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Diffstat (limited to 'macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-triangles.tex')
-rw-r--r-- | macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-triangles.tex | 192 |
1 files changed, 98 insertions, 94 deletions
diff --git a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-triangles.tex b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-triangles.tex index 7eb2875ec7..27d5304b6a 100644 --- a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-triangles.tex +++ b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-triangles.tex @@ -18,6 +18,7 @@ The following macros will allow you to define or construct a triangle from \tkzn \item \tkzname{cheops} determines a third point such that the triangle is isosceles with side measurements proportional to $2$, $\Phi$ and $\Phi$. \end{itemize} +\newpage \begin{NewMacroBox}{tkzDefTriangle}{\oarg{local options}\parg{A,B}}% The points are ordered because the triangle is constructed following the direct direction of the trigonometric circle. This macro is either used in partnership with \tkzcname{tkzGetPoint} or by using \tkzname{tkzPointResult} if it is not necessary to keep the name. @@ -34,16 +35,17 @@ options & default & definition \\ \TOline{pythagoras}{equilateral}{same as above} \TOline{egyptian}{equilateral}{same as above} \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{gold}{equilateral}{B rectangle and $AB/AC = \Phi$} +\TOline{euclid} {equilateral}{angles of 72, 72 and 36 degrees, $A$ is the apex} +\TOline{golden} {equilateral}{angles of 72, 72 and 36 degrees, $C$ is the apex} +\TOline{sublime} {equilateral}{angles of 72, 72 and 36 degrees, $C$ is the apex} \TOline{cheops} {equilateral}{AC=BC, AC and BC are proportional to $2$ and $\Phi$.} \TOline{swap} {false}{gives the symmetric point with respect to $AB$} \bottomrule \end{tabular} \medskip -\tkzcname{tkzGetPoint} allows you to store the point otherwise \tkzname{tkzPointResult} allows for immediate use. +\emph{\tkzcname{tkzGetPoint} allows you to store the point otherwise \tkzname{tkzPointResult} allows for immediate use.} \end{NewMacroBox} \subsubsection{Option \tkzname{equilateral}} @@ -94,6 +96,8 @@ The angles are 30, 60 and 90 degrees. \tkzLabelAngle[pos=0.8](A,C,B){$60^\circ$} \tkzDrawSegments(A,B) \tkzDrawSegments[new](A,C B,C) + \tkzLabelPoints(A,B) + \tkzLabelPoints[above](C) \end{tikzpicture} \end{tkzexample} @@ -108,10 +112,10 @@ This triangle has sides whose lengths are proportional to 3, 4 and 5. \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) + \tkzLabelPoints[above](A,B) + \tkzLabelPoints[new](C) \end{tikzpicture} \end{tkzexample} @@ -148,6 +152,23 @@ This triangle has sides whose lengths are proportional to 3, 4 and 5. \end{tikzpicture} \end{tkzexample} +\subsubsection{Option \tkzname{euclid}} +\tkzimp{Euclid} and \tkzimp{golden} 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(C) + \tkzLabelPoints[above](A,B) + \tkzLabelAngle[pos=0.8](A,B,C){$72^\circ$} + \tkzLabelAngle[pos=0.8](B,C,A){$72^\circ$} + \tkzLabelAngle[pos=0.8](C,A,B){$36^\circ$} +\end{tikzpicture} +\end{tkzexample} + \subsubsection{Option \tkzname{isosceles right}} \begin{tkzexample}[latex=7 cm,small] \begin{tikzpicture} @@ -158,7 +179,8 @@ This triangle has sides whose lengths are proportional to 3, 4 and 5. \tkzDrawPolygons(A,B,C) \tkzDrawPoints(A,B,C) \tkzMarkRightAngles(A,C,B) - \tkzLabelPoints(A,B,C) + \tkzLabelPoints(A,B) + \tkzLabelPoints[above](C) \end{tikzpicture} \end{tkzexample} @@ -170,31 +192,14 @@ This triangle has sides whose lengths are proportional to 3, 4 and 5. \tkzGetPoint{C} \tkzDrawPolygon(A,B,C) \tkzDrawPoints(A,B,C) - \tkzLabelPoints(B) \tkzLabelPoints[below](A,C) - \tkzLabelAngle[pos=0.8](C,A,B){$36^\circ$} - \tkzLabelAngle[pos=0.8](A,B,C){$72^\circ$} - \tkzLabelAngle[pos=0.8](B,C,A){$72^\circ$} -\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){$36^\circ$} + \tkzLabelPoints[above](A,B) + \tkzLabelPoints[below](C) + \tkzMarkRightAngle(A,B,C) + \tkzText(0,-2){$\dfrac{AC}{AB}=\varphi$} \end{tikzpicture} \end{tkzexample} +\clearpage \subsection{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. @@ -202,7 +207,6 @@ The centers of some triangles have been defined in the "points" section, here it \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 \begin{tabular}{lll}% \toprule @@ -223,7 +227,6 @@ options & default & definition \\ \TOline{name} {empty}{used to name the vertices} \midrule \end{tabular} - \end{NewMacroBox} \subsubsection{How to name the vertices} @@ -243,20 +246,22 @@ The geometric centroid of the polygon vertices of a triangle is the point $G$ ( In the following example, we obtain the Euler circle which passes through the previously defined points. \begin{tkzexample}[latex=7cm,small] -\begin{tikzpicture}[rotate=90,scale=.75] - \tkzDefPoints{0/0/A,6/0/B,0.8/4/C} - \tkzDefTriangleCenter[centroid](A,B,C) - \tkzGetPoint{M} - \tkzDefSpcTriangle[medial,name=M](A,B,C){_A,_B,_C} - \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} + \begin{tikzpicture}[rotate=90,scale=.75] + \tkzDefPoints{0/0/A,6/0/B,0.8/4/C} + \tkzDefTriangleCenter[centroid](A,B,C) + \tkzGetPoint{M} + \tkzDefSpcTriangle[medial,name=M](A,B,C){_A,_B,_C} + \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) + \tkzLabelPoints[above](B) + \tkzLabelPoints[below](A,C,M_B) + \tkzLabelPoints[right](M_C) + \tkzLabelPoints[left](M_A) + \tkzLabelPoints[font=\scriptsize](M) + \end{tikzpicture} \end{tkzexample} \subsubsection{Option \tkzname{in} or \tkzname{incentral} } @@ -270,18 +275,17 @@ respective opposite sides. \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture}[scale=1] - \tkzDefPoints{ 0/0/A,5/0/B,1/3/C} + \tkzDefPoints{ 0/0/A,5/0/B,2/3/C} \tkzDefSpcTriangle[in,name=I](A,B,C){_a,_b,_c} - \tkzInCenter(A,B,C)\tkzGetPoint{I} + \tkzDefCircle[in](A,B,C) \tkzGetPoints{I}{a} + \tkzDrawCircle(I,a) \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,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) + \tkzDrawPoints(A,B,C,I,I_a,I_b,I_c) + \tkzLabelPoints[below](A,B,I_c) + \tkzLabelPoints[above left](I_b) + \tkzLabelPoints[above right](C,I_a) \end{tikzpicture} \end{tkzexample} @@ -317,15 +321,16 @@ We obtain the intersections of the bisectors with the sides. \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} + \tkzDefCircle[in](A,B,C) \tkzGetPoints{I}{i} + \tkzDrawCircle(I,i) \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) - \tkzAutoLabelPoints[center=I,red,font=\scriptsize]% -(A,B,C) + \tkzLabelPoints[right](X_a) + \tkzLabelPoints[left](X_b) + \tkzLabelPoints[above](C) + \tkzLabelPoints[below](A,B,X_c) \end{tikzpicture} \end{tkzexample} @@ -355,10 +360,13 @@ We obtain the points of contact of the exinscribed circles as well as the triang \tkzDrawPolygon(A,B,C) \tkzDrawPolygon[new](T_a,T_b,T_c) \tkzDrawPoints(A,B,C,N_a) -\tkzLabelPoints(N_a) -\tkzAutoLabelPoints[center=N_a](A,B,C) -\tkzAutoLabelPoints[center=G,new, - dist=.4](T_a,T_b,T_c) +\tkzDrawPoints[new](T_a,T_b,T_c) +\tkzLabelPoints[below