18. Let the plane $K L M N$ be $\alpha$, and the line $A C$ be $l$.
(1) If $l / / a$ (as shown in Figure 6).
From $\frac{A N}{A D}=\frac{B L}{B C}$, we have $\frac{D N}{N A}=\frac{C L}{L B}$.
Since $l / / \alpha$, then $M N / / l, K L / / l$.
Thus, $\frac{D M}{M C}=\frac{D N}{N A}, \frac{A K}{K B}=\frac{C L}{L B}$, which means $\frac{D M}{M C}=\frac{A K}{K B}$.
(2) If $l$ intersects, let $l \cap_{\alpha}=P$ (as shown in Figure 7). Then point $P$ lies on both line $K L$ and line $M N$.

On $l$, take a point $Q$ such that $\frac{A Q}{A C}=\frac{A N}{A D}=\frac{B L}{B C}$, and connect $L Q$ and $N Q$. Then $L Q / / A B, N Q / / C D$.
Clearly, $\frac{D M}{M C}=\frac{A K}{K B} \Leftrightarrow \frac{M C}{D C}=\frac{K B}{A B}$.
And $\frac{M C}{D C}=\frac{M C}{N Q} \cdot \frac{N Q}{D C}=\frac{P C}{P Q} \cdot \frac{A Q}{A C}$, hence
$\frac{K B}{A B}=1-\frac{A K}{A B}=1-\frac{A K}{Q L} \cdot \frac{Q L}{A B}=1-\frac{A P}{P Q} \cdot \frac{Q C}{A C}$
$=\frac{1}{P Q \cdot A C}(P Q \cdot A G-A P \cdot Q C)$
$=\frac{1}{P O \cdot A C}[(P C+C Q) A C-(A C+C P) Q C]$
$=\frac{1}{P Q \cdot A C}(P C \cdot A C-P C \cdot Q C)$
$=\frac{P C \cdot A Q}{P Q \cdot A C}=\frac{M C}{D C}$,

which means $\frac{D M}{M C}=\frac{A K}{K B}$.