On considère un demi-cercle de centre O et de diamètre $[AB]$ tel que $AB=5$~cm.

Soit $C$ un point de ce demi-cercle. Le triangle $ABC$ est ainsi rectangle en $C$.

\medskip

Soit le carré $EFGH$ tel que:
\begin{itemize}
\item $E\in[BC]$ et $F\in[AC]$;
\item $G\in[AB]$ et $H\in[AB]$;
\item $(CG)$ coupe $[EF]$ en son milieu $I$.
\end{itemize}

On note $x$ la mesure des côtés de ce carré.

\bigskip


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\begin{center}
\begin{tikzpicture}
\draw[color=qqwuqq,fill=qqwuqq,fill opacity=0.10000000149011612] (2.6205266807797947,3.8102633403898967) -- (2.8102633403898976,3.430790021169691) -- (3.1897366596101033,3.620526680779794) -- (3,4) -- cycle; 
\fill[color=qqwwzz,fill=qqwwzz,fill opacity=0.1] (0.688,2.844) -- (3.578,2.844) -- (3.578,0) -- (0.688,0) -- cycle;
\draw (5,0) arc(0:180:5cm);
\draw (-5,0)-- (5,0);
\draw (-5,0)-- (3,4);
\draw (3,4)-- (5,0);
\draw (0.688,0)-- (3,4);
\draw [color=qqwwzz] (3.578,2.844)-- (3.578,0);
\draw [color=qqwwzz] (3.578,0)-- (0.688,0);
\draw [color=qqwwzz] (0.688,0)-- (0.688,2.844);
\draw [color=qqwwzz] (0.688,2.844)-- (2.3318320000000003,2.844);
\draw [color=qqwwzz] (1.4599160000000022,2.964) -- (1.4599160000000022,2.724);
\draw [color=qqwwzz] (1.5599160000000023,2.964) -- (1.5599160000000023,2.724);
\draw [color=qqwwzz] (2.3318320000000003,2.844)-- (3.578,2.844);
\draw [color=qqwwzz] (2.9049160000000027,2.964) -- (2.9049160000000027,2.724);
\draw [color=qqwwzz] (3.004916000000003,2.964) -- (3.004916000000003,2.724);
\draw  (0,0)-- ++(-2pt,0 pt) -- ++(4pt,0 pt) ++(-2pt,-2pt) -- ++(0 pt,4pt);
\draw (-0.01,-0.41) node {$O$};
\draw (-5.39,-0.11) node {$A$};
\draw (5.23,-0.05) node {$B$};
\draw (3.09,4.43) node {$C$};
\draw (3.77,3) node {$E$};
\draw (0.55,3) node {$F$};
\draw (2.2,3.1) node {$I$};
\draw (0.55,-0.25) node {$G$};
\draw (3.5,-0.25) node {$H$};
\end{tikzpicture}
\end{center}

\begin{enumerate}
\item Montrer que les triangles $AGC$ et $FIC$ sont semblables.

Exprimer alors $IC$, puis $IG$, en fonction de $x$.

\item En déduire que $x$ est solution de l'équation $x^2+5x-25=0$.

\item En déduire la valeur de $x$.
\end{enumerate}
\vspace*{-1.2em}

