10. In the figure, $A B C D$ is a trapezoid, and $A B / / C D$, the areas of triangle $A B O$ and triangle $O C D$ are 16 and 4, respectively. Find $\frac{\mathrm{DC}}{\mathrm{AB}}$.
【Analysis Based on $A B / / C D$, we can get $\frac{S \triangle_{\mathrm{BCD}}}{S \triangle \mathrm{ABO}}=\frac{S \triangle_{O C D}}{S_{\triangle A O D}}=\frac{O C}{A O}$, and also $s_{\triangle A O D}=S_{\triangle B C D}$, so $\frac{S_{\triangle B C O}}{16}$ $=\frac{4}{S \triangle_{\mathrm{BCO}}}$, thus $S_{\triangle B C O}=8$. Let the height of the trapezoid be $h$, and according to the formula for the area of a triangle, derive the result.
【Solution】Solution: From $A B / / C D$, we have
$$
\frac{S \triangle_{B C D}}{S \triangle A B O}=\frac{S \triangle_{O C D}}{S \triangle A O D}=\frac{O C}{A O}
$$

Also, $s_{\triangle A O D}=S_{\triangle B C D}$
So $\frac{S_{\triangle B C 0}}{16}=\frac{4}{S \triangle_{B O 0}}$
Thus $S_{\triangle B C O}=8$.
Let the height of the trapezoid be $h$, since $S_{\triangle A B C}=\frac{\mathrm{h} \times \mathrm{AB}}{2}, S_{\triangle D A C}=\frac{\mathrm{h} \times \mathrm{CD}}{2}$
So $\frac{S \triangle_{D A C}}{S_{\triangle A B C}}=\frac{C D}{A B}$
Also, $S_{\triangle A B C}=S_{\triangle A B O}+S_{\triangle B C D}=24, S_{\triangle D A C}=S_{\triangle D A C}+S_{\triangle A O D}=12$,
So $\frac{D C}{A B}=\frac{1}{2}$.