What is the easiest way to recreate the picture using tikz or tkz-euclide?

Question

I am trying to recreate the following picture in tkz-euclide. One big question I have is if there is a way to create a triangle using only the side lengths. In other words I would like to avoid using trigonometry or any other complication to draw the triangle accurately.

Here is my attempt:

\begin{tikzpicture}
    \tkzDefPoint (0,0){A}
    \tkzDefPoint (3.07,0){B}
    \tkzDefPoint (1.89,1.38){C}
    \tkzDrawPolygon (A,B,C)
    \tkzInCenter(A,B,C)\tkzGetPoint{G}
    \tkzDrawPoint(G)
    \tkzDrawCircle[in](A,B,C)
\end{tikzpicture}

Answer 1

Here's how this can be done using tkz-euclide:

\documentclass[border=5mm]{standalone}
\usepackage{tkz-euclide}
\usetkzobj{all}

\begin{document}
\begin{tikzpicture}[scale=2]

% Define the first two points
\tkzDefPoint(0,0){A}
\tkzDefPoint(3.07,0){B}

% Find the intersections of the circles around A and B with the given radii
\tkzInterCC[R](A,2.37cm)(B,1.82cm)
\tkzGetPoints{C}{C'}

% Draw the interior circle
\tkzDrawCircle[in](A,B,C)

% Reset the bounding box so we don't get empty space around our triangle
\pgfresetboundingbox

% Draw the triangle and the points
\tkzDrawPolygon(A,B,C)
\tkzDrawPoints(A,B,C)

% Label the sides
\tkzLabelSegment[below](A,B){3.07}
\tkzLabelSegment[above right](B,C){1.82}
\tkzLabelSegment[above left](A,C){2.37}

\end{tikzpicture}
\end{document}

Answer 2

Based on Jake's answer with some refinements

\documentclass[border=5mm]{standalone}
\usepackage{tkz-euclide}
\usetkzobj{all}
\begin{document}

    \pgfkeys{/pgf/decoration/.cd, distance/.initial = 10pt}  

    \pgfdeclaredecoration{add dim}{final}{
    \state{final}{% 
    \pgfmathsetmacro{\dist}{\pgfkeysvalueof{/pgf/decoration/distance}}
              \pgfpathmoveto{\pgfpoint{0pt}{0pt}}             
              \pgfpathlineto{\pgfpoint{0pt}{2*\dist}}   
              \pgfpathmoveto{\pgfpoint{\pgfdecoratedpathlength}{0pt}} 
              \pgfpathlineto{\pgfpoint{(\pgfdecoratedpathlength}{2*\dist}}     
              \pgfsetarrowsstart{latex}
              \pgfsetarrowsend{latex}
              \pgfpathmoveto{\pgfpoint{0pt}{\dist}}
              \pgfpathlineto{\pgfpoint{\pgfdecoratedpathlength}{\dist}} 
              \pgfusepath{stroke} 
              \pgfpathmoveto{\pgfpoint{0pt}{0pt}}
              \pgfpathlineto{\pgfpoint{\pgfdecoratedpathlength}{0pt}}
    }}

    \tikzset{
        dim/.style args={#1,#2,#3}{%
                    decoration = {add dim,distance=\ifx&#2&0pt\else#2\fi},
                    decorate,
                    postaction = {%
                       decorate,
                       decoration={%
                            raise=\ifx&#2&0pt\else#2\fi,
                            markings,
                            mark=at position .5 with {\node[inner sep=0pt,
                                                            font=\footnotesize,
                                                            fill=\ifx&#1&none\else white\fi,
                                                            #3] at (0,0) {#1};}
                       }
                    }
        },
        dim/.default={,0pt,}
    }       
\begin{tikzpicture}[scale=2]
\pgfkeys{/pgf/number format/.cd,fixed,precision=2}
% Define the first two points
\tkzDefPoint(0,0){A}
\tkzDefPoint(3.07,0){B}

% Find the intersections of the circles around A and B with the given radii
\tkzInterCC[R](A,2.37cm)(B,1.82cm)
\tkzGetPoints{C}{C'}

% Draw the interior circle
\tkzDrawCircle[in](A,B,C) \tkzGetPoint{G}
\tkzGetLength{rIn} 

% Reset the bounding box so we don't get empty space around our triangle
\pgfresetboundingbox

% Draw the triangle and the points
\tkzDrawPolygon(A,B,C)
\tkzDrawPoints(A,B,C)

% Label the sides
\tkzCalcLength[cm](A,B)\tkzGetLength{ABl}
\tkzCalcLength[cm](B,C)\tkzGetLength{BCl}
\tkzCalcLength[cm](A,C)\tkzGetLength{ACl}

% add dim
\tkzDrawSegment[dim={\pgfmathprintnumber\BCl,6pt,transform shape}](C,B)
\tkzDrawSegment[dim={\pgfmathprintnumber\ACl,6pt,transform shape}](A,C)
\tkzDrawSegment[dim={\pgfmathprintnumber\ABl,-6pt,transform shape}](A,B)

% Labels 
\tkzLabelPoints(A,B) \tkzLabelPoints[above](C)
\tkzDefShiftPoint[G](70:\rIn pt){g}  
\tkzDefShiftPoint[g](70: .5 cm){gg} \tkzDefShiftPoint[gg](0: .5 cm){ggg}  
\tkzDrawSegment[->](G,g) \tkzDrawSegment(g,gg) 
\tkzDrawLine[add=0 and 0,end={$r=?$}](gg,ggg) 
\end{tikzpicture}
\end{document}

Answer 3

This only shows only the drawing of the triangle and circle, not the arrows and labels.

I draw first one side, then make two circles starting at the end points of this line. The third point is the intersection of the two circles (there are two intersections).

