# How to draw incircle (inscribed circle)?

## First Question

\documentclass[pstricks,12pt,border=12pt]{standalone}
\usepackage{pst-eucl}
\begin{document}
\begin{pspicture}[showgrid](-3,-2)(2,2)
\psset{PointSymbol=none,CodeFig,CodeFigColor=red}
\pstTriangle(1,1){A}(-2,1){C}(-1,-1){B}
\pstProjection{A}{B}{C}[I]
\pstProjection{A}{C}{B}[J]
\pstProjection{C}{B}{A}[K]
\end{pspicture}
\end{document}


How to modify the size of right angle mark?

## Second Question

In TikZ, we can draw both excircle (escribed circle) and incircle (inscribed circle) of a triangle ABC by passing circum and in, respectively as follows.

\tkzDrawCircle[circum](A,B,C)
\tkzDrawCircle[in](A,B,C)


In PSTricks, what options do I have to pass to \pstCircleABC[<options>]{A}{B}{C}{O} to draw an incircle (of the triangle) ABC? I know that we can use \pstBissectBAC, \pstProjection and \pstCircleOA but it is too long.

## Third Question

For the following macros,

\pstLineAB[Options]{A}{B}
\pstCircleOA[Options]{O}{A}
\pstCircleAB[Options]{O}{A}
\pstDistAB[Options]{A}{B}
\pstDistVal[Options]{x}
\pstInterFF[Options]{f}{g}{x0}{M}
\pstInterFL[Options]{f}{A}{B}{x0}{M}
\pstInterFC[Options]{f}{O}{A}{x0}{M}
\psGetDistanceAB[Options](x1,y1)(x2,y2){<name>}
\psGetAngleABC[Options](x1,y1)(x2,y2)(x3,y3){<symbol>}


what are all possible optional arguments?

Please provide always complete examples!

1.

\documentclass[pstricks]{standalone}
\usepackage{pst-eucl}
\begin{document}

\begin{pspicture}[showgrid](-3,-2)(2,2)
\psset{PointSymbol=none,CodeFig,CodeFigColor=red,RightAngleSize=0.2}
\pstTriangle(1,1){A}(-2,1){C}(-1,-1){B}
\pstProjection{A}{B}{C}[I]
\pstProjection{A}{C}{B}[J]
\pstProjection{C}{B}{A}[K]
\end{pspicture}

\end{document}


2.Write an own macro \pstInnerCircle, which isn't too complicated

\documentclass[pstricks]{standalone}
\usepackage{pst-eucl}
\makeatletter
\def\pstTriangleIC{\pst@object{pstTriangleIC}}
\def\pstTriangleIC@i#1#2#3{%
\begingroup
\psset{PointSymbol=none,PointName=none,linestyle=none}
\pstBissectBAC{#3}{#1}{#2}{IC_BC}
\pstBissectBAC{#2}{#3}{#1}{IC_AB}
\pstInterLL{#3}{IC_AB}{#1}{IC_BC}{IC_M}
\pstProjection[CodeFig=false]{A}{B}{IC_M}[IC_MAB]
\endgroup
\begingroup
\use@par
\pstCircleOA{IC_M}{IC_MAB}
\endgroup
}
\makeatother
\begin{document}

\begin{pspicture}[showgrid](-3,-2)(2,2)
\psset{PointSymbol=none,CodeFig,CodeFigColor=red,RightAngleSize=0.2}
\pstTriangle(1,1){A}(-2,1){C}(-1,-1){B}
\pstProjection{A}{B}{C}[I]
\pstProjection{A}{C}{B}[J]
\pstProjection{C}{B}{A}[K]
\pstTriangleIC[linestyle=dashed,linecolor=blue]{A}{B}{C}
\end{pspicture}

\end{document}


• The macro \pstTriangleIC and \pstTriangleOC will be part of the next update of pst-eucl – user2478 Dec 11 '18 at 8:04
• Yes, except that the origin is internally named OC_O for \pstTriangleOC, because it has only three parameters and \pstCircleABC has four. – user2478 Dec 11 '18 at 8:34