54. As shown in the figure, in rectangle $A B C D$, $A B=10$. The paper is folded so that side $A D$ falls on side $A B$, with the fold line being $A E$. Then, $\triangle A E D$ is folded along $D E$, and the intersection of $A E$ and $B C$ is point $F$. Given that the area of $\triangle A B F$ is 2, then $B C=$ $\qquad$ -  

Answer: 6

Solution: It is easy to know that $\triangle A B F$ and $\triangle E C F$ are both isosceles right triangles, so $B F=2$. Let $B C=x$, then
$$
F C=x-2, \quad E C=10-x,
$$

By $F C=E C$, we get
$$
x=6 \text { . }
$$