Q:

How to draw a geometry triangle in LaTeX?

Name explicitly used properties from the sketch (for example, parallel lines, equilateral triangles, right angles), and give common calculation rules used.
Helpful calculation rules and methods can be depending on the example:

Area and Circumferential Formula for Rectangle, Triangle (Heron), Circle, Trapezoid • Angled Sets for Circle and Triangle, Pythagoras
Use of symmetries, similarity, proportionality
Setting up equations

A:

with pure tikz, using polar coordinates and intersections:
\documentclass[11pt, tikz, margin=3.141592]{standalone}
\usetikzlibrary{angles, intersections, quotes}

\begin{document}
    \begin{tikzpicture}[
myangle/.style args = {#1/#2}{angle radius=#1, angle eccentricity=#2, draw, font=\footnotesize},
myangle/.default=8mm/0.8
                        ]
\coordinate[label=$B$]      (b);
\coordinate[label=below:$C$](c) at (255:6);
\coordinate[label=below:$D$](d) at (285:6);
    \draw[name path=cbd]    (c) -- (b) -- (d) -- cycle;
    \draw[name path=A]      (b) ++ (250:6) arc (250:330:6);
    \path[name path=B]      (b) -- ++ (315:8.5);
    \draw[name path=ce,
          name intersections={of=A and B, by=e}] (c) -- (e) coordinate[label=right:E];
    \path[name path=C]      (d) -- ++ (0:4.5);
    \draw[name intersections={of=B and C, by=a}] (d) -- (a) coordinate[label=below:A];
    \draw   (b) -- (a);
% angles
\pic [myangle,"$\alpha$"] {angle = c--b--d};
\pic [myangle,"$\alpha$"] {angle = e--a--d};
    \path[name intersections={of=ce and cbd}] coordinate[label=75:F] (f) at (intersection-2);
\pic [myangle=4mm/0.5,"$\cdot$"] {angle = b--f--c};
% small circle
\draw[fill=white] (b) circle[radius=0.5mm];
    \end{tikzpicture}
\end{document}

A:

With just tkz-euclide (v3.01 here on CTAN
First method : we search the angle CBD todo this, we call Hb the projection of B on the line CD then it's easy to see that :
Ang(Hb,B,D)+Ang(D,B,A)+Ang(Hb,A,B) = 90 or Ang(Hb,B,D)=1/2 Ang(Hb,A,B).
We get 2.5x alpha = 90° so alpha=36°
\documentclass{standalone} 
\usepackage{tkz-euclide}
\begin{document} 
\begin{tikzpicture}

\tkzDefPoint(0,0){C} 
\tkzDefPoint(2,6){B}
% possible \tkzDefPoint[label=below:$C$](0,0){C} but don't do this
% We get D and E with a rotation
\tkzDefPointBy[rotation= center B angle 36](C) \tkzGetPoint{D} 
\tkzDefPointBy[rotation= center B angle 72](C) \tkzGetPoint{E} 
% Toget A we use an intersection of lines
\tkzInterLL(B,E)(C,D) \tkzGetPoint{A}
\tkzInterLL(C,E)(B,D) \tkzGetPoint{H}
% drawing
\tkzDrawArc[delta=10](B,C)(E)
\tkzDrawPolygon(C,B,D)
\tkzDrawSegments(D,A B,A C,E)
% angles 
\tkzMarkAngles(C,B,D E,A,D) %this is to draw the arcs
\tkzLabelAngles[pos=1.5](C,B,D E,A,D){$\alpha$}
\tkzMarkRightAngle(B,H,C)
\tkzDrawPoints(A,...,E)
% Label only now
\tkzLabelPoints[below left](C,A)
\tkzLabelPoints[below right](D)
\tkzLabelPoints[above](B,E)
\end{tikzpicture}
\end{document}

There is another possibility because the triangle CBD is a "Gold" triangle with angles 36°,72° and 72° and tkz-euclide knows how to define this kind of triangle.
\documentclass{standalone} 
\usepackage{tkz-euclide}
\begin{document}  
\begin{tikzpicture}
\tkzDefPoint(0,4){B}
\tkzDefPoint(8,0){A}
\tkzDefTriangle[gold](A,B) \tkzGetPoint{C}
\tkzDefTriangle[gold](B,C) \tkzGetPoint{D}
\tkzDuplicateSegment(B,C)(B,A)\tkzGetPoint{E}
\tkzInterLL(C,E)(B,D) \tkzGetPoint{F}
% drawing
\tkzDrawArc[delta=10](B,C)(E)
\tkzDrawPolygon(C,B,A)
\tkzDrawSegments(D,B C,E)
\tkzDrawPoints(A,...,F)
% % angles
\tkzFillAngles[fill=blue!10](C,B,D E,A,D)
\tkzMarkAngles(C,B,D E,A,D) %this is to draw the arcs
\tkzLabelAngles[pos=1.5](C,B,D E,A,D){$\alpha$}
\tkzMarkRightAngle[fill=blue!20](B,F,C)
% % Label only now
\tkzLabelPoints[below left](C,A)
\tkzLabelPoints[below right](D)
\tkzLabelPoints[above](B,E)
\end{tikzpicture}
\end{document}

