
Posons $x=AD$.

Dans le repère $\big(\text{A}~;~\vv{\text{AB}}~;~\frac{1}{x}\vv{\text{AD}}\big)$, on a :
\[
\vv{\text{AC}}\dbinom{1}{x}\qquad\text{et}\qquad\vv{\text{BR}}\dbinom{-1}{\frac{x}{2}}.
\]
Ainsi, $\begin{aligned}[t]
\prodscal{AC}{BR}=0 & \Longleftrightarrow -1+\dfrac{x^2}{2}=0\\[6pt]
& \Longleftrightarrow \dfrac{x^2}{2}=1\\[6pt]
& \Longleftrightarrow x^2=2\\[6pt]
& \Longleftrightarrow x=\sqrt{2}\qquad\text{car }x>0.
\end{aligned}$

\underline{Ainsi, $\text{AD}=\sqrt{2}$.}
