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}