6. (10 points) As shown in the figure, the side length of square $ABCD$ is $5$, and $E, F$ are two points outside the square, satisfying $AE=CF=4$, $BE$ $=DF=3$. Then $EF^{2}=$ $\qquad$

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
【Analysis】By extending $E A$, $F D$, $F C$, and $E B$, the shape can be expanded into a larger square. Using the Pythagorean theorem, it is not difficult to find $E F^{2}$.

【Solution】Solution: According to the analysis, as shown in the figure: Extend $E A$, $F D$, $F C$, and $E B$ to expand the shape into a larger square,
$$
\begin{array}{l}
\because A E=C F=4, \quad B E=D F=3, \\
\therefore C M=O A=D F=E B=3, \quad B M=O D=C F=A E=4 \\
\text { Also, } \because D F^{2}+C F^{2}=C D^{2}, A E^{2}+E B^{2}=A B^{2}, \\
O A^{2}+O D^{2}=A D^{2}, C M^{2}+B M^{2}=B C^{2} \\
\therefore \angle A E B=\angle D F C=\angle A O D=\angle B M C=90^{\circ}, \\
\therefore E O=F O=3+4=7 \\
\therefore E F^{2}=O E^{2}+O F^{2}=7^{2}+7^{2}=98
\end{array}
$$

Therefore, the answer is: 98