%!TEX root = /Users/ego/Boulot/TKZ/tkz-euclide/doc_fr/TKZdoc-euclide-main.tex \section{Gallery : Some examples} Some examples with explanations in english. %–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––> \subsection{White on Black} This example shows how to get a segment with a length equal at $\sqrt{a}$ from a segment of length $a$, only with a rule and a compass. \begin{center} \begin{tkzexample}[] \tikzset{background rectangle/.style={fill=black}} \begin{tikzpicture}[show background rectangle] \tkzInit[ymin=-1.5,ymax=7,xmin=-1,xmax=+11] \tkzClip \tkzDefPoint(0,0){O} \tkzDefPoint(1,0){I} \tkzDefPoint(10,0){A} \tkzDefPointWith[orthogonal](I,A) \tkzGetPoint{H} \tkzDefMidPoint(O,A) \tkzGetPoint{M} \tkzInterLC(I,H)(M,A)\tkzGetPoints{C}{B} \tkzDrawSegments[color=white,line width=1pt](I,H O,A) \tkzDrawPoints[color=white](O,I,A,B,M) \tkzMarkRightAngle[color=white,line width=1pt](A,I,B) \tkzDrawArc[color=white,line width=1pt,style=dashed](M,A)(O) \tkzLabelSegment[white,right=1ex,pos=.5](I,B){$\sqrt{a}$} \tkzLabelSegment[white,below=1ex,pos=.5](O,I){$1$} \tkzLabelSegment[pos=.6,white,below=1ex](I,A){$a$} \end{tikzpicture} \end{tkzexample} \end{center} \vfill\newpage %<–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––> \subsection{ Square root of the integers } How to get $1$, $\sqrt{2}$, $\sqrt{3}$ with a rule and a compass. \begin{center} \begin{tkzexample}[] \begin{tikzpicture}[scale=1.75] \tkzInit[xmin=-3,xmax=4,ymin=-2,ymax=4] \tkzGrid \tkzDefPoint(0,0){O} \tkzDefPoint(1,0){a0} \newcounter{tkzcounter} \setcounter{tkzcounter}{0} \newcounter{density} \setcounter{density}{20} \foreach \i in {0,...,15}{% \pgfmathsetcounter{density}{\thedensity+2} \setcounter{density}{\thedensity} \stepcounter{tkzcounter} \tkzDefPointWith[orthogonal normed](a\i,O) \tkzGetPoint{a\thetkzcounter} \tkzDrawPolySeg[color=Maroon!\thedensity,% fill=Maroon!\thedensity,opacity=.5](a\i,a\thetkzcounter,O)} \end{tikzpicture} \end{tkzexample} \end{center} %<–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––> \vfill\newpage %<–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––> % \subsection{How to construct the tangent lines from a point to a circle with a rule and a compass.} \begin{center} \begin{tkzexample}[] \begin{tikzpicture} \tkzPoint(0,0){O} \tkzPoint(9,2){P} \tkzDefMidPoint(O,P) \tkzGetPoint{I} \tkzDrawCircle[R](O,4cm) \tkzDrawCircle[diameter](O,P) \tkzCalcLength(I,P) \tkzGetLength{dIP} \tkzInterCC[R](O,4cm)(I,\dIP pt)\tkzGetPoints{Q1}{Q2} \tkzDrawPoint[color=red](Q1) \tkzDrawPoint[color=red](Q2) \tkzDrawLine(P,Q1) \tkzDrawLine(P,Q2) \tkzDrawSegments(O,Q1 O,Q2) \tkzDrawLine(P,O) \end{tikzpicture} \end{tkzexample} \end{center} % % %<–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––> \vfill\newpage %<–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––> \subsection{Circle and tangent} We have a point A $(8,2)$, a circle with center A and radius=3cm and a line $\delta$ $y=4$. The line intercepts the circle at B. We want to draw the tangent at the circle in