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I am having a difficult time drawing this complex geometry using LaTex enter image description here

I will be happy to get help. Thanks

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  • 1
    Draw it with GeoGebra... – projetmbc Apr 17 at 8:13
  • any particular angle for A and B and lengths – js bibra Apr 17 at 8:46
  • 2
    It's pretty straightforward with TikZ, did you try anything yet? Please feel free to share your code. – SebGlav Apr 17 at 9:00
  • Here is an example of triangles using tikz. With inspiration from this you should be able to draw you figure pretty easily. – John W. de Burgh Apr 17 at 9:55
  • @dupsy -- A solution with Tikz is added below -- please have a look -- for the sake of proof concept the angle measurement is done with tkz-euclide to show that the angles are same at P and Q – js bibra Apr 19 at 8:46
6

Here's a solution, using tkz-euclide, since you have several intersections and angles to take in account. This is fairly achievable with plain TikZ but here you can see the construction step by step and it's really understandable.

tkz-euclide

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{tkz-euclide}

\begin{document}
    \begin{tikzpicture}

        % Angles A & B (can be modified)
        \def\AngleA{20} \def\AngleB{40}

        % Base points
        \tkzDefPoints{0/0/P,4/0/R}

        % Aux points to create A and B angles and intersections
        \tkzDefShiftPoint[P](\AngleA:20){R1}
        \tkzDefShiftPoint[P](\AngleB+\AngleA:20){R2}

        % Perpendicular line to the base line through R
        \tkzDefLine[orthogonal=through R](P,R) \tkzGetPoint{r}      
        
        % Intersections X, Q, and N
        \tkzInterLL(P,R1)(R,r) \tkzGetPoint{X}
        \tkzInterLL(P,R2)(R,r) \tkzGetPoint{Q}
        
        % Perpendicular line to (PX) through Q to get N then O
        \tkzDefLine[orthogonal=through Q](P,R1) \tkzGetPoint{q}
        \tkzInterLL(P,R1)(Q,q) \tkzGetPoint{N}
        \tkzDefLine[orthogonal=through N](P,R) \tkzGetPoint{n}
        \tkzInterLL(P,R)(N,n) \tkzGetPoint{O}
        
        % Parallel to (PR) through N to get M
        \tkzDefLine[parallel=through N](R,P) \tkzGetPoint{n}
        \tkzInterLL(N,n)(R,r) \tkzGetPoint{M}
        
        % Now the drawings
        \tkzFillAngle[fill=orange!50,size=1.5cm,draw](R,P,X)
        \tkzLabelAngle[pos=1.8](R,P,X){$\alpha$}
        
        \tkzFillAngle[fill=pink!50,size=1cm,draw](X,P,Q)
        \tkzLabelAngle[pos=1.3](X,P,Q){$\beta$}
        
        \tkzFillAngle[fill=orange!50,size=1.5cm,draw](M,Q,N)
        \tkzLabelAngle[pos=1.8](M,Q,N){$\alpha$}
        
        \tkzMarkRightAngles[fill=blue!20,size=.3,draw](Q,R,P N,O,P N,M,Q Q,N,R1)
        
        \tkzDrawLine[add=0.1 and 0.1](P,O)
        \tkzDrawLine[add=0 and 0.1](R,Q)
        \tkzDrawLine[add=0 and 0.1](P,N)
        \tkzDrawSegments(O,N N,M P,Q Q,N)
        
        \tkzLabelPoints[below](P,R,O)
        \tkzLabelPoints[left](Q,M)
        \tkzLabelPoints[below right](X,N)
        
    \end{tikzpicture}
\end{document}
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  • right angle at N is required mark at outer side -- +1 for the nice answer – js bibra Apr 18 at 2:02
  • 1
    Even it was not that mandatory, I changed side for the right angle on point N ;) – SebGlav Apr 18 at 9:13
  • Thank you for the help. however, when I run it on WinEdt, the display was on two pages and part of the diagram itself is cut off – dupsy Apr 18 at 20:52
  • 1
    @dupsy You should try and compile on a more solid software, maybe. My code is also compiling fine on CoCalc, if you want to check. Try replacing the standalone class by another one (say article). The picture is less than 6 cm wide and less than 10 cm high so it shouldn't get out of your page, at least if you're on A4 paper with reasonable margins. – SebGlav Apr 18 at 22:29
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Here is a version in Metapost, wrapped up in luamplib. Compile with lualatex.

