
\begin{enumerate}
\item $\begin{aligned}[t]
\prodscal{AC}{DB} & = \big(\vv{AD}+\vv{DC}\big)\cdot\vv{DB}\\
& = \prodscal{AD}{DB} + \prodscal{DC}{DB}\\
& = -\prodscal{DA}{DB}  + \prodscal{DC}{DB}\\
& = -DH\times DB + DK \times DB\\
& = DB(DK-DH)\\
& = DB \times HK\\
& = \sqrt{7^2+3^2}\times HK\\
\Aboxed{\prodscal{AC}{DB} & = HK\sqrt{58}}
\end{aligned}$

\item $\begin{aligned}[t]
\prodscal{AC}{DB} & = \vv{AC}\cdot\big(\vv{DC} + \vv{CB}\big)\\
& = \prodscal{AC}{DC} + \prodscal{AC}{CB}\\
& = \prodscal{AC}{AB} + \prodscal{AC}{DA}\\
& = AB\times AB - \prodscal{AC}{AD}\\
& = AB^2 - AD^2\\
& = 7^2-3^3\\
\Aboxed{\prodscal{AC}{DB} & = 40}
\end{aligned}$

\item Des deux questions précédentes, on déduit que:
\[
\prodscal{AC}{DB} = 40 = HK\sqrt{58} \Rightarrow \boxed{HK = \dfrac{40}{\sqrt{58}}}
\]
On peut aussi mettre le résultat sous la forme:
\[
HK = \dfrac{40\sqrt{58}}{58}=\dfrac{20\sqrt{58}}{29}.
\]
\end{enumerate}
