# Drawing an obtuse triangle, and marking its height

I would like to have an obtuse triangle with vertices A, B, and C drawn. The lengths of the sides are to be labeled a, b, and c. If the angle at C is the obtuse angle, the height of the triangle is to be drawn as a dashed line segment from A, and the side BC is to be extended as a dashed line segment to the point P of its intersection with the dashed line segment indicating the height of the triangle.

I would also like the angle marked with one "|" to be an interior angle, and I would like the lengths of the sides to be outside the triangle. What does [auto] in \tkzLabelSegment[auto] tell TikZ to draw?

I guess that a command like \tkzDefPointWith[orthogonal normed,K=blah](B,C) is to be used. Is there a manual describing some of the commands of tkz-euclide? I just saw that there is a command \tkzDefTriangle[equilateral] for having TikZ calculate the coordinates of the third vertex of a triangle given its two other vertices so as to get an equilateral triangle.

\documentclass[10pt]{amsart}
\usepackage{tikz}
\usetikzlibrary{calc,angles,positioning,intersections,quotes,decorations.markings}
\usepackage{tkz-euclide}
\usetkzobj{all}

\begin{document}

\begin{tikzpicture}

\tkzDefPoint(0,-1){C}
\tkzDefPoint(4,1){B}
\tkzDefPoint(-1,3){A}

\tkzLabelPoint[left](A){$A$}
\tkzLabelPoint[right](B){$B$}
\tkzLabelPoint[below](C){$C$}

\tkzDrawSegment[green!60!black](A,C)
\tkzDrawSegment[green!60!black](C,B)
\tkzDrawSegment[green!60!black](B,A)

\tkzLabelSegment[auto](B,A){$c$}
\tkzLabelSegment[auto,swap](B,C){$a$}
\tkzLabelSegment[auto,swap](C,A){$b$}

\tkzMarkAngle[size=1cm,color=cyan,mark=|](C,B,A)
\tkzMarkAngle[size=1cm,color=cyan,mark=||](C,A,B)
\end{tikzpicture}

\end{document} • Providing MWE with an image of its output is recommended to ease us understanding your case at a glance. Aug 24, 2014 at 15:14
• What prompted you to choose "cyanide-based food" as your user name? How do I include the diagram that TikZ draws from my code on this site? I like the code that Mark Wibrow provided. It gives me the diagram that I wanted. I still have questions regarding my code. Why is an exterior angle drawn at C? Why are the lengths of the sides written inside the triangle? Aug 24, 2014 at 22:55
• Regarding the username, it is randomly chosen just for fun as usual. Regarding uploading your diagram, see my answer here (click). I am preparing my own answer for you. Aug 25, 2014 at 4:43

Something like this?

\documentclass[tikz, border=5]{standalone}
\begin{document}
\begin{tikzpicture}
\path (120:3) coordinate (A) (0:3) coordinate (B) (0:0) coordinate (C);
\draw (A)
-- (B) node [at start, above left] {$A$} node [midway, above] {$c$}
-- (C) node [at start, right]      {$B$} node [midway, below] {$a$}
-- (A) node [at start, below]      {$C$} node [midway, below] {$b$}
-- cycle;
\draw [dashed] (A) |- (C) node [midway, below left] {$P$};
\draw (0:.5) arc (0:120:.5) (60:.4) -- (60:.6);
\end{tikzpicture}
\end{document} or this...

