2. In trapezoid $A B C D$, $D C / / A B$ and $A B=2 D C$, point $P$ is on $A B$. If the perimeters of $\triangle P B C$, $\triangle P C D$, and $\triangle P D A$ are all equal, then $A P: P B=$ $\qquad$
2.1:1.

As shown in Figure 4, extend $C D$ to $E$ and extend $A B$ to $F$, such that $D E=D P$, $B F=B C$. Connect $E P$ and $C F$. Since $\triangle P B C$ and $\triangle P C D$ have equal perimeters, we have
$$
P B+B C=P D+D C .
$$

Thus, $P F=E C$.
Also, $D C \parallel A B$, which means $E C \parallel P F$, so quadrilateral $P F C E$ is a parallelogram.
Therefore, $\angle E=\angle F$.
It is easy to see that $\angle P D C=2 \angle E=2 \angle F=\angle P B C$. Then
$$
\begin{array}{l}
\angle D P B=180^{\circ}-\angle P D C \\
=180^{\circ}-\angle P B C=\angle D C B .
\end{array}
$$

Thus, quadrilateral $P B C D$ is a parallelogram.
Therefore, $P B=D C=\frac{1}{2} A B$.