Drawing a ray that bisects an angle

Question

A triangle is drawn via the following code. (In the display, only two of the vertices are labeled. In the code, the vertices are labeled A, B, and C.) I extend sides AC and BC. I want to draw a ray that bisects the (acute) angle made by these extended sides.

I have specified \coordinate (A) at (0,0); and \coordinate (C) at (290:3.25);; so, I know the extension of side AC is at an angle of 70 degrees below the horizontal line through C. The angle that the extension of side BC is above the horizontal line through C would have to be expressed in terms of an arc tangent of a messy expression! Can this ray be drawn using the calc package?

\documentclass{amsart}
\usepackage{tikz}
\usetikzlibrary{calc,angles,positioning,intersections}


\begin{tikzpicture}

%A is the vertex of an angle of 55 degrees; the sides of this angle are AB
%and AC. $AB = 22$ and $AC = 13$.
\coordinate (A) at (0,0);
\node (vertex_A) at ($(A) + (82.5:7.5pt)$){$A$};
\coordinate (B) at (235:5.5);
\node (vertex_B) at ($(B) + (225:7.5pt)$){$B$};
\coordinate (C) at (290:3.25);
\draw[name path=AB] (A) -- (B);
\draw[name path=AC] (A) -- (C);
\draw[name path=BC] (B) -- (C);


%These commands label the lengths of AB and of AC.
\coordinate (AB_midpoint) at ($(A)!0.5!(B)$);
\coordinate (AC_midpoint) at ($(A)!0.5!(C)$);


%These commands draw the altitude of the triangle from C. The foot of the altitude is
%labeled P.
\coordinate (P) at ($(A)!(C)!(B)$);
\draw[dashed] (C) -- (P);
\coordinate (PC_midpoint) at ($(P)!0.5!(C)$);

%The following commands make the right-angle mark.
\coordinate (U) at ($(P)!4mm!-45:(A)$);
\draw (U) -- ($(P)!(U)!(A)$);
\draw (U) -- ($(P)!(U)!(C)$);

\coordinate (S) at ($(B)!1.75!(C)$);
\coordinate (T) at ($(A)!2!(C)$);

\draw[-latex,loosely dashed,green] (C) -- (S);
\draw[-latex,loosely dashed,green] (C) -- (T);

\end{tikzpicture}


\end{document}

Answer 1

You can use calc to calculate the angle:

\draw let \p1=($(S)-(C)$), \p2=($(T)-(C)$), \n0={.5*atan2(\y1,\x1)+.5*atan2(\y2,\x2)} in
   (C) -- +(\n0:2);

You can construct the bisect:

\draw (C) -- ($($(C)!2cm!(S)$)!.5!($(C)!2cm!(T)$)$);

This uses layered calculation which might create issued, so you could just do the calculations beforehand with

… let \p1=($(C)!2cm!(S)$), \p2=($(C)!2cm!(T)$) in (C) -- ($(\p1)!.5!(\p2)$) …

However if you define S and T in equal distance from C, for example

\path ($(C)!-2cm!(B)$) coordinate (S)
      ($(C)!-2cm!(A)$) coordinate (T);

you could just do

\draw (C) -- ($(S)!.5!(T)$);

---

You could (should in my opinion) wrap this solution in an insert path, say something like

\tikzset{
  calc angle between/.style args={#1--#2--#3}{%
    insert path={let \p{@aux1}=($(#1)-(#2)$), \p{@aux2}=($(#3)-(#2)$),
      \n{angle}={.5*atan2(\y{@aux1},\x{@aux1})+.5*atan2(\y{@aux2},\x{@aux2})} in}}}

and then you can just do

\draw[calc angle between=S--C--T] (C) -- +(\n{angle}:2);

---

I also provide a PGFmath-powered solution which provides you with a function anglebisect(<p1>, <p2>, <p3>, <p4>) that calculates the directional angle for the lines (<p1>) -- (<p2>) and (<p3>) -- (<p4>) (in your case <p1> equals <p3>). Unfortunately, PGFmath needs " wrapped around such arguments as it should not evaluate them.