\documentclass[tikz, border=5]{standalone}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}
\path (80:3) coordinate (A) (-30:3) coordinate (B) (0:0) coordinate (C);
\draw (A)
-- (B) node [at start, above left] {$A$} node [midway, right] {$c$}
-- (C) node [at start, right]      {$B$} node [midway, below] {$a$}
-- (A) node [at start, below]      {$C$} node [midway, left]  {$b$}
-- cycle;
\coordinate (P) at ($(B)!(A)!(C)$);
\draw [dashed] (A) -- (P) node [below left] {$P$} -- (C);
\draw ($(C)!0.5cm!(B)$) arc (-30:80:.5cm) node [midway, sloped] {$|$};
\end{tikzpicture}
\end{document} • This is a diagram of an obtuse triangle indicating its height. This is the diagram that I said that I wanted. I should have specified that I do not want any sides of the triangle to be horizontal or vertical. To draw the dashed line segments indicating the height of such a triangle, you cannot use "\draw [dashed] (A) |- (C)node [midway, below left] {$P$};". Can you give me the code for instructing TikZ to draw such a triangle? Aug 24, 2014 at 23:45
• @user60254, The calc library can be used for this, see my updated answer. Note that the position of the labels has to be changed as this is not done automatically. Aug 25, 2014 at 6:00
• This is the diagram that I wanted! (You have an error in your code. You have "node [below left] {$P$} -- (C)". You are missing a comma between "below" and "left".) Aug 29, 2014 at 19:00
• This is the diagram that I wanted! (You have an error in your code. You are missing a comma between "below" and "left" in the command to typeset the letter "P".) Aug 29, 2014 at 20:00
• @user60254 actually below left is a valid option. If it produces an error for you, then you may need to update your PGF version. Aug 31, 2014 at 7:15

A PSTricks solution using the pst-eucl package:

\documentclass{article}

\usepackage{pst-eucl}

\def\Label[#1]#2#3#4{
\pcline[linestyle = none, offset = #1](#2)(#3)
\ncput*{$#4$}
}

\begin{document}

\psset{unit = 0.9}
\begin{pspicture}(-0.35,-0.5)(12.5,8.4)
\pnode(0,0){P}
\pstTriangle(0,8){A}(12,0){B}(4,0){C}
\psset{CodeFig, CodeFigColor = blue, LabelSep = 0.7}
\pstMarkAngle{C}{A}{B}{$\alpha$}
\pstMarkAngle{A}{B}{C}{$\beta$}
\pstMarkAngle{B}{C}{A}{$\gamma$}
\Label[8pt]{A}{B}{c}
\Label[8pt]{B}{C}{a}
\Label[-8pt]{A}{C}{b}
\pstProjection{P}{C}{A}[P]
\pstProjection[PointName = none]{A}{P}{C}[Q]
\end{pspicture}

\end{document} or

\documentclass{article}

\usepackage{pst-eucl}

\def\Label[#1]#2#3#4{
\pcline[linestyle = none, offset = #1](#2)(#3)
\ncput*{$#4$}
}

\begin{document}

\psset{unit = 0.9}
\begin{pspicture}(-0.35,-0.5)(12.5,8.4)
\pnode(0,0){P}
\pstTriangle(0,8){A}(12,0){B}(4,0){C}
\psset{CodeFig, CodeFigColor = blue, LabelSep = 0.8, MarkAngle = 90}
\pstMarkAngle[Mark = MarkHash]{C}{A}{B}{$\alpha$}
\pstMarkAngle[Mark = MarkHashh]{A}{B}{C}{$\beta$}
\pstMarkAngle[Mark = MarkHashhh]{B}{C}{A}{$\gamma$}
\Label[8pt]{A}{B}{c}
\Label[8pt]{B}{C}{a}
\Label[-8pt]{A}{C}{b}
\pstProjection{P}{C}{A}[P]
\pstProjection[PointName = none]{A}{P}{C}[Q]
\end{pspicture}

\end{document} Just for typing exercise with PSTricks.

\documentclass[pstricks,border=20pt,12pt]{standalone}
\usepackage{pst-eucl}

\begin{document}
\begin{pspicture}[CodeFigColor=.,shortput=nab](8,6)
%
% drawing the triangle
\pstTriangle(0,6){A}(8,0){B}(3,0){C}
%
% projecting the point A onto BC (extension)
\psline[linestyle=dashed](A|0,0)(C)
\pstProjection[CodeFig]{B}{C}{A}[P]
%
% labeling and marking angles
\pstMarkAngle[Mark=MarkHash]{C}{A}{B}{$\alpha$}
\pstMarkAngle[Mark=MarkHashh]{A}{B}{C}{$\beta$}
\pstMarkAngle[Mark=MarkHashhh]{B}{C}{A}{$\gamma$}
%
% labeling sides
\psset{linestyle=none}
\pcline(A)(B)^{$c$}
\pcline(A)(C)_{$b$}
\pcline(C)(B)_{$a$}
\end{pspicture}

