
\begin{enumerate}
\item $\begin{aligned}[t]
\prodscal{AE}{BD} & = \big(\vv{\text{AD}}+\vv{\text{DE}}\big)\cdot\big(\vv{\text{BC}}+\vv{\text{CD}}\big)\\
& = \prodscal{AD}{BC}+\prodscal{AD}{CD}+\prodscal{DE}{BC}+\prodscal{DE}{CD}\\
& = \prodscal{AD}{AD}+0+0+\prodscal{DE}{CD}\\
& = \|\vv{\text{AD}}\|^2+\|\vv{\text{DE}}\|\times\|\vv{\text{CD}}\|\times\cos\pi\\
& = 4-1\times4\\
\Aboxed{\prodscal{AE}{BD}& = 0}
\end{aligned}$

\uline{On en déduit alors que (AE) et (BD) sont perpendiculaires.}

\item $\begin{aligned}[t]
\prodscal{BA}{BD} & = \text{BH}\times\text{BD}\\
& = \text{BH}\times\sqrt{4^2+2^2}\qquad\text{(Théorème de Pythagore dans BAD rectangle en A)}\\
& = \text{BH}\times2\sqrt{5}
\end{aligned}$

De plus, on a aussi :

\medskip

$\begin{aligned}[t]
\prodscal{BA}{BD} & = \text{BA}\times\text{BD}\times\cos\angle{ABD}\\
& = \text{BA}\times\text{BD}\times\dfrac{\text{AB}}{\text{BD}}\qquad\text{(en se plaçant dans le triangle rectangle ABD)}\\
& = \text{AB}^2\\
& = 16.
\end{aligned}$

Finalement, on a donc :
\[
2\sqrt{5}\text{BH}=16\qquad\text{soit}\qquad\boxed{\text{BH}=\dfrac{8\sqrt{5}}{5}}
\]

\item $\begin{aligned}[t]
\vv{\text{HA}}+\vv{\text{HB}} & =\big(\vv{\text{HN}}+\vv{\text{NA}}\big)+\big(\vv{\text{HN}}+\vv{\text{NB}}\big)\\
& = 2\vv{\text{HN}}+\big(\vv{\text{NA}}+\vv{\text{NB}}\big)\\
& = 2\vv{\text{HN}}+\vv{\text{0}}\qquad\text{(car N est le milieu de [AB])}\\
\Aboxed{\vv{\text{HA}}+\vv{\text{HB}} & = 2\vv{\text{HN}}}
\end{aligned}$

\item\label{q4ex13prodscal} $\begin{aligned}[t]
\text{HA}^2 & = \text{AB}^2-\text{HB}^2\qquad\text{(théorème de Pythagore dans le triangle AHB rectangle en H)}\\
\text{HA}^2 & = 16-\dfrac{64}{5}\\
\text{HA}^2 & = \dfrac{16}{5}\\
\text{HA} & = \dfrac{4}{\sqrt{5}}\\
\Aboxed{\text{HA} & = \dfrac{4\sqrt{5}}{5}}
\end{aligned}$

Ainsi,

\medskip

$\begin{aligned}[t]
\prodscal{HN}{HA} & = \big(\vv{\text{HA}}+\vv{\text{AN}}\big)\cdot\vv{\text{HA}}\\
& = \vv{\text{HA}}^2+\prodscal{AN}{HA}\\
& = \dfrac{16}{5}-\prodscal{AN}{AH}
\end{aligned}$

$\begin{aligned}[t]
& = \dfrac{16}{5}-2\times\dfrac{4\sqrt{5}}{5}\times\cos\angle{HAN}\\
& = \dfrac{16}{5}-\dfrac{8\sqrt{5}}{5}\times\dfrac{\text{AH}}{\text{AB}}\qquad\text{(dans le triangle HAB rectangle en H)}\\
& = \dfrac{16}{5}-\dfrac{8\sqrt{5}}{5}\times\dfrac{4\sqrt{5}}{5}\times\dfrac{1}{4}\\
& = \dfrac{16}{5}-\dfrac{8}{5}\\
\Aboxed{\prodscal{HN}{HA} & = \dfrac{8}{5}}
\end{aligned}$

\item Dans le triangle BAH rectangle en H, [HN] est la médiane issue du sommet de l'angle droit, donc :
\[
\text{HN}=\dfrac{1}{2}\text{AB}\qquad\text{soit}\qquad\boxed{\text{HN}=2}
\]

\item $\begin{aligned}[t]
\prodscal{HN}{HA} & = \text{HN}\times\text{HA}\times\cos\angle{AHN}\\
\dfrac{8}{5} & = \dfrac{8\sqrt{5}}{5}\cos\angle{AHN}\qquad\text{(d'après la question \ref{q4ex13prodscal})}\\
\cos\angle{AHN} & = \dfrac{\sqrt{5}}{5}\\
\Aboxed{\angle{AHN} & \approx 63\degres}
\end{aligned}$
\end{enumerate}