Code

\documentclass[tikz]{standalone}
\usetikzlibrary{calc,arrows.meta}
\makeatletter
\newcommand*\pgfmathanglebetweenpointsNoCorrection[2]{%
  \begingroup%
    \pgf@process{\pgfpointdiff{#1}{#2}}%
    \edef\pgf@marshall{\expandafter\noexpand\csname pgfmathatan2@\endcsname
      {\expandafter\Pgf@geT\the\pgf@y}{\expandafter\Pgf@geT\the\pgf@x}}%
    \pgf@marshall%
  \pgfmath@smuggleone\pgfmathresult\endgroup}
\pgfmathdeclarefunction{anglebisect}{4}{%
  \begingroup
    \pgfmathanglebetweenpointsNoCorrection{\pgfpointanchor{\tikz@pp@name{#1}}{center}}
                                          {\pgfpointanchor{\tikz@pp@name{#2}}{center}}%
    \let\pgfmath@temp\pgfmathresult
    \pgfmathanglebetweenpointsNoCorrection{\pgfpointanchor{\tikz@pp@name{#3}}{center}}
                                          {\pgfpointanchor{\tikz@pp@name{#4}}{center}}%
    \pgfmathadd@{\pgfmath@temp}{\pgfmathresult}%
    \pgfmathmultiply@{.5}{\pgfmathresult}%
    \pgfmath@smuggleone\pgfmathresult\endgroup}
\makeatother
\tikzset{
  calc angle between/.style args={#1--#2--#3}{%
    insert path={let \p{@aux1}=($(#1)-(#2)$), \p{@aux2}=($(#3)-(#2)$),
      \n{angle}={.5*atan2(\y{@aux1},\x{@aux1})+.5*atan2(\y{@aux2},\x{@aux2})} in}}}
\begin{document}
\begin{tikzpicture}
\path[every label/.append style={circle,inner sep=1pt}]
      (0,0)      coordinate[label=$A$] (A)
    + (235:5.5)  coordinate[label=below left:$B$] (B)
    + (290:3.25) coordinate (C)
    ($(A)!.5!(B)$)  coordinate (AB_midpoint)
    ($(A)!.5!(C)$)  coordinate (AC_midpoint)
    ($(A)!(C)!(B)$) coordinate (P)
    ($(P)!.5!(C)$)  coordinate (PC_midpoint);
\draw (A) -- (B) -- (C) -- cycle;
% or: \draw plot coordinates {(A)(B)(C)} -- cycle;

\draw[dashed] (C) -- (P);
\draw coordinate (U) at ($(P)!4mm!-45:(A)$)
  ($(P)!(U)!(A)$) -- (U) -- ($(P)!(U)!(C)$);

\draw[Latex-Latex,loosely dashed,green]
  ($(B)!1.75!(C)$) coordinate (S)
  -- (C) --
  ($(A)!2!(C)$)    coordinate (T);

\draw[line width=.2cm] let \p1=($(S)-(C)$), \p2=($(T)-(C)$),
                 \n0={.5*atan2(\y1,\x1)+.5*atan2(\y2,\x2)} in
  (C) -- +(\n0:2);

\draw[line width=.15cm,red]    (C) -- ($($(C)!2cm!(S)$)!.5!($(C)!2cm!(T)$)$);
\draw[line width=.1cm,white] (C) -- +({anglebisect("C","S","C","T")}:1);

\draw[dashed,thick, calc angle between=S--C--T] (C) -- +(\n{angle}:2);

\path ($(C)!-2cm!(B)$) coordinate (S)
      ($(C)!-2cm!(A)$) coordinate (T);
\draw[thick,dashed,blue!50,dash phase=3pt] (C) -- ($(S)!.5!(T)$);
\end{tikzpicture}
\end{document}

Output

Answer 2

The angle between the horizontal line to BC can be calculated via \pgfmathanglebetweenpoints:

\pgfmathanglebetweenpoints{\pgfpointanchor{B}{center}}{\pgfpointanchor{C}{center
\let\AngleTmp\pgfmathresult
\pgfmathsetmacro\RayAngle{(-70 + \AngleTmp)/2}

Answer 3

If you're open to a new package, you can use tkz-euclide then add this to your preamble:

\usepackage{tkz-euclide}
\usetkzobj{all}

Then use the following commands:

\tkzDefLine[bisector](T,C,S)\tkzGetPoint{a}
\tkzDrawSegment[red, dotted](C,a)

Here's the output:

Answer 4

Here is simplified version of your code. The bisector is drawn using two points (S) and (T) positioned at the same distance from (C). Only the calc library is used.

\documentclass[tikz,border=7pt]{standalone}
\usetikzlibrary{calc}

\begin{document}
  \begin{tikzpicture}
    % draw the triangle
    \path (0,0) coordinate (A) node[above] {$A$}
        (235:5.5) coordinate (B) node[below left]{$B$}
        (290:3.25) coordinate (C);
    \draw (A) -- (B) -- (C) -- cycle;

    % draw dashed height
    \draw[dashed] (C) -- ($(A)!(C)!(B)$) coordinate (P);

    %The following commands make the right-angle mark.
    \draw ($(P)!4mm!(A)$) -- ([turn]90:4mm) -- ([turn]-90:4mm);

    % draw the opposit angle and the bisector in red
    \draw[-latex,loosely dashed,green] (C) -- ($(C)!-2cm!(B)$) coordinate (S) -- ([turn]0:15mm);
    \draw[-latex,loosely dashed,green] (C) -- ($(C)!-2cm!(A)$) coordinate (T) -- ([turn]0:5mm);
    \draw[thick, red] (C) -- ($(S)!.5!(T)$) -- ([turn]0:15mm);
  \end{tikzpicture}
\end{document}

Note: The first [turn] in the right-angle mark is a mystery for me, but it works.