Tikz intersection without extending the paths till they intersect

Question

The below MWE draws a certain isosceles triangle:

\documentclass[tikz, border=1cm]{standalone}
\usetikzlibrary{intersections, positioning}

\begin{document}%

\begin{tikzpicture}
\coordinate (A) at (0,0);
\coordinate (B) at (2,0);    
\path[name path=AC] (A)--++(80:6cm);
\path[name path=BC] (B)--++(100:6cm);
\path[name intersections={of=AC and BC, by=C}];
\draw (A)--(B)--(C)--cycle;
\end{tikzpicture}

\end{document}

However, if the named paths are not extended long enough, for example \path[name path=AC] (A)--++(80:1cm); I get an error. Though this is not a problem (that one just enters an arbitrary big number), is there a way (a notation perhaps, or whatever) to tell Tikz that the paths are intersecting (even if the defined path is not long enough)?

There is a (deprecated) intersection syntax that you could use, see Ulrike's answer to https://tex.stackexchange.com/questions/419504/tikz-extend-an-inclined-line/419508#419508 — Torbjørn T., Jun 09 '18 at 08:49

AndréC · Answer 1 · 2020-08-15T17:11:38.243

There is a syntax that still works for the intersection of two lines. It is no longer documented since version 3 of TikZ, but you will find its syntax in the old manuals version 1 and 2.

Its interest is that the intersection is found even if the segments are not intersecting, it is enough that the lines carrying them are not parallel.

It does not need any TikZ library to be used.

\documentclass[tikz, border=1cm]{standalone}
\begin{document}%
\begin{tikzpicture}
\coordinate (A) at (0,0);
\coordinate (B) at (2,0);

\path(A)--++(80:6cm)coordinate(u);
\path(B)--++(100:6cm)coordinate(v);
%\path[name intersections={of=AC and BC, by=C}];
\draw (A)--(B)--(intersection of A--u and B--v)--cycle;
\end{tikzpicture}
\end{document}

score 1 · Answer 2 · answered Aug 15 '20 at 10:21

Yes, you can, at least when the intersecting curves are lines. The idea is to use intersection in a let ... in action. I saw this construction in AndréC answer Drawing a circle through 3 non-collinear points.

I consider that two lines are given through a point and a vector for each---A and v for the first and B and w for the second. We need their meeting point C.

\documentclass[tikz, border=1cm]{standalone}
\usetikzlibrary{calc, intersections}
\begin{document}
\begin{tikzpicture}[every node/.style={scale=.8}]
  % points and directions
  \path
  (0, 0) coordinate (A) (80:1) coordinate (v)
  (2, 0) coordinate (B) (100:1) coordinate (w);
% intersection point
  \path let
  \p1 = (A),
  \p2 = ($(A)+(v)$),
  \p3 = (B),
  \p4 = ($(B)+(w)$),
  \p5 = (intersection of \p1--\p2 and \p3--\p4)
  in
  (\p5) coordinate (C);
\draw (A) -- (B) -- (C) -- cycle;
\draw[blue, very thick, ->] (A) -- +(v) node[above left] {$v$};
  \draw[blue, very thick, ->] (B) -- +(w) node[above right] {$w$};
\foreach \P/\pos in {A/below left, B/below right, C/above left}{%
    \draw[fill=white] (\P) circle (1.2pt) node[\pos] {$\P$};
  }
\end{tikzpicture}
\end{document}

Tikz intersection without extending the paths till they intersect

2 Answers2