Cho hình hộp chữ nhật ABCD.A'B'C'D' có AB = 3, BC = 2, AD' = căn bậc hai của 5

Gọi $O=AD\text{'}\cap A\text{'}D\Rightarrow I=A\text{'}D\cap \left(AD\text{'}I\right).$

Do đó $\frac{d\left(A\text{'}\left(AD\text{'}I\right)\right)}{d\left(D;\left(AD\text{'}I\right)\right)}=\frac{OA\text{'}}{OD}=1\Rightarrow d\left(A\text{'};\left(AD\text{'}I\right)\right)=d\left(D;\left(AD\text{'}I\right)\right)$.

Trong (ABCD) dựng $DM\perp AI,$ trong (DD'M) dựng $DH\perp D\text{'}M\left(H\in D\text{'}M\right)$ ta có:

$\left\{\begin{array}{l}AI\perp DM\\ AI\perp DD\text{'}\end{array}\right.\Rightarrow AI\perp \left(DD\text{'}M\right)\Rightarrow AI\perp DH$

$\left\{\begin{array}{l}DH\perp D\text{'}M\\ DH\perp AI\end{array}\right.\Rightarrow DH\perp \left(AD\text{'}I\right)\Rightarrow d\left(D;\left(AD\text{'}I\right)\right)=DH$

Ta có

$S_{ADI}=S_{ABCD}-S_{ABI}-S_{CDI}$

     $=AB.BC-\frac{1}{2}AB.\frac{1}{2}BC-\frac{1}{2}CD.\frac{1}{2}BC$

     $=\frac{1}{2}AB.BC=\frac{1}{2}\mathrm{.3.2}=3$

Lại có $S_{ADI}=\frac{1}{2}DM.AI\Rightarrow DM=\frac{2S_{ADI}}{AI}=\frac{2.3}{\sqrt{A{B}^2+B{I}^2}}=\frac{6}{\sqrt{3^2+1^2}}=\frac{6}{\sqrt{10}}$

Áp dụng định lí Pytago: $DD\text{'}=\sqrt{AD{\text{'}}^2-A{D}^2}=\sqrt{5-4}=1.$

Áp dụng hệ thức lượng trong tam giác vuông DD'M có: $DH=\frac{DD\text{'}.DM}{\sqrt{DD{\text{'}}^2+D{M}^2}}=\frac{1.\frac{6}{\sqrt{10}}}{\sqrt{1+\frac{18}{5}}}=\frac{3\sqrt{46}}{23}.$

Vậy $d\left(A\text{'};\left(AD\text{'}I\right)\right)=\frac{3\sqrt{46}}{23}.$