summaryrefslogtreecommitdiff
path: root/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-triangles.tex
diff options
context:
space:
mode:
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.tex192
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 $C$.