9. (12 points) As shown in the figure, in the right triangle $ABC$, the lengths of $AB$ and $BC$ are $15$ and $20$, respectively. Quadrilateral $BDEF$ is a square. If the height $EH$ of triangle $EMN$ is 2, then the area of square $BDEF$ is $\qquad$ .
【Solution】In the right triangle $ABC$, since the lengths of $AB$ and $BC$ are $15$ and $20$ respectively, we have $AC=25$. In $\triangle ABC$ and $\triangle EHM$,
$$
\begin{array}{l}
\because \frac{EH}{AB}=\frac{HM}{BC}=\frac{EM}{AC}, \\
\therefore \frac{2}{15}=\frac{HM}{20}=\frac{EM}{25}, \\
\therefore HM=\frac{8}{3}, EM=\frac{10}{3},
\end{array}
$$

Let the side length of the square $BDEF$ be $x$,
In $\triangle ADM$ and $\triangle EHM$,
$$
\begin{array}{l}
\because \frac{AD}{EH}=\frac{DM}{HM}, \\
\therefore \frac{15-x}{2}=\frac{x-\frac{10}{3}}{\frac{8}{3}},
\end{array}
$$

Solving for $x$ gives $x=10$,
$\therefore$ The area of the square $BDEF$ is 100,
Thus, the answer is 100.