10. (6 points) As shown in the figure, in square $A B C D$, point $E$ is on side $A D$, with $A E=3 E D$, and point $F$ is on side $D C$. When $S_{\triangle B E F}$ is minimized, the value of $S_{\triangle B E F}: S_{\text {square } A B C D}$ is $\qquad$ .
【Solution】Solution: According to the analysis, point $F$ moves along side $DC$. When point $F$ moves to point $D$, the area of triangle $BEF$ is minimized, hence

As shown in the figure:
$\because AE=3ED$
$\therefore S_{\triangle BEF}=S_{\triangle BDE}=\frac{1}{3} \times S_{\triangle ABE}=\frac{1}{1+3} S_{\triangle ABD}=\frac{1}{4} \times \frac{1}{2} S_{\text {square } ABCD}=\frac{1}{8} S_{\text {square } ABCD}$
$\therefore S_{\triangle BEF}: S_{\text {square } ABCD}=1: 8$
Therefore, the answer is: 1 : 8