# How can I draw a circle throught the vertices of a triangle? [closed]

i want to have the angles constructed ,but when clearly showing the arcs for the compass.so like the angle of 60 must be shown with its arcs ,and also the perpendicular bisectors must have the arcs as the meeting point on either sides

• What have you tried so far? Please add a minimal working example (MWE) of your current code. Commented Aug 19, 2020 at 11:28
• The tkz-euclide package allows you to do this natively. See 18.1.3 Circles inscribed and circumscribed for a given triangle on the manual. Commented Aug 19, 2020 at 12:05
• Another solution use pstricks, more precisely its pst-eucl module is done for that. Commented Aug 19, 2020 at 13:11
• thank you ,let me try out Commented Aug 19, 2020 at 13:13
• @kazibastephen please see if the answer meets the requirement Commented Aug 19, 2020 at 13:44

Please have a look and see if it meets the requirement

\documentclass{standalone}
\usepackage{tikz}

\usetikzlibrary{calc,positioning,angles,arrows.meta,quotes,intersections}
\usetikzlibrary{through}

\begin{document}
\begin{tikzpicture}[declare function={alpha=60; beta=75; sigma=110;}]

\path (0,0) coordinate (A)node[below]{A}  --  ++ (alpha:7cm) coordinate
(B)node[above]{B} --
([turn]alpha:-7cm) coordinate (C)node[below]{C};
\draw(A)--(B)--(C)--cycle;

\coordinate (D) at ($(A)!(B)!(C)$);
\draw[gray!50,dashed] (B)--(D);

\coordinate (E) at ($(C)!(A)!(B)$);
\draw[gray!50,dashed] (A)--(E);

\coordinate (F) at (intersection of B--D and A--E) ;
\node[fill=red,inner sep=1pt, circle] at(F){};
\node [draw,blue,thick,name path=circle,](c) at (F) [circle through={(A)}]
{};

\pic[ draw,<->,>=stealth,blue, "$60^0$"{fill=white},inner sep=1pt, circle,
draw,angle
eccentricity=1.1, angle radius = 10mm] {angle = C--A--B};

\pic[ draw,<->,>=stealth,blue, "$60^0$"{fill=white},inner sep=1pt, circle,
draw,angle
eccentricity=1.1, angle radius = 10mm] {angle = B--C--A};

\draw[line width=0.5pt] (0,-10mm) -- (0,2mm);
\draw[line width=0.5pt] (7cm,-10mm) -- (7cm,2mm);
\draw[<->,>=stealth,](0,-8mm) --node[midway,fill=white](){7cm}(7cm,-8mm);

\coordinate (G) at (0,5);
\path[name path=fg] (F)--(G);
\path [name intersections ={of=fg and circle,name=i}](i-1)  coordinate [];
\end{tikzpicture}
\end{document}


edit -- added another option with tkz euclide

\documentclass{article} % or another class
\usepackage{xcolor} % before tikz or tkz-euclide if necessary

\usepackage{tkz-euclide} % no need to load TikZ

\usetikzlibrary{babel} %if there are problems with the active characters
\begin{document}
\noindent\hspace{-4.5cm}
\begin{tikzpicture}
%equilateral triangle
\tkzDefPoint(0,0){A}
\tkzDefPoint(7,0){B}
%draw intersecting circles
\tkzInterCC(A,B)(B,A)\tkzGetPoints{C}{D}
\tkzDrawPoints[color=red](A,B,C)
\tkzDrawCircle[dashed, color=blue](A,B)
\tkzDrawCircle[dashed, color=blue](B,A)
\tkzCompass[color=blue,line width=2pt](A,C)
\tkzCompass[color=blue,line width=2pt](B,C)
%draw triangle
\tkzDrawPolygon[color=red, ](A,B,C)
\tkzMarkSegments[mark=s|,color=blue](A,C B,C)
\tkzLabelPoints[](A,B)
\tkzLabelPoints[above](C)
%circumscribed circle
\tkzDefMidPoint(A,B)\tkzGetPoint{P}
\tkzLabelPoints[](P)
\tkzDefLine[orthogonal =through P](A,B)\tkzGetPoint{X}
\tkzLabelPoints(X)
%orthogonal midpoints of two sides
\tkzDefMidPoint(B,C)\tkzGetPoint{Q}
\tkzLabelPoints[](Q)
\tkzDefLine[orthogonal =through Q](B,C)\tkzGetPoint{Y}
\tkzLabelPoints(Y)
%mark right angles
\tkzDrawLines[dashed,green](P,X Q,Y)
\tkzMarkRightAngles(B,Q,Y A,P,X)
%find center of circumscribe circle
\tkzInterLL(P,X)(Q,Y)\tkzGetPoint{Z}
\tkzLabelPoints(Z)
\tkzDrawPoint[green](Z)
\tkzCalcLength[cm](Z,A)\tkzGetLength{rZA}
\tkzDrawCircle[green!50!black,line width=2pt,R ](Z, \rZA cm)
%mark all three angles of equilateral triangle
\tkzMarkAngle[size=1cm,color=cyan,mark=||](B,A,C)
\tkzMarkAngle[size=1cm,color=cyan,mark=||](A,C,B)
\tkzMarkAngle[size=1cm,color=cyan,mark=||](C,B,A)
%find the angle of eq triangle
\tkzFindAngle(B,A,C)
\tkzGetAngle{angleBAC}
\edef\angleBAC{\fpeval{round(\angleBAC)}}
\tkzLabelAngle[pos=0.7](B,A,C){\angleBAC$^\circ$}
%draw radius of the circum circle
\tkzDefShiftPoint[Z](135:\rZA){z}
\tkzDrawSegments[arrows=-stealth](Z,z)

