2. In $\triangle A B C$, $A B=3, A C=4, B C=5$. Now fold it so that point $B$ coincides with point $C$. Then the length of the fold line is $\qquad$
2. $\frac{15}{8}$.

From the given information, $\triangle A B C$ is a right triangle with $\angle A=90^{\circ}$. As shown in Figure 4, the crease line $D E$ is the perpendicular bisector of segment $B C$. It is easy to see that $\mathrm{Rt} \triangle C D E \backsim \mathrm{Rt} \triangle C A B$. Therefore,
$$
\frac{D E}{A B}=\frac{C D}{A C},
$$

which means $D E=\frac{A B \cdot C D}{A C}=\frac{15}{8}$.