5. As shown in the figure, take any point $E$ on the side $B C$ inside the circumcircle of the equilateral $\triangle A B C$, and connect $A E$ to intersect $B C$ at $F$. Take a point $P$ on the segment $E F$, and connect $P B, P C$.

Then the relationship between $P A$ and $P B+P C$ is ( ).
(A) $P A>P B+P C$
(B) $P A=P B+P C$
(C) $P A<P B-P C$
(D) Cannot be determined
5. C.

Obviously, $\angle B C P < \angle B A P$, as shown in the figure. Now, rotate $\triangle B P C$ $60^{\circ}$ inward around point $B$ to get $\triangle B Q A$, such that $Q A = P C$, and $Q A$ is inside $\triangle B A P$. Connect $Q P$, then $\triangle B Q P$ is an equilateral triangle, $Q P = P B$.
In $\triangle A P Q$, we have
$$
P A < P Q + Q A = P B + P C .
$$