\end{tikzpicture}

\end{document}

• thank you very much Commented Aug 20, 2020 at 8:16
• thank you so much Commented Aug 23, 2020 at 7:50
• @kazibastephen -- in case the answer meets your requirement please accept by clicking on the tick/ checkmark on the left side of the answer Commented Aug 23, 2020 at 12:54

I assume that c, alpha and beta are given; then:

\documentclass[margin=5pt, tikz]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\usetikzlibrary{intersections}
\usetikzlibrary{backgrounds}
\usetikzlibrary{patterns}
\usetikzlibrary{positioning}
\usetikzlibrary{angles, quotes, babel}
\usepackage{amsmath, amssymb}

\begin{document}
% Gegebene Größen
\pgfmathsetmacro{\c}{7}
\pgfmathsetmacro{\Alpha}{60}
\pgfmathsetmacro{\Beta}{60}

% Seitenlängen
\pgfmathsetmacro{\Gamma}{180-\Alpha-\Beta}
\pgfmathsetmacro{\a}{\c*sin(\Alpha)/sin(\Gamma)} %
\pgfmathsetmacro{\b}{sqrt(\a*\a +\c*\c -2*\a*\c*cos(\Beta))} %

\pgfmathsetmacro{\R}{\a/(2*sin(\Alpha))} %
\pgfmathsetmacro{\McU}{\R*abs(cos(\Gamma))} %

\begin{tikzpicture}[%scale=0.7,
font=\footnotesize,
]

% Dreieckskonstruktion
%\pgfmathsetmacro{\Alpha}{acos((\b^2+\c^2-\a^2)/(2*\b*\c))} %
\coordinate[label=below:$A$] (A) at (0,0);
\coordinate[label=below:$B$] (B) at (\c,0);
\coordinate[label=$C$] (C) at (\Alpha:\b);
\draw[local bounding box=dreieck] (A) -- (B) -- (C) --cycle;

% Umkreis
\draw[red] ($(A)!0.5!(B)$) coordinate[label=-90:$M_c$] (Mc) -- +(90:\McU) coordinate[label=110:$U$](U) node[midway, right] {$|M_cU|$};
\draw[densely dashed, red] (U) -- (A) node[midway, above] {$R$};;
\draw pic [angle radius=3mm, %angle eccentricity=1.2,
draw,   "$\cdot$", red
] {angle =U--Mc--A};

% Annotationen - Dreieck
\draw[thick] (A) -- (B)  node[pos=0.25, below]{$c/2$} node[pos=0.75, below]{$c/2$};
\draw pic [angle radius=6mm, %angle eccentricity=1.2,
draw,   "$\alpha$", thick
] {angle =B--A--C};
\draw pic [angle radius=6mm, %angle eccentricity=1.2,
draw,   "$\beta$", thick
] {angle =C--B--A};

\draw[-latex] (U) -- +(44:\R) node[near end, above]{$R$};

% Annotationen - Aufgabe
\pgfmathsetmacro{\x}{max(\a, \b,\c)} %
\begin{scope}[shift={($(dreieck.north west)+(-\x cm-3mm,0)$)}]
% Strecken
\foreach[count=\y from 0] \s/\S in {c/c}{%%
\draw[|-|, yshift=-\y*5mm, local bounding box=strecken] (0,0) -- (\csname \s \endcsname,0) node[midway, above]{$\S$ %= \csname \s \endcsname cm
};}%%
\end{scope}
% Winkel
\pgfmathsetmacro{\Winkel}{\Alpha}
\pgfmathsetmacro{\WinkelXShift}{\Winkel > 90 ? -cos(\Winkel) : 0} %
\draw[shift={($(strecken.south west)+(\WinkelXShift,50mm)$)}] (\Winkel:1)  coordinate(P) -- (0,0) coordinate(Q) -- (1,0) coordinate(R);
\draw pic [draw, angle radius=7mm, %angle eccentricity=1.3,
% pic text={$\Winkel$}, pic text options={},
"$\alpha$",
] {angle =R--Q--P};
% Winkel 2
\pgfmathsetmacro{\Winkel}{\Beta}
\pgfmathsetmacro{\WinkelXShift}{\Winkel > 90 ? -cos(\Winkel) : 0} %
\draw[shift={($(strecken.south west)+(15mm+\WinkelXShift,50mm)$)}] (\Winkel:1)  coordinate(P) -- (0,0) coordinate(Q) -- (1,0) coordinate(R);
\draw pic [draw, angle radius=7mm, %angle eccentricity=1.3,
% pic text={$\Winkel$}, pic text options={},
"$\beta$",
] {angle =R--Q--P};

