Drawing complex geometry

Question

I am having a difficult time drawing this complex geometry using LaTex

I will be happy to get help. Thanks

Answer 1

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.

\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}

Answer 2

Here is a version in Metapost, wrapped up in luamplib. Compile with lualatex.

\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}

Answer 3

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

\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}