18. Given Rt $\triangle A B C$ and Rt $\triangle A D C$ share the hypotenuse $A C, M$ and $N$ are the midpoints of $A C$ and $B D$ respectively, and $M$ and $N$ do not coincide.
(1) Is line segment $M N$ perpendicular to $B D$? Please explain your reasoning.
(2) If $\angle B A C=30^{\circ}, \angle C A D=45^{\circ}, A C=4$, find the length of $M N$.
18. (1) As shown in Figures 7 and 8, connect $M B$ and $M D$.

Since $M$ is the midpoint of $A C$, $\angle A B C = \angle A D C = 90^{\circ}$, then $M B = M D = \frac{1}{2} A C$.
Also, $N$ is the midpoint of $B D$, by the property of isosceles triangles, $M N \perp B D$.
(2) Draw $C H \perp B D$, with the foot of the perpendicular at $H$.
Since $\angle C A D = 45^{\circ}$, hence $\angle C D M = 45^{\circ}$.
Thus, $D M \perp A C$.
Case 1: $B$ and $D$ are on opposite sides of $A C$, as shown in Figure 7.
Since $\angle M B C$
$$
= \angle B M C = 60^{\circ},
$$

then $\angle B M D$
$$
\begin{array}{l}
= 360^{\circ} - (\angle A M B + \\
\angle A M D) \\
= 360^{\circ} - (120^{\circ} + 90^{\circ}) \\
= 150^{\circ}, \\
\angle M B D = \angle M D B = \frac{1}{2} (180^{\circ} - \angle B M D) \\
= \frac{1}{2} (180^{\circ} - 150^{\circ}) = 15^{\circ}. \\
\text{Also, } \angle B D C = \angle M D C - \angle M D B = 45^{\circ} - 15^{\circ} = 30^{\circ}, \\
\angle C B D = \angle C B M - \angle M B D = 60^{\circ} - 15^{\circ} = 45^{\circ}, \\
M A = M B = M C = M D = \frac{1}{2} A C = 2,
\end{array}
$$

then $C D = \sqrt{M C^{2} + M D^{2}} = 2 \sqrt{2}, C H = \frac{1}{2} C D = \sqrt{2}$,
$$
\begin{array}{l}
D H = \sqrt{C D^{2} - C H^{2}} = \sqrt{6}, B H = C H = \sqrt{2}, \\
B N = \frac{1}{2} (B H + D H) = \frac{1}{2} (\sqrt{6} + \sqrt{2}).
\end{array}
$$

Thus, $M N^{2} = M B^{2} - N B^{2} = 2 - \sqrt{3}$, so $M N = \sqrt{2 - \sqrt{3}}$.
Case 2: $B$ and $D$ are on the same side of $A C$, as shown in Figure 8.
Since $\angle B M C = 60^{\circ}$,
then $\angle B M N = \angle D M N$
$$
\begin{array}{l}
= \frac{1}{2} (90^{\circ} - 60^{\circ}) \\
= 15^{\circ}.
\end{array}
$$

Thus, $\angle M B N = 75^{\circ}$.
And $\angle C B M = 60^{\circ}$, so
$$
\begin{array}{l}
\angle C B H = 180^{\circ} - (75^{\circ} + 60^{\circ}) = 45^{\circ}, \\
\angle C D H = \angle M D H - \angle M D C = 75^{\circ} - 45^{\circ} = 30^{\circ}.
\end{array}
$$

Also, $B C = \frac{1}{2} A C = 2$, then
$$
\begin{array}{l}
B H = C H = \frac{\sqrt{2}}{2} B C = \sqrt{2}, \\
C D = 2 C H = 2 \sqrt{2}, D H = \sqrt{C D^{2} - C H^{2}} = \sqrt{6}, \\
B N = \frac{1}{2} (D H - B H) = \frac{1}{2} (\sqrt{6} - \sqrt{2}), \\
M N^{2} = B M^{2} - B N^{2} = 2 + \sqrt{3}.
\end{array}
$$

Thus, $M N = \sqrt{2 + \sqrt{3}}$.
Note: For the proof in part (1), if only Figure 7 or only Figure 8 is drawn, 2 points will be deducted, giving a total of 5 points.