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\emph{par\\{\small Stéphane PASQUET}}
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\begin{document}
\pagestyle{fancy}
\chapHead{La longueur d'une\\bissectrice dans un triangle}

\section*{\'Enoncé du résultat}

Considérons un triangle quelconque ABC ; posons alors $\text{AB}=c$, $\text{AC}=b$ et $\text{BC}=a$.

Posons $\ell$ la longueur de la bissectrice issue de C ; alors, on a:

\begin{minipage}{0.5\linewidth}
\[ \ell=\sqrt{ab\left(1-\frac{c^2}{(a+b)^2}\right)}\]
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\section*{Démonstration (niveau : 1\iere{} S)}

\begin{itemize}
\item[$\bullet$] L'aire du triangle ABC est donnée par la formule :
\[ \mathscr{A}_{\text{ABC}}=\frac{1}{2}ab\sin(2\alpha). \]
\item[$\bullet$] L'aire du triangle ACI est donnée par la formule :
\[ \mathscr{A}_{\text{ACI}}=\frac{1}{2}b\ell\sin(\alpha). \]
\item[$\bullet$] L'aire du triangle BCI est donnée par la formule :
\[ \mathscr{A}_{\text{BCI}}=\frac{1}{2}a\ell\sin(\alpha). \]
\end{itemize}

On en déduit alors :
\begin{align*}
& \mathscr{A}_{\text{ABC}} =  \mathscr{A}_{\text{ACI}}+\mathscr{A}_{\text{BCI}}\\
\Leftrightarrow\ &
\dfrac{1}{2}ab\sin(2\alpha) = \dfrac{1}{2}b\ell\sin(\alpha) + \dfrac{1}{2}a\ell\sin(\alpha)\\
\Leftrightarrow\ & ab\sin(2\alpha) =\ell\sin(\alpha)(b + a)\\
\Leftrightarrow\ & \ell = \dfrac{ab}{a+b}\times\dfrac{\sin(2\alpha)}{\sin\alpha}.
\end{align*}

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Or,
\[ \sin(2\alpha)=2\sin\alpha\cos\alpha,\]
d'où :
\[\ell = \dfrac{ab}{a+b}\times\frac{2\sin\alpha\cos\alpha}{\sin\alpha},\]
et donc :
\[\ell = \dfrac{ab}{a+b}\times 2\cos\alpha.\]
On en déduit alors que :
\[ \ell^2 = \frac{(ab)^2}{(a+b)^2}\times 4\cos^2\alpha.\]
Or, 
\[ \cos^2\alpha = \frac{1+\cos(2\alpha)}{2},\]
d'où :
\begin{align*}
\ell^2 & = \frac{(ab)^2}{(a+b)^2}\times 4\times\frac{1+\cos(2\alpha)}{2}\\
& = \frac{(ab)^2}{(a+b)^2}\times 2(1+\cos(2\alpha)).
\end{align*}

La relation d'Al-Kashi dans le triangle ABC donne :
\[ c^2=a^2+b^2-2ab\cos(2\alpha).\]
Donc :
\[ \cos(2\alpha)=\frac{a^2+b^2-c^2}{2ab}.\]
On a alors :
\begin{align*}
\ell^2 & = \frac{(ab)^2}{(a+b)^2}\times 2\left(1+\frac{a^2+b^2-c^2}{2ab}\right)\\
& =  \frac{(ab)^2}{(a+b)^2}\times 2\left(\frac{2ab+a^2+b^2-c^2}{2ab}\right)\\
& = \dfrac{ab}{(a+b)^2}(a^2+2ab+b^2-c^2)\\
& = \dfrac{ab}{(a+b)^2}\left[(a+b)^2-c^2\right]\\
& = ab\left(1-\dfrac{c^2}{(a+b)^2}\right).
\end{align*}
On en déduit alors (car $\ell\geqslant0$):
\[ \ell=\sqrt{ab\left(1-\frac{c^2}{(a+b)^2}\right)}.\]
\end{document}