\documentclass{article}
\usepackage{tikz,amsmath}
\usepackage[paperwidth=11cm,paperheight=12cm,margin=0cm]{geometry}
\setlength{\parindent}{0pt}
\usepackage{tkz-euclide}
\newcommand{\fig}
{%
\tkzInit
\tkzDefPoint[label=below left:$A$](0,0){A}
\tkzDefPoint(4,0){B}
\tkzDefPoint[label=above:$C$](0,3){C}
\tkzDefPoint(0,-4){D}
\tkzDefPoint(4,-4){E}
\tkzDefPoint[label=below left:$G$](-3,0){G}
\tkzDefPoint[label=above left:$F$](-3,3){F}
\tkzDefPointBy[rotation= center C angle 90](B)
\tkzGetPoint{I}
\tkzDefPointBy[rotation= center B angle -90](C)
\tkzGetPoint{H}
\tkzDrawPoints(A,B,C,D,E,F,G,H,I)
\tkzLabelPoints(B,D,E,H)
\tkzLabelPoint[above](I){$I$}
\draw (A) -- (C) -- (F) -- (G) -- cycle;
\draw (A) -- (B) -- (E) -- (D) -- cycle;
\draw (B) -- (C) -- (I) -- (H) -- cycle;
\draw ($(A)+(0,.2)$) -- ($(A)+(.2,.2)$) -- ($(A)+(.2,0)$);
\tkzDefLine[orthogonal=through A](B,C) \tkzGetPoint{j}
\tkzInterLL(A,j)(I,H) \tkzGetPoint{K}
\tkzInterLL(A,j)(B,C) \tkzGetPoint{J}
\node[right,gray!50,scale=0.7] at (-3.5,-4.25) {\texttt{mathweb.fr}};
}
\begin{document}
% on trace la parallèle à (IC) passant ar A
\foreach \c in {0,0.1,0.2,...,1}
{
\begin{tikzpicture}
\fig
\draw[dashed] (A) -- (${1-\c}*(A)+\c*(J)$);
\node[below right,text width=5cm] at (-3.5,7.5) {On trace la parallèle à $(CI)$ passant par $A$\linebreak Elle coupe $(BC)$ en $J$...};
\end{tikzpicture}
}
\begin{tikzpicture}
\fig
\tkzLabelPoint[below](J){$J$}
\draw[dashed] (A) -- (J);
\node[below right,text width=5cm] at (-3.5,7.5) {On trace la parallèle à $(CI)$ passant par $A$\linebreak Elle coupe $(BC)$ en $J$...};
\end{tikzpicture}
\foreach \c in {0,0.1,0.2,...,1}
{
\begin{tikzpicture}
\fig
\tkzLabelPoint[below](J){$J$}
\draw[dashed] (A) -- (J);
\draw (J) -- (${1-\c}*(J)+\c*(K)$);
\node[below right,text width=5cm] at (-3.5,7.5) {On trace la parallèle à $(CI)$ passant par $A$\linebreak Elle coupe $(BC)$ en $J$ et $(IH)$ en $K$.};
\end{tikzpicture}
}
\begin{tikzpicture}
\fig
\tkzLabelPoint[below](J){$J$}
\draw[dashed] (A) -- (J);
\draw (J) -- (K);
\tkzLabelPoint[above](K){$K$}
\node[below right,text width=5cm] at (-3.5,7.5) {On trace la parallèle à $(CI)$ passant par $A$\linebreak Elle coupe $(BC)$ en $J$ et $(IH)$ en $K$.};
\end{tikzpicture}

%%% On trace le triangle ACI

\begin{tikzpicture}
\fig
\tkzLabelPoint[below](J){$J$}
\draw[dashed] (A) -- (J);
\tkzLabelPoint[above](K){$K$}
\draw (J) -- (K);
\filldraw[draw=orange,fill=orange!50,opacity=.5] (A) -- (C) -- (I) -- cycle;
\end{tikzpicture}

%%% On anime le triangle ACI qui se trabnsforme en JCI

\foreach\c in {0,0.1,0.2,...,1}
{
\begin{tikzpicture}
\fig
\tkzLabelPoint[below](J){$J$}
\draw[dashed] (A) -- (J);
\tkzLabelPoint[above](K){$K$}
\draw (J) -- (K);
\filldraw[draw=orange,fill=orange!50,opacity=.5] (${1-\c}*(A)+\c*(J)$) -- (C) -- (I) -- cycle;
\node[below right,text width=5cm] at (-3.5,7.5) {Les triangles $ACI$ et $JCI$ ont la même aire car ils ont la même base $[CI]$ et la même\linebreak hauteur $[CJ]$.};
\end{tikzpicture}
}

% On fait tourner ACI

\foreach\a in {0,10,20,...,90}
{
\begin{tikzpicture}
\fig
\tkzLabelPoint[below](J){$J$}
\draw[dashed] (A) -- (J);
\tkzLabelPoint[above](K){$K$}
\draw (J) -- (K);
%
\tkzDefPointBy[rotation= center C angle -\a](A)
\tkzGetPoint{M}
\tkzDefPointBy[rotation= center C angle -\a](I)
\tkzGetPoint{N}
\filldraw[draw=orange,fill=orange!50,opacity=.5] (A) -- (C) -- (I) -- cycle;
\filldraw[draw=blue,fill=blue!50,opacity=.5] (M) -- (C) -- (N) -- cycle;
\node[below right,text width=5cm] at (-3.5,7.5) {Les triangles $ACI$ et $FCB$ ont la même aire.};
\end{tikzpicture}
}

