7. In an equilateral triangle $\triangle A B C$ with side length $2 \sqrt{7}$, $M$ is the midpoint of side $B C$, and $P$ is a point on $A C$.
(1) For what value of $P C$ is $B P+P M$ minimized?
(2) Find the minimum value of $B P+P M$.
7. As shown in Figure 8, connect $A M$, and flip $\triangle A B C$ along the axis $A C$ by $180^{\circ}$ to get $\triangle A D C$. Let $N$ be the symmetric point of $M$ (about $A C$). At this point, we always have $P M = P N$. Since $B$ and $N$ are fixed points, $B P + P N \geqslant B N$, with equality holding if and only if point $P$ lies on the side $B N$.
To find the value of $P C$, extend $B C$ to $E$ such that $C E = C N$, forming an equilateral $\triangle C E N$. By the construction, we have
$$
P C \parallel N E, \frac{P C}{N E} = \frac{B C}{B E}, \text{ i.e., } \frac{P C}{\sqrt{7}} = \frac{2 \sqrt{7}}{3 \sqrt{7}}.
$$

Thus, when $P C = \frac{2 \sqrt{7}}{3}$, $B P + P M$ achieves its minimum value.
Draw $N F \perp C E$, with $F$ as the foot of the perpendicular. Then $N F = \frac{\sqrt{21}}{2}$, $B F = \frac{5 \sqrt{7}}{2}$,
$$
B N = \sqrt{B F^{2} + F N^{2}} = \sqrt{\frac{175}{4} + \frac{21}{4}} = 7
$$

Therefore, the minimum value of $B P + P M$ is 7.