left](A) +\tkzLabelPoints[below](N_a,B) +\tkzLabelPoints[above](C) +\tkzLabelPoints[new,below left](T_b) +\tkzLabelPoints[new,below right](T_c) +\tkzLabelPoints[new,right=6pt](T_a) \tkzMarkRightAngles[fill=gray!15](J_a,T_a,B J_b,T_b,C J_c,T_c,A) \end{tikzpicture} @@ -384,7 +392,6 @@ Given a triangle $ABC$, the triangle $H_AH_BH_C$ whose vertices are endpoints of \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) @@ -401,7 +408,7 @@ The Feuerbach triangle is the triangle formed by the three points of tangency of The points of tangency define the Feuerbach triangle. \begin{tkzexample}[latex=8cm,small] -\begin{tikzpicture}[scale=1.25] +\begin{tikzpicture}[scale=1] \tkzDefPoint(0,0){A} \tkzDefPoint(3,0){B} \tkzDefPoint(0.5,2.5){C} @@ -412,16 +419,18 @@ 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) + \tkzLabelPoints[below left](J_a,J_b,J_c) \tkzClipBB \tkzShowBB \tkzDrawCircle[purple](N,F_a) \tkzDrawPolygon(A,B,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) + \tkzDrawPoints[blue](J_a,J_b,J_c,% + F_a,F_b,F_c,A,B,C) + \tkzLabelPoints(A,B,F_c) + \tkzLabelPoints[above](C) + \tkzLabelPoints[right](F_a) + \tkzLabelPoints[left](F_b) \end{tikzpicture} \end{tkzexample} @@ -442,8 +451,11 @@ The tangential triangle is the triangle $T_aT_bT_c$ formed by the lines tangent \tkzDefCircle[circum](A,B,C) \tkzGetPoint{O} \tkzDrawCircle(O,A) - \tkzLabelPoints(A,B,C) - \tkzLabelPoints[new](T_a,T_b,T_c) + \tkzLabelPoints(A) + \tkzLabelPoints[above](B) + \tkzLabelPoints[left](C) + \tkzLabelPoints[new](T_b,T_c) + \tkzLabelPoints[new,left](T_a) \end{tikzpicture} \end{tkzexample} @@ -487,27 +499,27 @@ The Euler triangle of a triangle $ABC$ is the triangle $E_AE_BE_C$ whose vertice \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} + \tkzDefExCircle(A,B,C) \tkzGetPoints{I}{i} + \tkzDefExCircle(C,A,B) \tkzGetPoints{J}{j} + \tkzDefExCircle(B,C,A) \tkzGetPoints{K}{k} \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} + \tkzInterLC(I,C)(I,i) \tkzGetSecondPoint{Fc} + \tkzInterLC(J,B)(J,j) \tkzGetSecondPoint{Fb} + \tkzInterLC(K,A)(K,k) \tkzGetSecondPoint{Fa} \tkzDrawLines[add=1.5 and 1.5](A,B A,C B,C) - \tkzDrawCircle[euler,orange](A,B,C) \tkzGetPoint{E} + \tkzDefCircle[euler](A,B,C) \tkzGetPoints{E}{e} + \tkzDrawCircle[orange](E,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}) + \tkzDrawCircles(J,j I,i K,k) \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} - \subsubsection{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 $ABC$. @@ -521,7 +533,10 @@ The symmedial triangle$ K_AK_BK_C$ is the triangle whose vertices are the inters \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) +\tkzLabelPoints(A,B,K,K_C) +\tkzLabelPoints[above](C) +\tkzLabelPoints[right](K_A) +\tkzLabelPoints[left](K_B) \end{tikzpicture} \end{tkzexample} @@ -535,23 +550,12 @@ arguments & example & explanation \\ \midrule \end{tabular} +\medskip \emph{The triangle is unchanged.} \end{NewMacroBox} \subsubsection{Modification of the \tkzname{school} triangle} -This triangle is constructed from the segment $[AB]$ on $[A,x)$ -\begin{tkzexample}[latex=7cm,small] -\begin{tikzpicture} - \tkzDefPoints{0/0/A,4/0/B,6/0/x} - \tkzDefTriangle[school](A,B) - \tkzGetPoint{C} - \tkzDrawSegments(A,B B,x) - \tkzDrawSegments(A,C B,C) - \tkzDrawPoints(A,B,C) - \tkzLabelPoints(A,B,C,x) - \tkzMarkRightAngles(C,B,A) -\end{tikzpicture} -\end{tkzexample} +This triangle is constructed from the segment $[AB]$ on $[A,x)$. If we want the segment $[AC]$ to be on $[A,x)$, we just have to swap $B$ and $C$. |