1. In $\triangle A B C$, $A B=3, A C=4, B C=5$. Now fold $\triangle A B C$ so that point $B$ coincides with point $C$, then the length of the fold line is $\qquad$ .
II. $1 . \frac{15}{8}$.
As shown in Figure 5, it is easy to know that $\angle A=90^{\circ}$. Let $DE$ be the fold line, then $DE \perp BC$. Therefore, $\triangle CDE \backsim \triangle CAB$.

We have $\frac{CD}{AC}=\frac{DE}{AB}$,
which is $\frac{\frac{5}{2}}{4}=\frac{DE}{3}$.
Solving for $DE$ gives $DE=\frac{3 \times \frac{5}{2}}{4}=\frac{15}{8}$.