3. As shown in Figure 7, points $A$ and $C$ are on ray $O M$, and points $B$ and $D$ are on ray $O N$. Given that $O A=6, O D=16, \angle N O M$ $=20^{\circ}$. What is the shortest length of $A B+B C+C D$?
3. 14 .

As shown in Figure 11, reflect $O N$ about $O M$ to get ray $O P$ and its point $B$'s image $E$; then reflect $O M$ about $O P$ to get ray $O Q$ and its point $A$'s image $F$.
Thus, $O F=O A=6, \angle F O D=3 \times 20^{\circ}=60^{\circ}$.
According to the principle of reflection,
$F E=E A=A B, E C=B C$.
Therefore, the length of the broken line $A B C D$ is
$$
F E+E C+C D \geqslant F D \text {. }
$$

In $\triangle F O D$, by the cosine rule we have
$$
F D^{2}=O F^{2}+O D^{2}-2 O F \cdot O D \sin 60^{\circ}=196 \text {. }
$$

Thus, the shortest distance of the broken line is $\sqrt{196}=14$. The method to construct the shortest broken line is to connect $F D$, intersecting ray $O P$ and $O M$ at points $E$ and $C$ respectively, and then reflect point $E$ to get the position of point $B$.