2
\documentclass[pstricks,border=15pt]{standalone}
\usepackage{pstricks-add,pst-eucl}%
\begin{document}

\begin{pspicture}[showgrid](0,-3)(8,4)
\pnodes(3,3){A}(1,-1){B}(7,-1){C}

\pstMiddleAB[PosAngle=135]{A}{B}{M}
\pstMiddleAB{A}{C}{N}
\pstMiddleAB{M}{N}{I}
 \def\figA{\pspolygon(A)(M)(N)}
 \figA%
%\psrotate(I){-50}{\pspolygon(3,3)(2,1)(5,1)}%
\psrotate(I){-50}{\figA}
\end{pspicture}

\end{document}

With \psrotate(I){-50}{\pspolygon(3,3)(2,1)(5,1)} , I get

enter image description here

With \psrotate(I){-50}{\figA} , I don't get anything

enter image description here

Assume that A=(3.2783,3.5876), B=(1.07368,-1.235), C=(7.777,-1.3336).

How to get the output without calculating M and N ?

1

\psrotate doesn't work with nodes (Nodes are fix points in the current system).

\documentclass[pstricks,border=15pt]{standalone}
\usepackage{pstricks-add,pst-eucl}%
\begin{document}

\begin{pspicture}[showgrid](0,-3)(8,4)
    \pnodes(3,3){A}(1,-1){B}(7,-1){C}
    \pstMiddleAB[PosAngle=135]{A}{B}{M}
    \pstMiddleAB{A}{C}{N}
    \pstMiddleAB{M}{N}{I}
    \def\figA{\pspolygon(A)(M)(N)}
    \figA
    \pstRotation[RotAngle=-50]{I}{A,M,N}[a,m,n]
    \pspolygon[linecolor=red](a)(m)(n)
\end{pspicture}

\end{document}

enter image description here

0

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