2. In the rectangular paper piece $A B C D$, $A B=6, B C=8$. If the paper is folded so that $A$ coincides with $C$, then the length of the fold line $E F$ is
2. According to the problem, the fold line $E F$ is the perpendicular bisector of $A C$. Let $A F=x$, then $F C=x$. Since $A B$ $=6, B C=8$, so $A C=10, A O=O C=5$. In

Rt $\triangle C D F$, $6^{2}+(8-x)^{2}=x^{2}$, we get $x=\frac{25}{4}$. In Rt $\triangle A O F$, $\left(\frac{25}{4}\right)^{2}-5^{2}=O F^{2}, O F^{2}=\frac{25 \times 9}{16}, O F=$ $\frac{15}{4}$, thus $E F=\frac{15}{2}$.