[Tex/LaTex] Labeling sides and angles of a right triangle for an argument of the Pythagorean Theorem

tikz-pgf

I would like to draw a right triangle so that its sides are not vertical or horizontal. (No coordinate axes.) I would like the legs to be labeled a and b and the hypotenuse to be labeled c. I would like to drop a perpendicular, drawn as a dotted line segment, from the vertex across the hypotenuse to the hypotenuse. (I heard that there was a command to instruct TikZ to do this.) This creates two smaller triangles that are similar to each other. I would like the four acute angles of the two smaller triangles to be marked with arcs; one pair of equal angles marked with "|" through it, and the other pair of equal angles marked with "||" through it.

The only code that I could offer is the code for labeling the vertices of the triangle and drawing the line segments between them. I know that there is much more to code. I reckon that it would be more convenient for anybody responding to decide on the coordinates of the vertices himself/herself.

Best Answer

Here's one possibility using "pure" TikZ:

enter image description here

The image was produced using simply

\begin{tikzpicture}
\RectTri{(0,3)}{(1,0)}{6cm}
\begin{scope}[xshift=8.5cm]
\RectTri[black]{(0,0)}{(4,2)}{4cm}
\end{scope}
\end{tikzpicture}

\RectTri has three mandatory arguments; the first two are the coordinates for the vertices of one of the legs and the third one is the length of the second leg. The optional argument lets you customize the style used to draw the triangle.

The code:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc,angles,quotes,decorations.markings}

\newcommand\RectTri[4][thick,green!50!black,text=black]{%
\coordinate [label=left:$C$] (C) at #2;
\coordinate [label=below right:$B$] (B) at #3;
\coordinate (aux) at ($ #2 ! 1 ! 90:#3 $);
\coordinate [label=above:$A$] (A) at ($ #2 !#4!(aux) $);

\coordinate (perp) at ($(A)!(C)!(B)$);
\draw[purple!70!black,thick,dashed] (C) -- (perp);

\draw[#1] 
  (C) -- 
  node[auto] {$b$} (A) -- 
  node[auto] {$c$} (B) --
  node[auto] {$a$} 
  (C)
  pic ["$\alpha$",draw,cyan,thick,angle radius=1cm] {angle = C--A--B} 
  pic ["$\alpha$",draw,cyan,thick,angle radius=1cm] {angle = B--C--perp}
  pic ["$\beta$",draw,orange!70!black,thick,angle radius=1cm] {angle = A--B--C}
  pic ["$\beta$",draw,orange!70!black,thick,angle radius=1cm] {angle = perp--C--A};
}

\begin{document}

\begin{tikzpicture}
\RectTri{(0,3)}{(1,0)}{6cm}
\begin{scope}[xshift=8.5cm]
\RectTri[black]{(0,0)}{(4,2)}{4cm}
\end{scope}
\end{tikzpicture}

\end{document}

And here's an approach using tkz-euclide:

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

\begin{document}

\begin{tikzpicture}
\tkzDefPoint(0,1){A}
\tkzDefPoint(2,4){C}
\tkzDefPointWith[orthogonal normed,K=7](C,A)
\tkzGetPoint{B}

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

\tkzMarkRightAngle(A,C,B)

\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){$b$}

\tkzDrawAltitude[dashed,color=magenta](A,B)(C)
\tkzGetPoint{D}

\tkzMarkAngle[size=1cm,color=cyan,mark=|](C,B,A)
\tkzMarkAngle[size=1cm,color=cyan,mark=|](A,C,D)
\tkzMarkAngle[size=0.75cm,color=orange,mark=||](D,C,B)
\tkzMarkAngle[size=0.75cm,color=orange,mark=||](B,A,C)
\end{tikzpicture}

\end{document}

enter image description here

Related Solutions

[Tex/LaTex] Hexagonal diagrams

I try to find another way in the spirit of TikZ without complicated macros.

First part : the method

First I define vertices. Some of them are on circle and define hexagons. Circles or hexagons are numeroted from 0 to 3. 0 is the center. Then on each hexagons, I place vertices numeroted from 0 to 5 for the first; 0 to 11 for the second and 0 to 17 for the third one. In a first time,I use nodes but for the final drawing, I will use coordinates. A vertice is defined by h;i h for hexagon and i for indice.

