# Tag Info

6

\documentclass[12pt,border=15pt,pstricks]{standalone} \usepackage{pst-eucl} \begin{document} \begin{pspicture}[showgrid,PointSymbol=none](-1,-3)(11,5) \pstGeonode[PointName=none](2,4){A}(0,0){B}(6,0){C} \pstCircleABC[PosAngle=180]{A}{B}{C}{O} \pstGeonode[PosAngle=60](10,2){M} \pstMiddleAB[PointName=none]{O}{M}{M'} \pstInterCC[DiameterB=\pstDistAB{O}{M},...

6

Edit (based on @chishimotoji's comment) Only PointSymbolA=none is enough to delete the black bullet. But, that is strange to me since the bullet is at point C in coordinates (0,8) (note the change of units). Pass the options PointSymbolA=none (and for other 2 points) to the \pstTriangle command. \documentclass[border=15pt,pstricks,12pt]{standalone} \...

5

Step 1 \documentclass[pstricks,12pt,border=15pt]{standalone} \usepackage{pst-eucl} \begin{document} \begin{pspicture}[showgrid](-1,-3)(7,5) \pstTriangle(2,4){A}(0,0){B}(6,0){C} \pstCircleABC[PosAngle=60]{A}{B}{C}{O} \pstHomO[HomCoef=0.6666]{A}{B}[M]% 2/3 --> 0.6666 \end{pspicture} \end{document} Note: HomCoef cannot accept RPN 2 3 div so I have to ...

5

Not sure if this is the easiest. But it works. \documentclass[pstricks,border=1cm]{standalone} \usepackage{pst-eucl} \begin{document} \begin{pspicture}(8,-6) \pstTriangle(0,-6){B}(8,-6){A}(2,0){C} \pstMarkAngle{A}{B}{C}{} \pstGeonode[PosAngle=180]([nodesep=4]{B}C){E} \pstInterLC[PointSymbol=none,PointName=none]{C}{A}{C}{E}{G}{F} \...

4

I think making double arcs for angle marks has not been implemented yet. You can mimic it by making two angle marks in which one is without label as follows. You can rotate the angle label (about its center of gravity) with \rput because LabelMarkOffset is not for rotating angle label (about its center of gravity). \documentclass[border=15pt,pstricks,12pt]{...

4

The simpler the code the more challenging it becomes. \documentclass[pstricks,12pt]{standalone} \usepackage{pst-eucl} \begin{document} \begin{pspicture}[PointSymbolB=none,PointNameB=](7,5) \pstTriangle(1,1){B}(2.5,4){C}(6,1){A} \pstTranslation[PointName=none,PointSymbol=none]{B}{A}{C}[C'] \pstInterLC[PosAngle=45]{A}{C}{C}{B}{N}{x} \pstInterLC[PosAngle=-90]{...

4

\documentclass[12pt,pstricks]{standalone} \usepackage{pst-eucl} \begin{document} \begin{pspicture}[showgrid](-4,-2)(4,6) \pstGeonode[PosAngle={180,0,-90,90}](-3,0){A}(3,0){C}(-1,-1){B}(0,5){S} \pstTranslation[PosAngle=45]{B}{A}{C}[D] \pstCircleABC[linestyle=none]{S}{A}{C}{I} \pstInterLL[PosAngle=-60]{A}{C}{B}{D}{H} \psline(S)(A)(B)(S)(C)(B) \bgroup \...

3

An alternative tikz version: \documentclass[tikz,border=12pt]{standalone} \usetikzlibrary{calc,bending,arrows.meta,quotes,angles} \usepackage{textcomp} \begin{document} \begin{tikzpicture}[dot/.style={fill,circle,inner sep=1.2pt},>= Stealth]\footnotesize \pgfmathsetmacro\BC{sqrt(1.5*1.5+3*3)} \pgfmathsetmacro\AC{sqrt(3.5*3.5+3*3)} \draw (1,2)coordinate[...

3

\documentclass[pstricks,12pt]{standalone} \usepackage{pst-eucl} \def\b{3 } \begin{document} \begin{pspicture}(-.6,-1)(4.85,4) \pstGeonode[PosAngle={-135,135,-45,45}](0,0){A}(!0 \b){B}(!\b 2 sqrt mul 0){D}(D|B){C} \pstMiddleAB{B}{C}{M} \pstInterLL[PointName=none]{A}{C}{M}{D}{E} \pspolygon[fillstyle=solid,fillcolor=green](E)([nodesep=6pt]{C}E)([offset=6pt,...

3

Cache\pnode(IC_O){I} for I before IC_O gets overridden for J. \documentclass[pstricks,12pt]{standalone} \usepackage{pst-eucl} \begin{document} \begin{pspicture}(-1,-2)(8,6) \psset{linejoin=2,dotsize=2pt,labelsep=3pt,PointNameSep=9pt} \pstGeonode[PointSymbol={*,*,none},PointName={B,C,none},PosAngle={-110,-60}](0,0){B}(7,0){C}(-1,0){Z}(5,0){Y} \pstInterCC[...

2

You should consider the following. Edit Responding to your comment below. \documentclass[pstricks,border=12pt,12pt]{standalone} \usepackage{pst-eucl} \begin{document} \multido{\i=0+30}{12}{% \begin{pspicture}(5,5) \pstGeonode(0,0){O}(4,4){A} \uput[180](A){Angle = $\i^\circ$} \pstSegmentMark[SegmentSymbol=MarkHashhh,MarkAngle=\i]{O}{A} \end{...

2

Just for comparison, anyone wrestling with the pst-eucl syntax and documentation, might like to try this type of thing in Metapost, using the elegant implicit definition of linear variables. \documentclass[border=5mm]{standalone} \usepackage{luatex85} \usepackage{luamplib} \begin{document} \mplibtextextlabel{enable} \begin{mplibcode} vardef angle_mark(expr ...

2

\documentclass[12pt,pstricks,border=12pt]{standalone} \usepackage{pst-eucl} \begin{document} \begin{pspicture}[showgrid,PosAngle=30](-1,-3)(7,5) \pstTriangle(2,4){A}(0,0){B}(6,0){C} \pstCircleABC[PosAngle=60]{A}{B}{C}{O} { \psset{linestyle=none,PointName=none,PointSymbol=none} \pstBissectBAC{B}{A}{C}{A'} \pstBissectBAC{A}{C}{B}{C'} } { \...

1

Herbert's solution can be generalized by replacing his \rput{(C)}(E){\psframe[fillcolor=red,fillstyle=solid](0.2,0.2)}%%%%% with \pnode(C){c}% temporary \pnode(E){e}% temporary \rput{!N-c.y N-e.y sub N-c.x N-e.x sub atan}(E){\psframe[fillcolor=red,fillstyle=solid](0.2,0.2)}%%%%% Red alert: the temporary nodes are important and needed because it is one ...

1

My own solution. \documentclass[pstricks,border=15pt]{standalone} \usepackage{pst-eucl} \begin{document} \begin{pspicture}(6,-4) \pstTriangle(0,-4){B}(6,-4){A}(2,0){C} \pstMarkAngle{A}{B}{C}{} \pstGeonode[PosAngle=180]([nodesep=3]{B}C){E} \pstRotation[RotAngle=\pstAngleAOB{B}{A}{C},PointName=none,PointSymbol=none]{E}{C}[C'] \pstInterLL[...

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