B. \begin{center} \begin{tkzexample}[] \begin{tikzpicture} \tkzInit[xmax=14,ymin=-2,ymax=6] \tkzDrawX[noticks,label=$(d)$] \tkzPoint[pos=above right](8,2){A}; \tkzPoint[color=red,pos=above right](0,0){O}; \tkzDrawCircle[R,color=blue,line width=.8pt](A,3 cm) \tkzHLine[color=red,style=dashed]{4} \tkzText[above](12,4){$\delta$} \FPeval\alphaR{arcsin(2/3)}% on a les bonnes valeurs \FPeval\xB{8-3*cos(\alphaR)} \tkzPoint[pos=above left](\xB,4){B}; \tkzDrawSegment[line width=1pt](A,B) \tkzDefLine[orthogonal=through B](A,B) \tkzGetPoint{b} \tkzDefPoint(1,0){i} \tkzInterLL(B,b)(O,i) \tkzGetPoint{B'} \tkzDrawPoint(B') \tkzDrawLine(B,B') \end{tikzpicture} \end{tkzexample} \end{center} \vfill\newpage %<–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––> \subsection{About right triangle} We have a segment $[AB]$ and we want to determine a point $C$ such as $AC=8 cm$ and $ABC$ is a right triangle in $B$. \begin{center} \begin{tkzexample}[] \begin{tikzpicture} \tkzInit \tkzClip \tkzPoint[pos=left](2,1){A} \tkzPoint(6,4){B} \tkzDrawSegment(A,B) \tkzDrawPoint[color=red](A) \tkzDrawPoint[color=red](B) \tkzDefPointWith[orthogonal,K=-1](B,A) \tkzDrawLine[add = .5 and .5](B,tkzPointResult) \tkzInterLC[R](B,tkzPointResult)(A,8 cm) \tkzGetPoints{C}{J} \tkzDrawPoint[color=red](C) \tkzCompass(A,C) \tkzMarkRightAngle(A,B,C) \tkzDrawLine[color=gray,style=dashed](A,C) \end{tikzpicture} \end{tkzexample} \end{center} %<–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––> \vfill\newpage %<–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––> \subsection{Archimedes} This is an ancient problem proved by the great Greek mathematician Archimedes . The figure below shows a semicircle, with diameter $AB$. A tangent line is drawn and touches the semicircle at $B$. An other tangent line at a point, $C$, on the semicircle is drawn. We project the point $C$ on the segment$[AB]$ on a point $D$ . The two tangent lines intersect at the point $T$. Prove that the line $(AT)$ bisects $(CD)$ \begin{center} \begin{tkzexample}[] \begin{tikzpicture}[scale=1.25] \tkzInit[ymin=-1,ymax=7] \tkzClip \tkzDefPoint(0,0){A}\tkzDefPoint(6,0){D} \tkzDefPoint(8,0){B}\tkzDefPoint(4,0){I} \tkzDefLine[orthogonal=through D](A,D) \tkzInterLC[R](D,tkzPointResult)(I,4 cm) \tkzGetFirstPoint{C} \tkzDefLine[orthogonal=through C](I,C) \tkzGetPoint{c} \tkzDefLine[orthogonal=through B](A,B) \tkzGetPoint{b} \tkzInterLL(C,c)(B,b) \tkzGetPoint{T} \tkzInterLL(A,T)(C,D) \tkzGetPoint{P} \tkzDrawArc(I,B)(A) \tkzDrawSegments(A,B A,T C,D I,C) \tkzDrawSegment[color=orange](I,C) \tkzDrawLine[add = 1 and 0](C,T) \tkzDrawLine[add = 0 and 1](B,T) \tkzMarkRightAngle(I,C,T) \tkzDrawPoints(A,B,I,D,C,T) \tkzLabelPoints(A,B,I,D) \tkzLabelPoints[above right](C,T) \tkzMarkSegment[pos=.25,mark=s|](C,D) \tkzMarkSegment[pos=.75,mark=s|](C,D) \end{tikzpicture} \end{tkzexample} \end{center} \subsection{Example from Dimitris Kapeta} You need in this example to use \tkzname{mkpos=.2} with \tkzcname{tkzMarkAngle} because the measure of $ \widehat{CAM}$ is too small. Another possiblity