Tikz 2 triangles picture

Question

I am working upon an illustration for a groups theory article. I want to draw a process of reflection of a triangle across a ray coming out of one of its vertices.
So far I have the following:
documentclass{standalone}
usepackage{tikz}
usetikzlibrary{shapes.geometric,decorations.markings,arrows,positioning}

tikzset{
    buffer/.style={
        draw,
        regular polygon,
        regular polygon sides=3,
        node distance=3cm,
        minimum height=6em
    }
}

begin{document}
begin{tikzpicture}
  node[buffer] (T) {};
  coordinate [label=left:B] (B) at (-0.9cm, -0.6);
  coordinate [label=above:A] (A) at (0,1.04cm);
  node at (3.3em, -0.5) {C};
  draw (A) -- (B-|A) -- (-90:1.5cm) node[above right]{$l_1$};
  node[right = 1cm of T] (Arr) {$Longrightarrow$};
  node[buffer, right = 0.5cm of Arr] (T1) {};
end{tikzpicture}
end{document}

How can I get the coordinates of the vertices of the triangle? Now I am calculating them manually, which doesn't look good.

How to position left and right triangles, so they would be equidistant from the arrow without setting the manual padding?

Is it possible to move the arrow a upwards, so it would be at the vertical centre of the triangles?

ferahfeza · Accepted Answer

1-) You can use the corners option and get the corners as a node for regular polygon.
2-) Draw second triangle within scope with xshift=Xcm. Then use the x coordinate for arrow as (X/2,0).
3-) Can be manually adjusted giving appropriate y value.
documentclass{standalone}
usepackage{tikz}
usetikzlibrary{shapes.geometric,decorations.markings,arrows,positioning}

tikzset{
    buffer/.style={
        draw,
        regular polygon,
        regular polygon sides=3,
        node distance=3cm,
        minimum height=6em
    }
}

begin{document}
begin{tikzpicture}
  node[buffer] (T) {};
  
  begin{scope}[xshift=4cm];
   node[buffer] (T1) {};
  end{scope}

node at (2,0.25) (Arr) {$Longrightarrow$};
  node at (T.corner 1)(A)[above]{A};  
  node at (T.corner 2)(B)[left]{B};
  node at (T.corner 3)(C)[right]{C};  
  draw (A) -- (B-|A) -- (-90:1.5cm) node[above right]{$l_1$}; 
 
end{tikzpicture}
end{document}

Bernard · Answer

For fun, a simple code with pstricks, more specifically with pst-eucl.  I drew two pairs of points, symmetric w.r.t. the origin, and asked to build two equilateral triangles from these segments, then drew the altitude of the first triangle through $A$ and finally placed the implication symbol between the triangles.
 documentclass[border=6pt]{standalone}
 usepackage{pst-eucl}%

begin{document}

begin{pspicture}(-4.5,-1)(4,3)%
    psset{PointSymbol=none, PtNameMath=false, linejoin=1}
    pstGeonode[PosAngle={180,0}](-4,0){B}(-1,0){C} pstETriangleAB[PosAngle=90]{B}{C}{A}
    rput(0.1,1.3 ){$Longrightarrow$}
    psset{PointName=none}
    pstGeonode[PointName=none](1,0){E}(4,0){F}pstETriangleAB[PointName=none] {E}{F}{D}
    pstProjection{B}{C}{A}[H]
    pstLineAB[nodesepB=-1, linewidth=0.6pt]{A}{H}naput[npos=0.9, labelsep=2pt]{$l_1 $}
    end{pspicture}

end{document}

Tikz 2 triangles picture

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