% Annotationen - Rechnung
\tikzset{PosUnten/.style={below=5mm of dreieck, anchor=north,}}
\tikzset{PosLinks/.style={shift={($(dreieck.north)+(-44mm,-17mm)$)}, anchor=north east,}}
\node[yshift=-0mm, draw, align=left, fill=lightgray!50,
%PosUnten,
] (Rechnung){
$\begin{array}{l l} c = \c \text{ cm} & \\ \alpha = \Alpha^\circ & \\ \beta = \Beta^\circ & \\ \hline \gamma = 180^\circ-\alpha-\beta &=\Gamma^\circ \\ a = c\cdot\dfrac{\sin(\alpha)}{\sin(\gamma)} & =\a \text{ cm} \\[1em] b = \sqrt{a^2+c^2-2ac\cos(\beta)} & =\b \text{ cm} \\ R = \dfrac{a}{2\sin(\alpha)} &=\R \text{ cm} \\[1em] |M_cU| = R|\cos(\gamma)| &=\McU \text{ cm} \\ \end{array}$
};

\node[anchor=north west, yshift=-3mm, inner sep=0pt, draw=none] at (Rechnung.south west){
$\begin{array}{l l} \text{Hint: } |M_cU|^2 \hspace{-3mm}& =R^2-\left(\dfrac{c}{2}\right)^2 = R^2 - \bigl( R \sin (\gamma)\bigr)^2 \\[0.75em] &= R^2 \bigl(1-\sin^2(\gamma) \bigr) =R^2\cos^2(\gamma)\\[1em] \multicolumn{2}{l}{\Rightarrow |M_cU| = R|\cos(\gamma)|} \end{array}$
};

%% Punkte
\foreach \P in {U, Mc}
\draw[fill=black!1, draw=red] (\P) circle (1.75pt);
\end{tikzpicture}
\end{document}


Hint: The vector-computational way from the side lengths and corner points to the centerpoint (U) of the circumscribed circle

\pgfmathsetmacro{\Da}{\a^2*(\b^2+\c^2-\a^2)} %
\pgfmathsetmacro{\Db}{\b^2*(\a^2+\c^2-\b^2)} %
\pgfmathsetmacro{\Dc}{\c^2*(\a^2+\b^2-\c^2)} %
\pgfmathsetmacro{\D}{\Da+\Db+\Dc} %
\pgfmathsetmacro{\au}{\Da/\D} %
\pgfmathsetmacro{\bu}{\Db/\D} %
\pgfmathsetmacro{\cu}{\Dc/\D} %

\coordinate[] (U) at ($\au*(A)+\bu*(B)+\cu*(C)$);


is at wikipedia.

MWE:

\documentclass[margin=5pt, tikz]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\usetikzlibrary{intersections}
\usetikzlibrary{backgrounds}
\usetikzlibrary{patterns}
\usetikzlibrary{positioning}
\usetikzlibrary{angles, quotes, babel}
\usepackage{amsmath, amssymb}

\begin{document}

% Gegebene Größen
\pgfmathsetmacro{\c}{7}
\pgfmathsetmacro{\Alpha}{60}
\pgfmathsetmacro{\Beta}{60}

% Seitenlängen
\pgfmathsetmacro{\Gamma}{180-\Alpha-\Beta}
\pgfmathsetmacro{\a}{\c*sin(\Alpha)/sin(\Gamma)} %
\pgfmathsetmacro{\b}{sqrt(\a*\a +\c*\c -2*\a*\c*cos(\Beta))} %

\pgfkeys{/tikz/savevalue/.code 2 args={\global\edef#1{#2}}}

\begin{tikzpicture}[%scale=0.7,
font=\footnotesize,
background rectangle/.style={draw=none, fill=black!1, rounded corners}, show background rectangle,
Punkt/.style 2 args={  label={[#1]:$#2$}   },
Dreieck/.style={thick},
]

