Draw height in Tikz triangle

Question

I am trying to draw the height line in Tikz. It should go from B and right down to the lower part of the triangle. But that isn't happening with the following MWE:

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{positioning, calc}

\begin{document}

\begin{tikzpicture}[scale=1.25]%,cap=round,>=latex]
   \coordinate [label=left:$A$] (A) at (-2cm,-1.cm);
   \coordinate [label=right:$C$] (C) at (2.2cm,-1.0cm);
   \coordinate [label=above:$B$] (B) at (1cm,1.0cm);
   \draw (A) -- node[sloped,above] {c} (B) -- node[sloped,above,] {a} (C) -- node[below] {b} (A);
   \draw[dashed] (B) -- (b) ;
\end{tikzpicture}

\end{document}

You can use let to do this: \draw[dashed] let \p1 = (B) in (B) -- (\x1, -1cm) ;. — osjerick, Nov 10 '15 at 14:07

score 3 · Answer 1 · answered Nov 10 '15 at 14:10

Here a MWE using the calc-library for drawing the perpendicular on the b-edge.

For the dashed line (A)!(B)!(C) means: project point B on the line AC. For the dotted line you need to give the node with label b a name.

\documentclass[tikz, border=5mm]{standalone}
\usetikzlibrary{calc}

\begin{document}
  \begin{tikzpicture}
    \coordinate [label=left:$A$] (A) at (-2, -1);
    \coordinate [label=above:$B$] (B) at (1, 1);
    \coordinate [label=right:$C$] (C) at (2.2, -1);
    \draw (A) -- node [sloped, above] (c) {c} (B) -- node [sloped, above] (a) {a} (C) -- node [below] (b) {b} cycle;        
    \draw [dashed] (B) -- ($(A)!(B)!(C)$);
    \draw [dotted] (B) -- (b);
  \end{tikzpicture}
\end{document}

score 2 · Accepted Answer · answered Nov 10 '15 at 14:10

When line A--C is horizontal, like in this case, command

\draw[dashed] (B) -- (B|-A);

will also draw triangle height without loading calc library.

\documentclass[a4paper,10pt]{article}
\usepackage[utf8]{inputenc}
\usepackage{tikz}
\usetikzlibrary{positioning, calc}

\begin{document}

\begin{tikzpicture}[scale=1.25]%,cap=round,>=latex]
\coordinate [label=left:$A$] (A) at (-2cm,-1.cm);
\coordinate [label=right:$C$] (C) at (2.2cm,-1.0cm);
\coordinate [label=above:$B$] (B) at (1cm,1.0cm);
\draw (A) -- node[sloped,above] {c} (B) -- node[sloped,above,] {a} (C) -- node[below] {b} (A);
\draw[dashed] (B) -- (A-|B) ;

\end{tikzpicture}

\end{document}

score 0 · Answer 3 · answered Nov 10 '15 at 14:07

Note the error that says no shape b is known.

Your node with the text b has no name, i.e. it is made with node[below]{node text} rather than node[below](node name) {node text}.

If you want a line down to the b, make it into node[below](b){b}, but that is not a normal. To get a normal you can do \draw (B) -- (B|-A);, where B|-A is the point with the x-coordinate of B and the y-coordinate of A.

\documentclass[a4paper,10pt]{article}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}[scale=1.25]%,cap=round,>=latex]
\coordinate [label=left:$A$] (A) at (-2cm,-1.cm);
\coordinate [label=right:$C$] (C) at (2.2cm,-1.0cm);
\coordinate [label=above:$B$] (B) at (1cm,1.0cm);
\draw (A) -- node[sloped,above] {c} (B) -- node[sloped,above,] {a} (C) -- node[below] (b) {b} (A);
\draw[dashed] (B) -- (b) ;
\draw[blue,dashed] (B) -- (B|-A) ;
\end{tikzpicture}
\end{document}

Draw height in Tikz triangle

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