\end{document} ## Animated version

\documentclass[pstricks,border=20pt,12pt]{standalone}
\usepackage{pst-eucl}

\def\Atom#1{%
\begin{pspicture}[CodeFigColor=.,shortput=nab,saveNodeCoors,NodeCoorPrefix=N](8,6)
%
% drawing the triangle
\pstTriangle(8,0){B}(3,0){C}(6;#1){A}
%
% projecting the point A onto BC (extension)
\psline[linestyle=dashed](A|0,0)(C)
\pstProjection[CodeFig]{B}{C}{A}[P]
%
% labeling and marking angles
\pstMarkAngle[Mark=MarkHash]{C}{A}{B}{$\alpha$}
\pstMarkAngle[Mark=MarkHashh]{A}{B}{C}{$\beta$}
\pstMarkAngle[Mark=MarkHashhh]{B}{C}{A}{$\gamma$}
%
% labeling sides
\psset{linestyle=none}
\pcline(A)(B)^{$c$}
\pcline(A)(C)_{$b$}
\pcline(C)(B)_{$a$}
\end{pspicture}}

\begin{document}
\multido{\i=30+5}{13}{\Atom{\i}}
\end{document} What does [auto] in \tkzLabelSegment[auto] tell TikZ to draw?

auto is a tikz option to put the labels to the segments/curves such that they are not on the curve but next to it. If you add swap option additionally, tikz will mirror the label with respect to the line. See this example:

\documentclass{article}
\usepackage{tikz}
\begin{document}
Without any option to label node:

\begin{tikzpicture}
\draw (0,0) -- node {1} (2,3);
\end{tikzpicture}

With the option \verb|auto|:

\begin{tikzpicture}
\draw (0,0) -- node[auto] {1} (2,3);
\end{tikzpicture}

With the option \verb|auto| and \verb|swap|:

\begin{tikzpicture}
\draw (0,0) -- node[auto,swap] {1} (2,3);
\end{tikzpicture}
\end{document} To draw the perpendicular you can use

\tkzDrawAltitude[dashed,color=magenta](C,B)(A)


which tells tikz to draw a perpendicular from A onto the segment CB.

Now the code with comments in it for explanation:

\documentclass[10pt]{amsart}
\usepackage{tikz}
\usetikzlibrary{calc,angles,positioning,intersections,quotes,decorations.markings}
\usepackage{tkz-euclide}
\usetkzobj{all}

\begin{document}

\begin{tikzpicture}

\tkzDefPoint(0:0){C}
\tkzDefPoint(0:3){B}
\tkzDefPoint(120:3){A}

\tkzLabelPoint[left](A){$A$}
\tkzLabelPoint[right](B){$B$}
\tkzLabelPoint[below](C){$C$}

\tkzDrawSegment[green!60!black](A,C)
\tkzDrawSegment[green!60!black](C,B)
\tkzDrawSegment[green!60!black](A,B)

\tkzDrawAltitude[dashed,color=magenta](C,B)(A)   %% draw the perpendicular
\tkzGetPoint{P}                                  %% get the point P
\tkzLabelPoint[below](P){$P$}                    %% label the point P
\tkzDrawSegment[dashed,color=magenta](C,P)       %% draw CP

\tkzLabelSegment[auto,swap](B,A){$c$}
\tkzLabelSegment[auto](B,C){$a$}
\tkzLabelSegment[auto](C,A){$b$}

%% here use the correct order. For example, BCA is inner angle and ACB is outer angle
\tkzMarkAngle[size=0.5cm,color=cyan,mark=|](B,C,A)
\tkzMarkAngle[size=1cm,color=cyan,mark=||](C,A,B)
\tkzMarkAngle[size=1cm,color=cyan,mark=|||](A,B,C)

\end{tikzpicture}

\end{document} Is there a manual describing some of the commands of tkz-euclide?

tkz-euclide-screen is the manual for tkz-euclide and unfortunately it is in French (Allain is French). But it has sample codes to illustrate almost all the commands. You can get it by texdoc tkz-euclide from your command line/terminal or from texdoc.net. You have to read tikz manual too along with tkz-euclide-screen.

• Thanks for answering my questions regarding syntax and commands. Maybe I will use a French-to-English translation web site to read the manual that you suggested. Aug 29, 2014 at 19:09