\foreach\i in {1,2,...,10}
{
\begin{tikzpicture}
\fig
\tkzLabelPoint[below](J){$J$}
\draw[dashed] (A) -- (J);
\tkzLabelPoint[above](K){$K$}
\draw (J) -- (K);
\filldraw[draw=orange,fill=orange!50,opacity=.5] (J) -- (C) -- (I) -- cycle;
\filldraw[draw=blue,fill=blue!50,opacity=.5] (F) -- (C) -- (B) -- cycle;
\node[below right,text width=5cm] at (-3.5,7.5) {Les triangles $JCI$ et $FCB$ ont donc la même aire.};
\end{tikzpicture}
}

\foreach\c in {0,0.1,0.2,...,1}
{
\begin{tikzpicture}
\fig
\tkzLabelPoint[below](J){$J$}
\draw[dashed] (A) -- (J);
\tkzLabelPoint[above](K){$K$}
\draw (J) -- (K);
\filldraw[draw=orange,fill=orange!50,opacity=.5] (J) -- (C) -- (I) -- cycle;
\filldraw[draw=blue,fill=blue!50,opacity=.5] (F) -- (C) -- (${1-\c}*(B)+\c*(A)$) -- cycle;
\node[below right,text width=5cm] at (-3.5,7.5) {Les triangles $FCB$ et $ACB$ ont donc la même aire car ils ont la même base $[FC]$ et la même hauteur $[AC]$.};
\end{tikzpicture}
}

\foreach\i in {1,2,...,10}
{
\begin{tikzpicture}
\fig
\tkzLabelPoint[below](J){$J$}
\draw[dashed] (A) -- (J);
\tkzLabelPoint[above](K){$K$}
\draw (J) -- (K);
\filldraw[draw=orange,fill=orange!50,opacity=.5] (J) -- (C) -- (I) -- cycle;
\filldraw[draw=blue,fill=blue!50,opacity=.5] (F) -- (C) -- (A) -- cycle;
\draw[very thick,blue] (A)--(G)--(F)--(C)--cycle;
\draw[very thick,orange] (C)--(I)--(K)--(J)--cycle;
\node[below right,text width=5cm] at (-3.5,7.5) {$AGFC$ et $CIKJ$ ont donc la même aire.};
\end{tikzpicture}

\begin{tikzpicture}
\fig
\tkzLabelPoint[below](J){$J$}
\draw[dashed] (A) -- (J);
\tkzLabelPoint[above](K){$K$}
\draw (J) -- (K);
\filldraw[draw=orange,fill=orange!50,opacity=.5] (J) -- (C) -- (I) -- cycle;
\filldraw[draw=blue,fill=blue!50,opacity=.5] (F) -- (C) -- (A) -- cycle;
\node[below right,text width=5cm] at (-3.5,7.5) {$AGFC$ et $CIKJ$ ont donc la même aire.};
\end{tikzpicture}
}

\foreach\i in {1,2,...,10}
{
\begin{tikzpicture}
\fig
\tkzLabelPoint[below](J){$J$}
\draw[dashed] (A) -- (J);
\tkzLabelPoint[above](K){$K$}
\draw (J) -- (K);
\filldraw[draw=orange,fill=orange!50,opacity=.5] (J) -- (C) -- (I) -- cycle;
\filldraw[draw=blue,fill=blue!50,opacity=.5] (F) -- (C) -- (A) -- cycle;
\filldraw[very thick,draw=purple,fill=purple!20] (A)--(B)--(E)--(D)--cycle;
\filldraw[very thick,draw=green!50!black,fill=green!20] (B)--(J)--(K)--(H)--cycle;
\node[below right,text width=5cm] at (-3.5,7.5) {On démontre de même que $ABED$ et $BJKH$ ont la même aire.};
\end{tikzpicture}

\begin{tikzpicture}
\fig
\tkzLabelPoint[below](J){$J$}
\draw[dashed] (A) -- (J);
\tkzLabelPoint[above](K){$K$}
\draw (J) -- (K);
\filldraw[draw=orange,fill=orange!50,opacity=.5] (J) -- (C) -- (I) -- cycle;
\filldraw[draw=blue,fill=blue!50,opacity=.5] (F) -- (C) -- (A) -- cycle;
\node[below right,text width=5cm] at (-3.5,7.5) {On démontre de même que $ABED$ et $BJKH$ ont la même aire.};
\end{tikzpicture}
}

\foreach \in in {1,2,...,10}
{
\begin{tikzpicture}
\fig
\tkzLabelPoint[below](J){$J$}
\draw[dashed] (A) -- (J);
\tkzLabelPoint[above](K){$K$}
\draw (J) -- (K);
\filldraw[draw=blue,fill=blue!50,opacity=.5] (F) -- (C) -- (A) -- (G) -- cycle;
\filldraw[draw=blue,fill=blue!50,opacity=.5] (C) -- (I) -- (K) -- (J) -- cycle;
\filldraw[draw=purple,fill=purple!20] (A)--(B)--(E)--(D)--cycle;
\filldraw[draw=purple,fill=purple!20] (B)--(J)--(K)--(H)--cycle;
\node[below right,text width=5cm] at (-3.5,7.5) {Donc:\linebreak $\text{Aire}(FCAG)+\text{Aire}(ABED)=\text{Aire}(BHIC)$, soit:\linebreak $AC^2+AB^2=BC^2$.};
\end{tikzpicture}
}
\end{document}