% Dreieckskonstruktion
%\pgfmathsetmacro{\Alpha}{acos((\b^2+\c^2-\a^2)/(2*\b*\c))} %
\coordinate[Punkt={below}{A}] (A) at (0,0);
\coordinate[Punkt={below}{B}] (B) at (\c,0);
\coordinate[Punkt={above}{C}] (C) at (\Alpha:\b);
\draw[local bounding box=dreieck] (A) -- (B) -- (C) --cycle; % Dreieck zeichnen

% Umkreis
\pgfmathsetmacro{\s}{0.5*(\a+\b+\c)} %
\pgfmathsetmacro{\F}{sqrt(\s*(\s-\a)*(\s-\b)*(\s-\c))} %

\pgfmathsetmacro{\Da}{\a^2*(\b^2+\c^2-\a^2)} %
\pgfmathsetmacro{\Db}{\b^2*(\a^2+\c^2-\b^2)} %
\pgfmathsetmacro{\Dc}{\c^2*(\a^2+\b^2-\c^2)} %
\pgfmathsetmacro{\D}{\Da+\Db+\Dc} %
\pgfmathsetmacro{\au}{\Da/\D} %
\pgfmathsetmacro{\bu}{\Db/\D} %
\pgfmathsetmacro{\cu}{\Dc/\D} %

\coordinate[Punkt={below}{U}] (U) at ($\au*(A)+\bu*(B)+\cu*(C)$);

\pgfmathsetmacro{\R}{(\a*\b*\c)/(4*\F)} %

% Annotationen - Dreieck
\draw[thick] (A) -- (B)  node[midway, below]{$c$};
\draw pic [angle radius=6mm, %angle eccentricity=1.2,
draw,   "$\alpha$", thick
] {angle =B--A--C};
\draw pic [angle radius=6mm, %angle eccentricity=1.2,
draw,   "$\beta$", thick
] {angle =C--B--A};

\draw[-latex] (U) -- +(33:\R) node[near end, above]{$R$};

% Annotationen - Aufgabe
\pgfmathsetmacro{\x}{max(\a, \b,\c)} %
\begin{scope}[shift={($(dreieck.north west)+(-\x cm-3mm,0)$)}]
% Strecken
\foreach[count=\y from 0] \s/\S in {c/c}{%%
\draw[|-|, yshift=-\y*5mm, local bounding box=strecken] (0,0) -- (\csname \s \endcsname,0) node[midway, above]{$\S$ %= \csname \s \endcsname cm
};}%%
\end{scope}
% Winkel
\pgfmathsetmacro{\Winkel}{\Alpha}
\pgfmathsetmacro{\WinkelXShift}{\Winkel > 90 ? -cos(\Winkel) : 0} %
\draw[shift={($(strecken.south west)+(\WinkelXShift,50mm)$)}] (\Winkel:1)  coordinate(P) -- (0,0) coordinate(Q) -- (1,0) coordinate(R);
\draw pic [draw, angle radius=7mm, %angle eccentricity=1.3,
% pic text={$\Winkel$}, pic text options={},
"$\alpha$",
] {angle =R--Q--P};
% Winkel 2
\pgfmathsetmacro{\Winkel}{\Beta}
\pgfmathsetmacro{\WinkelXShift}{\Winkel > 90 ? -cos(\Winkel) : 0} %
\draw[shift={($(strecken.south west)+(15mm+\WinkelXShift,50mm)$)}] (\Winkel:1)  coordinate(P) -- (0,0) coordinate(Q) -- (1,0) coordinate(R);
\draw pic [draw, angle radius=7mm, %angle eccentricity=1.3,
% pic text={$\Winkel$}, pic text options={},
"$\beta$",
] {angle =R--Q--P};

% Annotationen - Rechnung
\tikzset{PosUnten/.style={below=5mm of dreieck, anchor=north,}}
\tikzset{PosLinks/.style={shift={($(dreieck.north)+(-44mm,-20mm)$)}, anchor=north east,}}
\node[yshift=-0mm, draw, align=left, fill=lightgray!50,
%PosUnten,
$\begin{array}{l l} c = \c \text{ cm} & \\ \alpha = \Alpha^\circ & \\ \beta = \Beta^\circ & \\ \hline \gamma = 180^\circ-\alpha-\beta &=\Gamma^\circ \\ a = c\cdot\dfrac{\sin(\alpha)}{\sin(\gamma)} & =\a \text{ cm} \\[1em] b = \sqrt{a^2+c^2-2ac\cos(\beta)} & =\b \text{ cm} \\ R = \dfrac{a}{2\sin(\alpha)} &=\R \text{ cm} \\ %\beta = \Beta^\circ & (5) \\ %\gamma = \Gamma^\circ & (2) \\ %\multicolumn{2}{l}{s_{a, \text{max}} = \saMax \text{ cm}} \\ \end{array}$