enter image description here

\documentclass[border=5mm]{standalone}
\usepackage{luamplib}
\begin{document}
\mplibtextextlabel{enable}
\begin{mplibcode}
beginfig(1);

numeric alpha, beta;
alpha = 28;
beta = 36;

pair M, N, O, P, Q, R, X;
P = origin;
Q = 240 dir (alpha + beta);
R = (xpart Q, ypart P);
N = whatever * dir alpha;
Q - N = whatever * dir (alpha + 90);
M = (xpart R, ypart N);
O = (xpart N, ypart R);
X = whatever[P, N] = whatever[R, Q];

drawoptions(withpen pencircle scaled 1/4 withcolor 1/2);
draw unitsquare scaled 5 shifted R;
draw unitsquare scaled 5 shifted O;
draw unitsquare scaled 5 shifted M;
draw unitsquare scaled 5 rotated alpha shifted N;

drawoptions(withpen pencircle scaled 3/8 withcolor 2/3 red);
draw fullcircle scaled 32 cutafter (P--N);
draw fullcircle scaled 32 rotated 270 shifted Q cutafter (Q--N);
label("$\alpha$", 22 dir 1/2 alpha);
label("$\alpha$", 22 dir (270 + 1/2 alpha) shifted Q);

drawoptions(withpen pencircle scaled 3/8 withcolor 2/3 blue);
draw fullcircle scaled 28 rotated alpha cutafter (P--Q);
label("$\beta$", 21 dir (alpha + 1/2 beta));

drawoptions();

vardef through(expr a, b, o) = (1+o/abs(a-b))[b, a] -- (1+o/abs(a-b))[a, b] enddef;
vardef half_through(expr a, b, o) = a -- (1+o/abs(a-b))[a, b] enddef;

draw M--N--O;
draw P--Q--N;
draw through(P, O, 16);
draw half_through(P, N, 12);
draw half_through(R, Q, 12);

interim dotlabeldiam := 2;
dotlabel.bot("$P$", P);
dotlabel.bot("$R$", R);
dotlabel.bot("$O$", O);
dotlabel.rt("$Q$", Q);
dotlabel.lrt("$M$", M);
dotlabel.lrt("$N$", N);
dotlabel.lrt("$X$", X);

endfig;
\end{mplibcode}
\end{document}
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  • 1
    As usual, very educational code. I like the trick of using whatever rather than intersectionpoint to find X! – Aditya Apr 24 at 5:18
0

A solution with Tikz -- for the sake of proof concept the angle measurement is done with tkz-euclide to show that the angles are same at P and Q

enter image description here

\documentclass[tikz,border=11pt]{standalone}
\usetikzlibrary{calc} 
\usetikzlibrary{angles,intersections,quotes}
\usepackage{tkz-euclide}


\newcommand{\MarkRightAngle}[4][.3cm]% #1=size (optional), #2-#4 three points: \angle #2#3#4
{\coordinate (tempa) at ($(#3)!#1!(#2)$);
    \coordinate (tempb) at ($(#3)!#1!(#4)$);
    \coordinate (tempc) at ($(tempa)!0.5!(tempb)$);%midpoint
    \draw (tempa) -- ($(#3)!2!(tempc)$) -- (tempb);
}
\begin{document}
\begin{tikzpicture}
    \draw [help lines] (0,0) grid (10,12);
    \draw [blue!70!black, thick](0,0) -- (10,0)coordinate(p10);
    \coordinate[label=-90:P](B) at (1,0);
    
    \draw [blue!70!black, thick,name path=rvert](6,0)coordinate[label=-90:R](r)--+(90:11cm);
    
    \path[name path=pr](B)--++(60:12)coordinate(p12);
    \path [name intersections={of=pr and rvert,by={D}}];
    \draw[blue!70!black, thick](B)--(D)coordinate[label=135:Q](D);
    \draw[blue!70!black, thick,name path=pn](B)--+(20:9cm)coordinate(p20);
    \draw[red!70!black,ultra thick](D)--($(B)!(D)!(p20)$)coordinate[label=-45:N](A);
    \draw[red!70!black,ultra thick](A)--($(p10)!(A)!(B)$)coordinate[label=-90:O](C);
    \draw[red!70!black,ultra thick](A)--($(D)!(A)!(r)$)coordinate[label=180:M](M);
    \path [name intersections={of=pn and rvert,by={pr2}}];
    \node[label=-45:X] at(pr2){};
    
    \MarkRightAngle{p20}{A}{D}
    \MarkRightAngle{A}{M}{D}
    \MarkRightAngle{B}{r}{D}
    \MarkRightAngle{B}{C}{A}
    \pic[ draw,,<->,>=stealth,red!60!black, "$\alpha$"{fill=red!20},inner sep=1pt, circle, angle eccentricity=1.1, angle radius = 20mm] {angle = M--D--A}; 
    \pic[ draw,,<->,>=stealth,red!60!black, "$\alpha$"{fill=red!20},inner sep=1pt, circle, angle eccentricity=1.1, angle radius = 20mm] {angle = C--B--A}; 
    \pic[ draw,,<->,>=stealth,red!60!black, "$\beta$"{fill=red!20},inner sep=1pt, circle, angle eccentricity=1.1, angle radius = 20mm] {angle = A--B--D}; 
    
    
    \tkzFindAngle(C,B,A)
    \tkzGetAngle{angleCBA}
    \tkzLabelAngle[yshift=-1cm,below](C,B,A){\angleCBA$^\circ$}
    
    \tkzFindAngle(M,D,A)
    \tkzGetAngle{angleMDA}
    \tkzLabelAngle[xshift=1cm,right](M,D,A){\angleMDA$^\circ$}
    
\end{tikzpicture}
\end{document}

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