\documentclass{article}

\usepackage{tikz}
\usetikzlibrary{calc,arrows}

\begin{document}
\begin{tikzpicture}
  %%%  define vertices with coordinates     
\node (00) at (0,0) {00}; 
\foreach \c in {1,...,3}{%  
\foreach \i in {0,...,5}{% 
\pgfmathtruncatemacro\j{\c*\i}
 % little trick \c*\i  gives 0,1,2,3,4,5 for this first circle
 % 0,2,4 etc.. for the second one and 0,3,6 for the third one.
 % the for the second hexagon, i define the midpoints ith indices : 1,3,5,..
 % and for the third hexagon , two points betweens 3;0 and 3;3 etc...

 %  pgfmathtruncatemacro is used because we can't accept 3;2.0 for a name
\node[circle,minimum width=4pt,inner sep=0pt] (\c;\j) at (60*\i:\c){\c;\j};  
} }
% now on the second hexagon we need to place between each corners, a midpoint
% to finish we need to change 12 in 0 so I use  mod 
\foreach \i in {0,2,...,10}{%
 % perhaps foreach now gives some possibilities to avoid  \pgfmathtruncatemacro
\pgfmathtruncatemacro\j{mod(\i+2,12)}% 
\pgfmathtruncatemacro\k{\i+1}
 % midpoint I use the same method further 
\node (2\k) at ($(2;\i)!.5!(2;\j)$)  {2;\k} ;    }

% now on the third hexagon we need to place between each corners, two points
% to finish we need to change 18 in 0 so I use  mod
%  ($(3;\i)!1/3!(3;\j)$) is a barycenter coeff 1 and 2
\foreach \i in {0,3,...,15}{% 
\pgfmathtruncatemacro\j{mod(\i+3,18)} 
\pgfmathtruncatemacro\k{\i+1} 
\pgfmathtruncatemacro\l{\i+2}
\node (3\k) at ($(3;\i)!1/3!(3;\j)$)  {3;\k} ;
\node (3\l) at ($(3;\i)!2/3!(3;\j)$)  {3;\l} ;
 }     
\end{tikzpicture} 
\end{document}

I get this picture. You can see the names of the nodes.

enter image description here

Second part : Drawing the graph

I change node with coordinates

\documentclass{article}

\usepackage{tikz}
\usetikzlibrary{calc,arrows}


\begin{document}

\begin{tikzpicture}
%%%  define vertices with coordinates
\coordinate (0;0) at (0,0); 
\foreach \c in {1,...,3}{%  
\foreach \i in {0,...,5}{% 
\pgfmathtruncatemacro\j{\c*\i}
\coordinate (\c;\j) at (60*\i:\c);  
} }
\foreach \i in {0,2,...,10}{% 
\pgfmathtruncatemacro\j{mod(\i+2,12)} 
\pgfmathtruncatemacro\k{\i+1}
\coordinate (2;\k) at ($(2;\i)!.5!(2;\j)$) ;}

\foreach \i in {0,3,...,15}{% 
\pgfmathtruncatemacro\j{mod(\i+3,18)} 
\pgfmathtruncatemacro\k{\i+1} 
\pgfmathtruncatemacro\l{\i+2}
\coordinate (3;\k) at ($(3;\i)!1/3!(3;\j)$)  ;
\coordinate (3;\l) at ($(3;\i)!2/3!(3;\j)$)  ;
 }

