Tukukkk

This code does not work with named coordinates (such as the following code). How can I reflect the blue line over the red line by using coordinate names.

documentclass[tikz]{standalone}



begin{document}

begin{tikzpicture}[scale=0.55]

coordinate (A) at (0,0);

coordinate (B) at (1,1);

coordinate (C) at (1,2);

coordinate (D) at (2,0);

coordinate (E) at (2,3);



draw[blue] (B)--(A)--(C);

draw[red] (D)--(E);

end{tikzpicture}

end{document}

One possibility is using the tkz-euclide package.

To define A1 the mirror image of the point A with respect to the line DE use: tkzDefPointBy[reflection=over D--E](A) tkzGetPoint{A1}

documentclass[border=1cm,tikz]{standalone}

usepackage{tkz-euclide}

begin{document}

begin{tikzpicture}

draw[help lines,dashed](0,0)grid(4,4);

coordinate (A) at (0,0);

coordinate (B) at (1,1);

coordinate (C) at (1,2);

coordinate (D) at (2,0);

coordinate[label=E] (E) at (2,3);



tkzDefPointBy[reflection=over D--E](A) tkzGetPoint{A1}

tkzDefPointBy[reflection=over D--E](B) tkzGetPoint{B1}

tkzDefPointBy[reflection=over D--E](C) tkzGetPoint{C1}



draw[blue] (B)--(A)--(C);

draw[red] (D)--(E);



draw [green] (B1)--(A1)--(C1);

end{tikzpicture}

end{document}

UPDATE: Here is a simple style reflect at that allows you to reflect straight lines at whatever line you are interested in. No auxiliary coordinates and the like are needed. (Note, however, that you need to use to instead of -- since this is a transformation that depends on the point, of course.)

documentclass[tikz]{standalone}

usetikzlibrary{calc}

tikzset{reflect at/.style args={#1--#2}{to path={%

($2*($(#1)!(tikztostart)!(#2)$)-(tikztostart)$)

-- ($2*($(#1)!(tikztotarget)!(#2)$)-(tikztotarget)$)

}}}

begin{document}

begin{tikzpicture}[scale=0.55]

coordinate (A) at (0,0);

coordinate (B) at (1,1);

coordinate (C) at (1,2);

coordinate (D) at (2,0);

coordinate (E) at (2,3);



draw[blue] (B)--(A)--(C);

draw[red] (D)--(E);

draw[blue,reflect at=D--E] (B) to (A) (A) to (C);

end{tikzpicture}

end{document}

OLD ANSWER: Paul Gaborit's solution seems to work.

documentclass[tikz]{standalone}

usetikzlibrary{spy,decorations.fractals}

tikzset{

  mirror scope/.is family,

  mirror scope/angle/.store in=mirrorangle,

  mirror scope/center/.store in=mirrorcenter,

  mirror setup/.code={tikzset{mirror scope/.cd,#1}},

  mirror scope/.style={mirror setup={#1},spy scope={

      rectangle,lens={rotate=mirrorangle,yscale=-1,rotate=-1*mirrorangle},size=80cm}},

}

newcommandmirror[1]{spy[overlay,#1] on (mirrorcenter) in node at (mirrorcenter)}



begin{document}

begin{tikzpicture}

coordinate (A) at (0,0);

coordinate (B) at (1,1);

coordinate (C) at (1,2);

coordinate (D) at (2,0);

coordinate (E) at (2,3);

  draw [help lines] (0,0) grid (4,3);

  begin{scope}[mirror scope={center={2,0},angle=90}]

    draw[blue] (B) -- (A) -- (C);

    draw[red]  (D) -- (E);

    mirror;

  end{scope}

end{tikzpicture}

end{document}

ADDENDUM: A style that computes the reflected coordinates. Unfortunately, the syntax in this version requires to specify the coordinate twice, e.g. there are two Bs in ([reflect=B at D--E]B), and it does not work well with global transformations like scale=0.55. Other than that it uses this answer which shows how to compute the orthogonal projection of a point on a line. (Of course, I could write that I got it from the pgfmanual, but the truth is that I got it from Jake's nice answer...)

documentclass[tikz]{standalone}

usetikzlibrary{calc}

tikzset{reflect/.style args={#1 at #2--#3}{shift={%

($2*($(#2)!(#1)!(#3)$)-2*(#1)$)

}}}

begin{document}

begin{tikzpicture}

coordinate (A) at (0,0);

coordinate (B) at (1,1);

coordinate (C) at (1,2);

coordinate (D) at (2,0);

coordinate (E) at (2,3);



draw[blue] (B)--(A)--(C);

draw[red] (D)--(E);



draw[orange] ([reflect=B at D--E]B) -- ([reflect=A at D--E]A)

-- ([reflect=C at D--E]C);

end{tikzpicture}

end{document}