is to use \tkzcname{tkzFillAngle}. \begin{center} \begin{tkzexample}[] \begin{tikzpicture}[scale=1.25] \tkzInit[xmin=-5.2,xmax=3.2,ymin=-3.2,ymax=3.3] \tkzClip \tkzDefPoint(0,0){O} \tkzDefPoint(2.5,0){N} \tkzDefPoint(-4.2,0.5){M} \tkzDefPointBy[rotation=center O angle 30](N) \tkzGetPoint{B} \tkzDefPointBy[rotation=center O angle -50](N) \tkzGetPoint{A} \tkzInterLC(M,B)(O,N) \tkzGetFirstPoint{C} \tkzInterLC(M,A)(O,N) \tkzGetSecondPoint{A'} \tkzMarkAngle[fill=blue!25,mkpos=.2, size=0.5](A,C,B) \tkzMarkAngle[fill=green!25,mkpos=.2, size=0.5](A,M,C) \tkzDrawSegments(A,C M,A M,B) \tkzDrawCircle(O,N) \tkzLabelCircle[above left](O,N)(120){$\mathcal{C}$} \tkzMarkAngle[fill=red!25,mkpos=.2, size=0.5cm](C,A,M) \tkzDrawPoints(O, A, B, M, B, C) \tkzLabelPoints[right](O,A,B) \tkzLabelPoints[above left](M,C) \tkzLabelPoint[below left](A'){$A'$} \end{tikzpicture} \end{tkzexample} \end{center} \newpage \subsection{Example 1 from John Kitzmiller } This figure is the last of beamer document. You can find the document on my site Prove $\bigtriangleup LKJ$ is equilateral \begin{center} \begin{tkzexample}[vbox] \begin{tikzpicture}[scale=1.5] \tkzDefPoint[label=below left:A](0,0){A} \tkzDefPoint[label=below right:B](6,0){B} \tkzDefTriangle[equilateral](A,B) \tkzGetPoint{C} \tkzMarkSegments[mark=|](A,B A,C B,C) \tkzDefBarycentricPoint(A=1,B=2) \tkzGetPoint{C'} \tkzDefBarycentricPoint(A=2,C=1) \tkzGetPoint{B'} \tkzDefBarycentricPoint(C=2,B=1) \tkzGetPoint{A'} \tkzInterLL(A,A')(C,C') \tkzGetPoint{J} \tkzInterLL(C,C')(B,B') \tkzGetPoint{K} \tkzInterLL(B,B')(A,A') \tkzGetPoint{L} \tkzLabelPoint[above](C){C} \tkzDrawPolygon(A,B,C) \tkzDrawSegments(A,J B,L C,K) \tkzMarkAngles[fill= orange,size=1cm,opacity=.3](J,A,C K,C,B L,B,A) \tkzLabelPoint[right](J){J} \tkzLabelPoint[below](K){K} \tkzLabelPoint[above left](L){L} \tkzMarkAngles[fill=orange, opacity=.3,thick,size=1,](A,C,J C,B,K B,A,L) \tkzMarkAngles[fill=green, size=1, opacity=.5](A,C,J C,B,K B,A,L) \tkzFillPolygon[color=yellow, opacity=.2](J,A,C) \tkzFillPolygon[color=yellow, opacity=.2](K,B,C) \tkzFillPolygon[color=yellow, opacity=.2](L,A,B) \tkzDrawSegments[line width=3pt,color=cyan,opacity=0.4](A,J C,K B,L) \tkzDrawSegments[line width=3pt,color=red,opacity=0.4](A,L B,K C,J) \tkzMarkSegments[mark=o](J,K K,L L,J) \end{tikzpicture} \end{tkzexample} \end{center} \newpage \subsection{Example 2 from John Kitzmiller } Prove $\dfrac{AC}{CE}=\dfrac{BD}{DF} \qquad$ Another interesting example from John, you can see how to use some extra options like \tkzname{decoration} and \tkzname{postaction} from \TIKZ\ with \tkzname{tkz-euclide}. \begin{center} \begin{tkzexample}[vbox] \begin{tikzpicture}[scale=1.5,decoration={markings, mark=at position 3cm with {\arrow[scale=2]{>}};}] \tkzInit[xmin=-0.25,xmax=6.25, ymin=-0.5,ymax=4] \tkzClip \tkzDefPoints{0/0/E, 6/0/F, 0/1.8/P, 6/1.8/Q, 0/3/R, 6/3/S} \tkzDrawLines[postaction={decorate}](E,F P,Q R,S) \tkzDefPoints{3.5/3/A, 5/3/B} \tkzDrawSegments(E,A F,B) \tkzInterLL(E,A)(P,Q) \tkzGetPoint{C} \tkzInterLL(B,F)(P,Q) \tkzGetPoint{D} \tkzLabelPoints[above