 %%%%%%%%% draw lines %%%%%%%%
 \foreach \i in {0,...,6}{% 
 \pgfmathtruncatemacro\k{\i}
 \pgfmathtruncatemacro\l{15-\i}
 \draw[thin,gray] (3;\k)--(3;\l);
 \pgfmathtruncatemacro\k{9-\i} 
 \pgfmathtruncatemacro\l{mod(12+\i,18)}   
 \draw[thin,gray] (3;\k)--(3;\l); 
 \pgfmathtruncatemacro\k{12-\i} 
 \pgfmathtruncatemacro\l{mod(15+\i,18)}   
 \draw[thin,gray] (3;\k)--(3;\l);}    
%%%%%%%%% some specific lines %%%%%%%%%% 
 \foreach \i in {0,2,...,10} {
   \pgfmathtruncatemacro\j{mod(\i+2,12)} 
   \draw[thick,dashed] (2;\i)--(2;\j);}     
%%%%%%%%% draw points %%%%%%%% 
\fill [gray] (0;0) circle (2pt);
 \foreach \c in {1,...,3}{%
 \pgfmathtruncatemacro\k{\c*6-1}    
 \foreach \i in {0,...,\k}{% 
   \fill [gray] (\c;\i) circle (2pt);}}  
%%%%%%%%% some specific points %%%%%%%%%%  
 \foreach \n in {0,3,...,15}{% 
   \draw (3;\n) circle (4pt);}
 \foreach \n in {1,3,...,11}{% 
   \draw (2;\n) circle (4pt);}
%%%%%%%%%% arrows %%%%%%%%%%%%
\draw[->,red,thick,shorten >=4pt,shorten <=2pt](0;0)--(2;3);
\draw[->,red,thick,shorten >=4pt,shorten <=2pt](0;0)--(2;1); 
\end{tikzpicture}  
\end{document}

The result :

enter image description here

[Tex/LaTex] Marking angles in a triangle

Two simple possibilities using TikZ:

For version 3.0, using the angles and quotes libraries:

enter image description here

The code:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{angles,quotes}

\tikzset{
mydot/.style={
  fill,
  circle,
  inner sep=1.5pt
  }
}

\begin{document}

\begin{tikzpicture}[>=latex]
% the coordinates of the vertices
\coordinate (O) at (0,0);
\coordinate (A) at (2,1);
\coordinate (B) at (-3,5);
\coordinate (E) at (2,0);

% the axis
\draw[help lines,->] (-3.5,0) -- (2.5,0);
\draw[help lines,->] (0,-0.5) -- (0,5.5);

% the edges of the triangle    
\draw (O) -- (A) -- (B) -- cycle;

% labelling the vertices
\node[mydot,label={right:$A$}] at (A) {};
\node[mydot,label={left:$B$}] at (B) {};
\node[mydot,label={below:$O$}] at (O) {};

% the arcs for the angles
\path[gray]
  pic["$\alpha$" shift={(23pt,3pt)},draw,->,angle radius=1.5cm] {angle = E--O--A}
  pic["$\beta$" above=6pt,draw,->,angle radius=0.75cm] {angle = E--O--B};
\end{tikzpicture}

\end{document}

For version 2.10, without libraries and using the arc path:

enter image description here

The code:

\documentclass{amsart}
\usepackage{tikz}

\tikzset{
mydot/.style={
  fill,
  circle,
  inner sep=1.5pt
  }
}

\begin{document}

\begin{tikzpicture}[>=latex]
% the coordinates of the vertices
\coordinate (O) at (0,0);
\coordinate (A) at (2,1);
\coordinate (B) at (-3,5);

% the axis
\draw[help lines,->] (-3.5,0) -- (2.5,0);
\draw[help lines,->] (0,-0.5) -- (0,5.5);

% the edges of the triangle    
\draw (O) -- (A) -- (B) -- cycle;

% labelling the vertices
\node[mydot,label={right:$A$}] at (A) {};
\node[mydot,label={left:$B$}] at (B) {};
\node[mydot,label={below:$O$}] at (O) {};

% the arcs for the angles    
\begin{scope}[gray]
\draw[->] 
  (1,0) +(0:0.5cm) arc [radius=1cm,start angle=0,end angle=41] node[midway,right] {$\alpha$};
\draw[->] 
  (0.5,0) +(0:0.25cm) arc [radius=0.75cm,start angle=0,end angle=122] node[midway,above] {$\beta$};
\end{scope}
\end{tikzpicture}

\end{document}

Best Answer

Related Solutions

[Tex/LaTex] Hexagonal diagrams

[Tex/LaTex] Marking angles in a triangle

Related Question