There are some calculations to get the radius of the incircle, the center of which is found with barycentric coordinates. I realize tkz-euclide simplifies this considerably, but there are enough examples of that in other answers, so I thought I'd add a pure TikZ suggestion for that part.

\documentclass[border=3mm]{standalone}
\usepackage{tikz}
\usetikzlibrary{intersections}
\begin{document}
\begin{tikzpicture}%[rotate=-20]

% save side lenghths in macros
\pgfmathsetmacro{\sideA}{3.07}
\pgfmathsetmacro{\sideB}{1.82}
\pgfmathsetmacro{\sideC}{2.35}

% calculate semiperimeter
\pgfmathsetmacro{\semip}{(\sideA+\sideB+\sideC)/2}
% calculate radius of incircle: area (given by Heron's formula) divided by semiperimeter
\pgfmathsetmacro{\inrad}{sqrt(\semip*(\semip-\sideA)*(\semip-\sideB)*(\semip-\sideC))/\semip}

% define coordinates for points A and B
\coordinate (A) at (0,0);
\coordinate (B) at (\sideA,0);

% define circular paths centered on A and B, with radius as side AC and BC respectively
\path [name path=AC] (A) circle[radius=\sideC];
\path [name path=BC] (B) circle[radius=\sideB];

% calculate the intersection(s) of the two circles
% by default the coordinates are called intersection-N, name=C means they're called C-1 and C-2
\path [name intersections={of=AC and BC,name=C}];

% reset bounding box to avoid extra whitespace
\pgfresetboundingbox

% draw the triangle
\draw (A) -- (B) -- (C-1) -- cycle;
%\draw [help lines,dashed] (A) -- (B) -- (C-2) -- cycle;

% center of inradius given in barycentric coordinates
\coordinate (center) at (barycentric cs:A=\sideB,B=\sideC,C-1=\sideA);
\draw (center) circle[radius=\inrad];

\end{tikzpicture}
\end{document} 

Answer 4

Here's how I'd do it. There's some Tikz too in this code, but I suppose it's not a problem. The commented out code was not necessary so I replaced it with different code.

The lengths are automatically calculated.

Output

Code

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

\usetikzlibrary{calc, arrows.meta}

\begin{document}
\begin{tikzpicture}[rotate=-30, transform shape]
\tkzDefPoint(0,0){A}
\tkzDefPoint(3.07,0){B}
\tkzDefPoint(1.89,1.38){C}

\tkzDrawPolygon (A,B,C)

%\tkzInCenter(A,B,C)\tkzGetPoint{G}
%\tkzDrawPoint(G)
\tkzDrawCircle[in](A,B,C) \tkzGetPoint{G}
\tkzGetLength{rIn} 

\tkzCalcLength[cm](A,B)\tkzGetLength{ABl}
\tkzCalcLength[cm](B,C)\tkzGetLength{BCl}
\tkzCalcLength[cm](A,C)\tkzGetLength{ACl}

\foreach \lef/\rig/\len/\pos in {A/B/\ABl/below, B/C/\BCl/above, C/A/\ACl/above}{
    \draw[gray] ($(\lef)!0!90:(\rig)$) -- ($(\lef)!-.3!90:(\rig)$);
    \draw[gray] ($(\rig)!0!-90:(\lef)$) -- ($(\rig)!-.3!-90:(\lef)$);
    \draw[{Latex}-{Latex}] ($(\lef)!-.2!90:(\rig)$) -- ($(\rig)!-.2!-90:(\lef)$) node[\pos, midway, sloped] {\pgfmathprintnumber[fixed,precision=2]{\len}};
}

\draw[-{Latex}] (G) --++ (90:\rIn pt) coordinate (a);
\draw (G) -- ($(G)!3!(a)$) --++ (.2,0) node[right] {$r = ?$};

\end{tikzpicture}
\end{document}

Answer 5

Perhaps you would also like a Metapost answer. This one calculates the radius of the incircle for you. Just set a, b, c to the three lengths required at the top of the figure (with a<b+c of course).

prologues := 3;
outputtemplate := "%j%c.eps";

vardef incircle(expr A,B,C) = 
  save i; pair i;
  i = whatever[A,A+unitvector(C-A)+unitvector(B-A)] 
    = whatever[B,B+unitvector(C-B)+unitvector(A-B)];
  fullcircle scaled 2 abs ypart((i-A) rotated -angle(C-A)) shifted i
enddef;

beginfig(1);

u = 128; % or whatever unitsize you like...

a = 3.07u;
b = 2.35u;
c = 1.82u;

z0 = origin;
z1 = right scaled a rotated -23;
z2 = fullcircle scaled 2b shifted z0
     intersectionpoint
     fullcircle scaled 2c shifted z1;

path c, t, r;
c = incircle(z0,z1,z2);
t = z0--z1--z2--cycle;     
r = center c -- point 0 of c;

answer = round(arclength r / u * 100)/100;

draw      c withcolor .768 red;
drawarrow r withcolor .768 red; label.top(decimal answer, point 1/2 of r);
draw t;

path m[]; picture n;
for i=0 upto 2:
  m[i] = subpath (i,i+1) of t shifted (unitvector(point i of t - point i+1 of t) rotated 90 scaled 12);
  drawdblarrow m[i] withcolor .6 white;
  draw (down--up) scaled 6 rotated angle direction 0 of m[i] shifted point 0 of m[i] withcolor .6 white;
  draw (down--up) scaled 6 rotated angle direction 1 of m[i] shifted point 1 of m[i] withcolor .6 white;
  n := thelabel(decimal (round(arclength m[i] / u * 100)/100), point 1/2 of m[i]);
  unfill bbox n; draw n withcolor .5 white;
endfor

endfig;
end.

Most of the effort here is to make the nice length labels for the sides of the triangle.