right](A,B) \tkzLabelPoints[below](E,F) \tkzLabelPoints[above left](C) \tkzDrawSegments[style=dashed](A,F) \tkzInterLL(A,F)(P,Q) \tkzGetPoint{G} \tkzLabelPoints[above right](D,G) \tkzDrawSegments[color=teal, line width=3pt, opacity=0.4](A,C A,G) \tkzDrawSegments[color=magenta, line width=3pt, opacity=0.4](C,E G,F) \tkzDrawSegments[color=teal, line width=3pt, opacity=0.4](B,D) \tkzDrawSegments[color=magenta, line width=3pt, opacity=0.4](D,F) \end{tikzpicture} \end{tkzexample} \end{center} \newpage \subsection{Example 3 from John Kitzmiller } Prove $\dfrac{BC}{CD}=\dfrac{AB}{AD} \qquad$ (Angle Bisector) \begin{center} \begin{tkzexample}[vbox] \begin{tikzpicture}[scale=1.5] \tkzInit[xmin=-4,xmax=5,ymax=4.5] \tkzClip[space=.5] \tkzDefPoints{0/0/B, 5/0/D} \tkzDefPoint(70:3){A} \tkzDrawPolygon(B,D,A) \tkzDefLine[bisector](B,A,D) \tkzGetPoint{a} \tkzInterLL(A,a)(B,D) \tkzGetPoint{C} \tkzDefLine[parallel=through B](A,C) \tkzGetPoint{b} \tkzInterLL(A,D)(B,b) \tkzGetPoint{P} \begin{scope}[decoration={markings, mark=at position .5 with {\arrow[scale=2]{>}};}] \tkzDrawSegments[postaction={decorate},dashed](C,A P,B) \end{scope} \tkzDrawSegment(A,C) \tkzDrawSegment[style=dashed](A,P) \tkzLabelPoints[below](B,C,D) \tkzLabelPoints[above](A,P) \tkzDrawSegments[color=magenta, line width=3pt, opacity=0.4](B,C P,A) \tkzDrawSegments[color=teal, line width=3pt, opacity=0.4](C,D A,D) \tkzDrawSegments[color=magenta, line width=3pt, opacity=0.4](A,B) \tkzMarkAngles[size=0.7](B,A,C C,A,D) \tkzMarkAngles[size=0.7, fill=green, opacity=0.5](B,A,C A,B,P) \tkzMarkAngles[size=0.7, fill=yellow, opacity=0.3](B,P,A C,A,D) \tkzMarkAngles[size=0.7, fill=green, opacity=0.6](B,A,C A,B,P B,P,A C,A,D) \tkzLabelAngle[pos=1](B,A,C){1} \tkzLabelAngle[pos=1](C,A,D){2} \tkzLabelAngle[pos=1](A,B,P){3}) \tkzLabelAngle[pos=1](B,P,A){4} \tkzMarkSegments[mark=|](A,B A,P) \end{tikzpicture} \end{tkzexample} \end{center} \newpage \subsection{Example 4 from John Kitzmiller } Prove $\overline{AG}\cong\overline{EF} \qquad$ (Detour) \begin{center} \begin{tkzexample}[vbox] \begin{tikzpicture}[scale=2] \tkzInit[xmax=5, ymax=5] \tkzDefPoint(0,3){A} \tkzDefPoint(6,3){E} \tkzDefPoint(1.35,3){B} \tkzDefPoint(4.65,3){D} \tkzDefPoint(1,1){G} \tkzDefPoint(5,5){F} \tkzDefMidPoint(A,E) \tkzGetPoint{C} \tkzFillPolygon[yellow, opacity=0.4](B,G,C) \tkzFillPolygon[yellow, opacity=0.4](D,F,C) \tkzFillPolygon[blue, opacity=0.3](A,B,G) \tkzFillPolygon[blue, opacity=0.3](E,D,F) \tkzMarkAngles[size=0.6,fill=green](B,G,A D,F,E) \tkzMarkAngles[size=0.6,fill=orange](B,C,G D,C,F) \tkzMarkAngles[size=0.6,fill=yellow](G,B,C F,D,C) \tkzMarkAngles[size=0.6,fill=red](A,B,G E,D,F) \tkzMarkSegments[mark=|](B,C D,C) \tkzMarkSegments[mark=s||](G,C F,C) \tkzMarkSegments[mark=o](A,G E,F) \tkzMarkSegments[mark=s](B,G D,F) \tkzDrawSegment[color=red](A,E) \tkzDrawSegment[color=blue](F,G) \tkzDrawSegments(A,G G,B E,F F,D) \tkzLabelPoints[below](C,D,E,G) \tkzLabelPoints[above](A,B,F) \end{tikzpicture} \end{tkzexample} \end{center} \endinput