$17 \cdot 121$ In $\triangle A B C$, $\angle B A C=120^{\circ}$, point $P$ is a point inside $\triangle A B C$, then
(A) $P A+P B+P C < A B+A C$.
(C) $P A+P B+P C=A B+A C$.
(D) The size relationship between $P A+P B+P C$ and $A B+A C$ depends on the position of point $P$. (China Guangzhou, Wuhan, Fuzhou, etc. five cities junior high school mathematics league;
China some provinces and cities junior high school mathematics correspondence competition, 1984)
[Solution] Rotate $\triangle A P C$ counterclockwise by $60^{\circ}$ around point $A$ to get $\triangle A P^{\prime} C^{\prime}$. Connect $P P^{\prime}, C C^{\prime}$, then
$$
\begin{array}{l}
\text { then } \triangle A P C \cong \triangle A P^{\prime} C^{\prime}, \\
\because \quad \angle B A C=120^{\circ}, \angle C^{\prime} A C=60^{\circ},
\end{array}
$$
$\therefore B, A, C^{\prime}$ are collinear.
Also, $\because \angle P A P^{\prime}=60^{\circ}, A P=A P^{\prime}$,
$\therefore \triangle P A P^{\prime}$ is an equilateral triangle.
$\therefore \quad P A=P^{\prime} A=P P^{\prime}$.
$\because \quad A C^{\prime}=A C, \quad P C=P^{\prime} C^{\prime}$,
$$
\begin{array}{ll}
\therefore & P A+P B+P C=P P^{\prime}+P B+P^{\prime} C^{\prime} . \\
\because & A B+A C=A B+A C^{\prime}A B+A C .
\end{array}
$$

Therefore, the correct choice is (B).