9. Given a square piece of paper $A B C D$ with an area of $2007 \mathrm{~cm}^{2}$. Now fold the paper along a line segment (as shown in Figure 1), so that point $D$ lands on point $D^{\prime}$ on side $B C$, and point $A$ lands on point $A^{\prime}$, with $A^{\prime} D^{\prime}$ intersecting $A B$ at point $E$. Then the perimeter of $\triangle B D^{\prime} E$ is $\mathrm{cm}$.
$9.6 \sqrt{223}$.
Let the side length of the square be $a=\sqrt{2007}, \angle D^{\prime} D C=\alpha$.
Then $\angle B D^{\prime} E=2 \alpha, C D^{\prime}=a \tan \alpha$, $B D^{\prime}=a(1-\tan \alpha)$.
Therefore, the perimeter of $\triangle B D^{\prime} E$ is
$$
\begin{array}{l}
a(1-\tan \alpha)(1+\tan 2 \alpha+\sec 2 \alpha) \\
=a \cdot \frac{\cos \alpha-\sin \alpha}{\cos \alpha} \cdot \frac{\cos 2 \alpha+\sin 2 \alpha+1}{\cos 2 \alpha} \\
=a \cdot \frac{\cos \alpha-\sin \alpha}{\cos \alpha} \cdot \frac{2 \cos ^{2} \alpha+2 \sin \alpha \cdot \cos \alpha}{\cos ^{2} \alpha-\sin ^{2} \alpha} \\
=2 a=6 \sqrt{223} .
\end{array}
$$