \begin{tabular}{lll}% arguments & example & explication \\ \midrule \TAline{(pt1,pt2)\{name of macro\}} {\tkzcname{tkzCalcLength}(A,B)\{dAB\}}{\tkzcname{dAB} donne $AB$ en pt} \bottomrule \end{tabular} \medskip Une seule option \begin{tabular}{lll}% options & default & example \\ \midrule \TOline{cm} {false}{\tkzcname{tkzCalcLength}[cm](A,B)\{dAB\} \tkzcname{dAB} gives AB en cm} \end{tabular} The result is stored in a macro. \end{NewMacroBox} \subsubsection{Compass square construction} \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture}[scale=1] \tkzDefPoint(0,0){A} \tkzDefPoint(4,0){B} \tkzDrawLine[add= .6 and .2](A,B) \tkzCalcLength[cm](A,B)\tkzGetLength{dAB} \tkzDefLine[perpendicular=through A](A,B) \tkzDrawLine(A,tkzPointResult) \tkzGetPoint{D} \tkzShowLine[orthogonal=through A,gap=2](A,B) \tkzMarkRightAngle(B,A,D) \tkzVecKOrth[-1](B,A)\tkzGetPoint{C} \tkzCompasss(A,D D,C) \tkzDrawArc[R](B,\dAB)(80,110) \tkzDrawPoints(A,B,C,D) \tkzDrawSegments[color=gray,style=dashed](B,C C,D) \tkzLabelPoints(A,B,C,D) \end{tikzpicture} \end{tkzexample} \subsection{Transformation from pt to cm or cm to pt} Not sure if this is necessary and it is only a division by 28.45274 and a multiplication by the same number. The macros are: \begin{NewMacroBox}{tkzpttocm}{\parg{nombre}\marg{name of macro}}% The result is stored in a macro. \medskip \begin{tabular}{lll}% \toprule arguments & example & explication \\ \midrule \TAline{(nombre){name of macro}} {\tkzcname{tkzpttocm}(120)\{len\}}{\tkzcname{len} donne un nombre de tkzname{cm}} \bottomrule \end{tabular} \medskip You'll have to use \tkzcname{len} along with \tkzname{cm}. \end{NewMacroBox} \subsection{Change of unit} \begin{NewMacroBox}{tkzcmtopt}{\parg{nombre}\marg{name of macro}}% The result is stored in a macro. \medskip \begin{tabular}{lll} \toprule arguments & example & explication \\ \midrule \TAline{(nombre)\{name of macro\}}{\tkzcname{tkzcmtopt}(5)\{len\}}{\tkzcname{len} longueur en \tkzname{pts}} \bottomrule \end{tabular} \medskip \noindent{The result can be used with \tkzcname{len}\tkzname{pt}} \end{NewMacroBox} \subsubsection{Example} The macro \tkzcname{tkzDefCircle[radius](A,B)} defines the radius that we retrieve with \tkzcname{tkzGetLength}, but this result is in \tkzname{pt}. \begin{tkzexample}[latex=6cm,small] \begin{tikzpicture}[scale=.5] \tkzDefPoint(0,0){A} \tkzDefPoint(3,-4){B} \tkzDefCircle[through](A,B) \tkzGetLength{rABpt} \tkzpttocm(\rABpt){rABcm} \tkzDrawCircle(A,B) \tkzDrawPoints(A,B) \tkzLabelPoints(A,B) \tkzDrawSegment[dashed](A,B) \tkzLabelSegment(A,B){$\pgfmathprintnumber{\rABcm}$} \end{tikzpicture} \end{tkzexample} %<--------------------------------------------------------------------------–> % Coordonnées d'un point % result in #2x et #2y #1 est le point et on récupère ses coordonnées % usage soit A un point \tkzGetPointCoord(A){V} alors \Vx = xA et \Vy = yA % en cm % tkzGetPointCoord avec [#1] cm ou bien pt ?? todo %<--------------------------------------------------------------------------–> \begin{NewMacroBox}{tkzGetPointCoord}{\parg{$A$}\marg{name of macro}}% Stores in two macros the coordinates of a point \medskip \begin{tabular}{lll} \toprule arguments & example & explanation \\ \midrule \TAline{(point)\{name of macro\}} {\tkzcname{tkzGetPointCoord}(A)\{A\}}{\tkzcname{Ax} and \tkzcname{Ay} give the coordinates of $A$} \end{tabular} \medskip If the name of the macro is \tkzname{p}, then \tkzcname{px} and \tkzcname{py} give the coordinates of the chosen point with the cm as. \end{NewMacroBox} \subsubsection{Coordinate transfer with \tkzcname{tkzGetPointCoord}} \begin{tkzexample}[width=8cm,small] \begin{tikzpicture} \tkzInit[xmax=5,ymax=3] \tkzGrid[sub,orange] \tkzAxeXY \tkzDefPoint(1,0){A} \tkzDefPoint(4,2){B} \tkzGetPointCoord(A){a} \tkzGetPointCoord(B){b} \tkzDefPoint(\ax,\ay){C} \tkzDefPoint(\bx,\by){D} \tkzDrawPoints[color=red](C,D) \end{tikzpicture} \end{tkzexample} \subsubsection{Sum of vectors with \tkzcname{tkzGetPointCoord}} \begin{tkzexample}[width=6cm,small] \begin{tikzpicture}[>=latex] \tkzDefPoint(1,4){a} \tkzDefPoint(3,2){b} \tkzDefPoint(1,1){c} \tkzDrawSegment[->,red](a,b) \tkzGetPointCoord(c){c} \draw[color=blue,->](a) -- ([shift=(b)]\cx,\cy) ; \draw[color=purple,->](b) -- ([shift=(b)]\cx,\cy) ; \tkzDrawSegment[->,blue](a,c) \tkzDrawSegment[->,purple](b,c) \end{tikzpicture} \end{tkzexample} \endinput