# Labeling angles in a triangle

I've tried looking at past answers to this question but it won't seem to solve my problem.

I want to label the angles ABD and DBC in this triangle:

\documentclass{article}
\usepackage{tikz,amsmath,amssymb,tkz-euclide}
\begin{document}
\begin{tikzpicture}

\tkzInit[xmin=-0.5,xmax=6.7,ymin=-0.7,ymax=3.5]
\tkzClip[space=0.25]

\coordinate (A) at (0,0); \tkzLabelPoints[left](A)
\coordinate (B) at (2.25,3.320718914); \tkzLabelPoints[above](B)
\coordinate (C) at (6,0); \tkzLabelPoints[right](C)
\coordinate (E) at (3,0); \tkzLabelPoints[below](E)
\draw (A)--(B)--(C)--cycle;
\draw (B)--(E);

\tkzDefLine[bisector](A,B,C)\tkzGetPoint{a}
\tkzInterLL(A,C)(B,a) \tkzGetPoint{D} \tkzLabelPoints[below](D)
\draw (B)--(D);

\tkzDefMidPoint(A,B) \tkzGetPoint{12}
\tkzLabelPoints[left](12)
\tkzDefMidPoint(B,C) \tkzGetPoint{15}
\tkzLabelPoints[right=0.1cm](15)
\tkzDefMidPoint(A,C) \tkzGetPoint{18}
\tkzLabelPoints[below=0.4cm](18)

\tkzMarkSegment[mark=|](A,E)
\tkzMarkSegment[mark=|](C,E)

\tkzLabelSegment(A,B)
\tkzLabelSegment(B,C)
\tkzLabelSegment(C,A)

\end{tikzpicture}
\end{document}


You can use angles library for that.

\documentclass{article}
\usepackage{tikz,amsmath,amssymb,tkz-euclide}
\usetikzlibrary{angles}

\begin{document}
\begin{tikzpicture}

\tkzInit[xmin=-0.5,xmax=6.7,ymin=-0.7,ymax=3.5]
\tkzClip[space=0.25]

\coordinate (A) at (0,0); \tkzLabelPoints[left](A)
\coordinate (B) at (2.25,3.320718914); \tkzLabelPoints[above](B)
\coordinate (C) at (6,0); \tkzLabelPoints[right](C)
\coordinate (E) at (3,0); \tkzLabelPoints[below](E)
\draw (A)--(B)--(C)--cycle;
\draw (B)--(E);

\tkzDefLine[bisector](A,B,C)\tkzGetPoint{a}
\tkzInterLL(A,C)(B,a) \tkzGetPoint{D} \tkzLabelPoints[below](D)
\draw (B)--(D);

\tkzDefMidPoint(A,B) \tkzGetPoint{12}
\tkzLabelPoints[left](12)
\tkzDefMidPoint(B,C) \tkzGetPoint{15}
\tkzLabelPoints[right=0.1cm](15)
\tkzDefMidPoint(A,C) \tkzGetPoint{18}
\tkzLabelPoints[below=0.4cm](18)

\tkzMarkSegment[mark=|](A,E)
\tkzMarkSegment[mark=|](C,E)

%    \tkzLabelSegment(A,B)
%    \tkzLabelSegment(B,C)
%    \tkzLabelSegment(C,A)
\draw pic[draw=blue, angle radius=5mm] {angle = A--B--D};
\draw pic[draw=red, angle radius=7mm] {angle = D--B--C};

\end{tikzpicture}
\end{document}

• Thanks a lot, that's great. I was wondering if there was a way to make a dash going through the angles similar to the one going through the AE and AC segments. – KLG May 31 '15 at 5:03
• @SamFisher I don't know a fast way know. You can make a follow-up question. By the way, tkz-euclide has \tkzMarkAngle which is equivalent to angle pic. And for dashes, you can find the angle midpoint and draw a little segment on it. – Ignasi May 31 '15 at 7:59

You can use \tkzFindAngle, tkzGetAngle, tkzMarkAngle and tkzLabelAngle.

By the way, doing

\tkzDefMidPoint(A,B) \tkzGetPoint{12}
\tkzLabelPoints[left](12)


And doing

\tkzLabelSegment[left](A,B){12}


is the same (in this case at least). But as you can see the command is much shorter. I would use that if I was you. I had corrected it from your code because \tkzLabelSegment(A,B) was missing the name between curly brackets.

## Code

\documentclass[margin=10pt]{standalone}
\usepackage{tikz,amsmath,amssymb,tkz-euclide}
\usetkzobj{all}

\begin{document}
\begin{tikzpicture}

\tkzInit[xmin=-0.5,xmax=6.7,ymin=-0.7,ymax=3.5]
\tkzClip[space=0.25]

\coordinate (A) at (0,0); \tkzLabelPoints[left](A)
\coordinate (B) at (2.25,3.320718914); \tkzLabelPoints[above](B)
\coordinate (C) at (6,0); \tkzLabelPoints[right](C)
\coordinate (E) at (3,0); \tkzLabelPoints[below](E)
\draw (A)--(B)--(C)--cycle;
\draw (B)--(E);

\tkzDefLine[bisector](A,B,C)\tkzGetPoint{a}
\tkzInterLL(A,C)(B,a) \tkzGetPoint{D} \tkzLabelPoints[below](D)
\draw (B)--(D);

\tkzDefMidPoint(A,B) \tkzGetPoint{12}
\tkzLabelPoints[left](12)
\tkzDefMidPoint(B,C) \tkzGetPoint{15}
\tkzLabelPoints[right=0.1cm](15)
\tkzDefMidPoint(A,C) \tkzGetPoint{18}
\tkzLabelPoints[below=0.4cm](18)

\tkzMarkSegment[mark=|](A,E)
\tkzMarkSegment[mark=|](C,E)

\tkzFindAngle(A,B,D)
\tkzFindAngle(D,B,C)
\tkzGetAngle{angleABD}; \FPround\angleABD\angleABD{0}
\tkzGetAngle{angleDBC}; \FPround\angleDBC\angleDBC{0}

\tkzMarkAngle[draw=black,opacity=.5,fill=green!10](A,B,D)
\tkzMarkAngle[draw=black,opacity=.5,fill=red!10](D,B,C)
\tkzLabelAngle[dist=1.2](A,B,D){\tiny $\angleABD^\circ$}
\tkzLabelAngle[dist=1.2](D,B,C){\tiny $\angleDBC^\circ$}

%\tkzLabelSegment[left](A,B){AB}
%\tkzLabelSegment[right](B,C){BC}
%\tkzLabelSegment[below](C,A){CA}

\end{tikzpicture}
\end{document}