\documentclass[12pt]{article}\usepackage[Aymar]{ammaths}\en

\begin{document}

\module{Vectors and configurations}{01}{14}{1 period}

\prereq{Basic notions about vectors.}

\object{\begin{itemize}
        \item Discover some simple but important configurations involving vectors.
	\item Practise the vocabulary about geometry.
	\end{itemize}}

\mater{\begin{itemize}
	\item Hand-outs with all the configurations and properties.
	\item Properties involving vectors and figures.
	\end{itemize}}

\modpart{Matching game}{10 mins}

Each student is handed out a card with either a property involving vectors or a figure. Students mingle to find who has the right property for their figure and vice-versa.

\modpart{Writing a property}{15 mins}

Each pair has to write a precise rule about the property and figure they have. The teacher checks each one of them.

\modpart{Presentations}{30 mins}

Each pair goes to the board to explain their property. A hand-out with all properties is given at the end of the
session.

\pagebreak

\moddocdis{Vectors and configurations}{01}{14}{Summary}


\begin{multicols}{2}

A quadrilateral $ABCD$ is a parallelogram if and only if
$$\vect{AB}=\vect{DC}.$$

\vfill

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\end{multicols}

\begin{multicols}{2}
A quadrilateral $ABDC$ is a parallelogram if and only if
$$\vect{AB}=\vect{CD}.$$

\vfill

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\end{multicols}

\begin{multicols}{2}
For any three points $A$, $B$ and $C$,
$$\vect{AB}+\vect{BC}=\vect{AC}.$$

\vfill

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\end{multicols}

\begin{multicols}{2}
A quadrilateral $ABCD$ is a parallelogram if and only if
$$\vect{AB}+\vect{AD}=\vect{AC}.$$

\vfill

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\end{multicols}

\begin{multicols}{2}
A point $A$ is the midpoint of a segment $BC$ if and only if
$$\vect{AB}+\vect{AC}=\vect{0}.$$

\vfill

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\end{multicols}

\begin{multicols}{2}
A point $C$ is the midpoint of a segment $AB$ if and only if
$$\vect{AB}=2\vect{AC}.$$

\vfill

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\end{multicols}

\begin{multicols}{2}
The fact that the lines $CD$ and $AB$ are parallel, that the vectors have the same direction and that $CD=2AB$ can
be written as the equality
$$\vect{CD}=\vect{2AB}.$$

\vfill

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\end{multicols}

\begin{multicols}{2}
If point $B$ is the midpoint of a segment $AC$, then for any point $M$,
$$\vect{MA}+\vect{MC}=2\vect{MB}.$$

\vfill

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\end{multicols}

\begin{multicols}{2}
If the points $B$ and $C$ are the midpoints of the legs of a triangle $ADE$, then
$$\vect{DE}=2\vect{BC}.$$

\vfill

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\end{multicols}

\begin{multicols}{2}
If $A$ is the intersection of two segments $BC$ and $DE$ such that $AB=\frac13AC$ and $AD=\frac13AE$ then
$$\vect{BD}=-\frac13\vect{CE}.$$

\vfill

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\pagebreak

\moddoc{Properties and figures}

{\Large

\begin{center}
\begin{tabular}{|p{1cm}cp{1cm}|p{1cm}cp{1cm}|}
\hline
& $\vect{AB}=\vect{DC}$ &&& $\vect{AB}=\vect{CD}$ & \rule{0pt}{35pt}\\[10pt]
\hline
& $\vect{AB}+\vect{BC}=\vect{AC}$ &&& $\vect{AB}+\vect{AD}=\vect{AC}$ & \rule{0pt}{35pt}\\[10pt]
\hline
& $\vect{AB}+\vect{AC}=\vect{0}$ &&& $\vect{AB}=2\vect{AC}$ & \rule{0pt}{35pt}\\[10pt]
\hline
& $\vect{CD}=\vect{2AB}$ &&& $\vect{MA}+\vect{MC}=2\vect{MB}$ &  \rule{0pt}{35pt}\\[10pt]
\hline
& $\vect{DE}=2\vect{BC}$ &&& $\vect{BD}=-\frac13\vect{CE}$& \rule{0pt}{35pt}\\[10pt]
\hline
\end{tabular}
\end{center}
}

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\hline
& \newrgbcolor{zwzwzw}{0.59 0.59 0.59}
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\hline
\end{tabular}
\end{center}



\end{document}
