Similar Triangle

Question

I am trying to draw something like this in overleaf using Tikzpicture, I need assistance

Answer 1

This is relative simple problem. You only need to make yourself a little bit familiar with TikZ package. It has very detailed (consequently huge) documentation, but for start is sufficient to read the first Tutorial or A very minimal introduction to TikZ and see examples of its use TeXample

With use of the TikZ library angles and scale of picture:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{angles}
\begin{document}
    \begin{figure}
    \centering
\begin{tikzpicture}[
myangles/.style = {draw, 
                   angle radius = #1 mm},
myangles/.default = 4
                    ]
%%%% first triangle
\draw   (0,0) coordinate[label= left:A] (a) -- 
        (3,0) coordinate[label=right:B] (b) -- 
        (1,2) coordinate[label=above:C] (c) -- cycle;
\pic [myangles]     {angle = b--a--c};
\pic [myangles]     {angle = a--c--b};
\pic [myangles=5]   {angle = a--c--b};
\pic [myangles]     {angle = c--b--a};
\pic [myangles=5]   {angle = c--b--a};
\pic [myangles=6]   {angle = c--b--a};
%%%% second triangle: scaled of the first
    \begin{scope}[xshift=44mm, scale=2]
\draw   (0,0) coordinate[label= left:D] (a) --
        (3,0) coordinate[label=right:E] (b) --
        (1,2) coordinate[label=above:F] (c) -- cycle;
\pic [myangles]     {angle = b--a--c};
\pic [myangles]     {angle = a--c--b};
\pic [myangles=5]   {angle = a--c--b};
\pic [myangles]     {angle = c--b--a};
\pic [myangles=5]   {angle = c--b--a};
\pic [myangles=6]   {angle = c--b--a};
    \end{scope}
\end{tikzpicture}
    \caption{Triangles in Tikz}
    \end{figure}
\end{document}

Answer 2

One solution with tkz-euclide:

\documentclass{article}
\usepackage{tkz-euclide}
\begin{document}
    \begin{figure}
        \centering
        \begin{tikzpicture}[double distance=3pt]
            \clip(-.5,-.5) rectangle (12,5);
            \tkzDefPoints{-5/0/S,0/0/A,3/0/B,1/2/C}
            \tkzDefPointsByhomothety=center S ratio 2{D,E,F}
        \tkzDrawPolygon(A,B,C)
        \tkzDrawPolygon(D,E,F)
        \tkzLabelPoints[left](A,D)
        \tkzLabelPoints[right](B,E)
        \tkzLabelPoints[above](C,F)
        \tkzMarkAngle[size=5mm,mark=|,arc=l](B,A,C)
        \tkzMarkAngle[size=5mm,mark=none,arc=l](C,B,A)
        \tkzMarkAngle[size=5mm,mark=s,arc=l](A,C,B)
        \tkzMarkAngle[size=10mm,mark=|,arc=l](E,D,F)
        \tkzMarkAngle[size=10mm,mark=none,arc=l](F,E,D)
        \tkzMarkAngle[size=10mm,mark=s,arc=l](D,F,E)
    \end{tikzpicture}
    \caption{Triangles in Tikz}
\end{figure}

\end{document}

Answer 3

Here is something to start with.

How to properly make the right angle mark see here.

\documentclass[border=3mm]{article}
\usepackage{tikz}
\usepackage{caption}
\newcommand*{\rechterWinkel}[3]{% #1 = point, #2 = start angle, #3 = radius
    \draw[shift={(#2:#3)}] (#1) arc[start angle=#2, delta angle=90, radius = #3];
}
\begin{document}
\begin{figure}
    \centering

\begin{tikzpicture}     

    \draw (0,0) -- (1,1) -- (3,0) -- cycle;
\node[xshift=-2mm](A) at (0,0) {A};
\node[yshift=2mm](C) at (1,1) {C};
\node[xshift=2mm](B) at (3,0) {B};
\rechterWinkel{0.2,0}{1}{.2} %A

\rechterWinkel{1,1.05}{-127}{.2} %B
\rechterWinkel{1,1.04}{-127}{.25} %B

\rechterWinkel{2.8,0}{90}{.1} %C   
\rechterWinkel{2.75,0}{90}{.128} %C
\rechterWinkel{2.7,0}{90}{.155} %C




    \end{tikzpicture}

\begin{tikzpicture}     

\draw (0,0) -- (2,2) -- (5,0) -- cycle;
\node[xshift=-2mm](D) at (0,0) {D};
\node[yshift=2mm](F) at (2,2) {F};
\node[xshift=2mm](E) at (5,0) {E};

\end{tikzpicture}

\caption{Triangle in Tikz}
\end{figure}


\end{document}