2. In an acute triangle $ABC$, $AD$, $DE$, $CF$ are altitudes, $H$ is the orthocenter of the triangle. If $EF$ bisects the area of triangle $ABC$, then $DE^2 + EF^2 + FD^2 =$ $\qquad$ (Given the side lengths of the triangle are $3$, $2\sqrt{2}$, $\sqrt{5}$)
2. 5 .
$$
\begin{array}{l}
\text { Sol: } \triangle B E F \sim \triangle A B C \Rightarrow \angle A=45^{\circ}, \\
\text { and } \angle E D F=90^{\circ} \text {; } \\
\therefore a=\sqrt{5}, D E^{2}+E F^{2}+F D^{2}=2 E F^{2}=a^{2}=5 .
\end{array}
$$