Four. (20 points) Given in the rectangular prism $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$, the edge lengths are $A B=B C=3, B B_{1}=4$. Connect $B_{1} C$, and draw a perpendicular from point $B$ to $B_{1} C$ intersecting $C C_{1}$ at point $E$ and $B_{1} C$ at point $F$.
(1) Prove: $A_{1} C \perp$ plane $E B D$;
(2) Let $A_{1} C \cap$ plane $E B D=K$, find the length of segment $A_{1} K$.
(1) Taking $D$ as the origin, the lines $D A$, $D C$, and $D D_{1}$ as the $x$-axis, $y$-axis, and $z$-axis respectively, we establish a spatial rectangular coordinate system, then
$$
\begin{array}{l}
D(0,0,0), A(3,0,0), C(0,3,0), B(3,3,0), \\
A_{1}(3,0,4), D_{1}(0,0,4), C_{1}(0,3,4), B_{1}(3,3,4).
\end{array}
$$

Let $E(0,3, z)$. Then $B E \cdot B_{1} C=0$,
$$
B E=(-3,0, z), B_{1} C=(-3,0,-4).
$$

Thus, $B E \cdot B_{1} C=(-3,0, z) \cdot(-3,0,-4)=9-4 z=0$.
Therefore, $z=\frac{9}{4}$. Hence, $E\left(0,3, \frac{9}{4}\right)$.
Then, $\boldsymbol{A}, \boldsymbol{C} \cdot \boldsymbol{D B}=-3 \times 3+3 \times 3=0$,
$$
A_{1} C \cdot D E=3 \times 3-4 \times \frac{4}{9}=0.
$$

Thus, $A_{1} C \perp D B, A_{1} C \perp D E$.
Therefore, $A_{1} C \perp$ plane $E B D$.
(2) By $D K=m D B+n D E$
$$
=m(3,3,0)+n\left(0,3, \frac{9}{4}\right)=\left(3 m, 3 m+3 n, \frac{9}{4} n\right),
$$

we know $K\left(3 m, 3 m+3 n, \frac{9}{4} n\right)$.
Thus, $A_{1} K=\left(3 m-3,3 m+3 n, \frac{9}{4} n-4\right)$.
Since $A_{1} K \perp D B$, we have
$A_{1} \boldsymbol{K} \cdot D B=\left(3 m-3,3 m+3 n, \frac{9}{4} n-4\right) \cdot(3,3,0)$ $=0$.
Then $2 m+n-1=0$.
Similarly, since $A_{1} K \perp D E$, we have
$16 m+25 n-16=0$.
Solving equations (1) and (2), we get $m=\frac{9}{34}, n=\frac{8}{17}$.
Thus, $A_{1} K=\left(-\frac{75}{34}, \frac{75}{34},-\frac{50}{17}\right)$.
Therefore, $\left|A_{1} K\right|=\frac{25 \sqrt{34}}{34}$ is the length of $A_{1} K$ we are looking for.