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\usepackage{lipsum} % Generate placeholder text, you can remove this
\usepackage{multicol} % For multiple columns
\usepackage{amsmath,amsfonts,amssymb}
\usepackage{tikz}
\usepackage{tabularx,tabulary}
\usepackage{array} % Add the array package for 'm' column type
\usepackage{graphicx}
\usepackage{booktabs}
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\newcommand\ddfrac[2]{\frac{\displaystyle #1}{\displaystyle #2}}
\begin{document}
\thispagestyle{empty}
\setlength{\columnsep}{1cm} % Adjust the column separation here
\setlength{\columnseprule}{1pt}
\newcolumntype{M}[1]{>{\centering\arraybackslash}m{#1}}
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\noindent
\begin{tabularx}{\linewidth}{|M{0.30\linewidth}|M{0.25\linewidth}|M{0.35\linewidth}|}
\hline
Collège Tariq Ben Ziad 2
&
Devoir N 2 S 02
&
Année Scolaire 2023-2024
\\ \hline
Prof : MOSAID
&
1APIC-A
&
durée : 1h30min
\\ \hline
\end{tabularx}
\begin{tabularx}{\linewidth}{|p{0.965\linewidth}|}
Nom et prénom : \\
\hline
\end{tabularx} \\
\noindent
\underline{\textbf{Exercice 1:}}\\
Soit la figure suivante telle que $(AB)\parallel(CD)$
\begin{center}
\begin{tikzpicture}[scale=1]
% Draw the lines
\draw[thick] (-2,-2) -- (2,2);
\draw[thick] (1,-2) -- (5,2);
\draw[thick] (-3,0) -- (6,0);
% Define the points A, B, C, and D
\coordinate (A) at (-1.5,-1.5);
\coordinate (B) at (1.5,1.5);
\coordinate (C) at (1.5,-1.5);
\coordinate (D) at (4.5,1.5);
\coordinate (E) at (-2.5,0);
\coordinate (F) at (5.5,0);
\coordinate (I) at (0,0);
\coordinate (J) at (3,0);
\draw[fill=red] (I) -- ++(0.5,0) arc (0:45:0.5) node[pos=0.3,anchor=south west ] {\tiny 45};
% Draw ticks at points A, B, C, and D
\foreach \point/\name/\pos in {A/A/above, B/B/above, C/C/above, D/D/above,
E/E/above, F/F/above right, I/I/above left, J/J/above}
\draw[rotate=-90, \pos] (\point) +(0.1,0.1) -- +(-0.1,-0.1) node {\name};
\end{tikzpicture}
\end{center}
Calculer les mesures des angles suivants:\\
\hspace*{0.4cm}$A\widehat I E$ , $I\widehat J C$ , $E\widehat I B$ , $I\widehat J D$ , $D\widehat J F$ et $C\widehat J F$\\
\noindent
\underline{\textbf{Exercice 2:}}\\
Résoudre les équations suivantes:\\
\(x+3=4\quad\);\(\quad 2x-3=4+x\quad\);
\(\quad 4x+3=2(x-1)\quad\)\\
$\quad6x-7=4\quad$; $\quad-2(3-x)=3(4x-7)\quad$;
$\dfrac{1}{2}x-3=\dfrac{2}{3}x-4$
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\end{multicols} % End two-column layout
\end{document}
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