6. As shown in Figure $1, P$ is a point on the arc $\overparen{A B}$ of the circumcircle of the equilateral $\triangle A B C$. Then the relationship between $B C$ and $P C$ is ( ).
(A) $B C \leqslant \frac{\sqrt{3}}{2} P C$
(B) $B C \leqslant \frac{\sqrt{3}}{3} P C$
(C) $B C \geqslant \frac{\sqrt{3}}{2} P C$
(D) $B C \geqslant \sqrt{3} P C$
6. C.

First, prove that in $\triangle ABC$, if $\angle A=60^{\circ}$, then $BC^{2}=AB^{2}+AC^{2}-AB \cdot AC$.

As shown in Figure 6, draw $CD \perp AB$, then $AD=\frac{1}{2} AC$,
$$
CD=\frac{\sqrt{3}}{2} AC .
$$

Thus, $BC^{2}=BD^{2}+CD^{2}$
$=(AB-AD)^{2}+CD^{2}$
$=\left(AB-\frac{1}{2} AC\right)^{2}+\left(\frac{\sqrt{3}}{2} AC\right)^{2}$
$=AB^{2}+\frac{1}{4} AC^{2}-AB \cdot AC+\frac{3}{4} AC^{2}$
$$
=AB^{2}+AC^{2}-AB \cdot AC \text {. }
$$

In $\triangle PAC$ and $\triangle PBC$, given $\angle APC=$ $\angle BPC=60^{\circ}$, we have
$$
\begin{array}{l}
AC^{2}=PA^{2}+PC^{2}-PA \cdot PC, \\
BC^{2}=PB^{2}+PC^{2}-PB \cdot PC, \\
\text { i.e., } PA^{2}-PC \cdot PA+PC^{2}-AC^{2}=0 \text {, } \\
PB^{2}-PC \cdot PB+PC^{2}-BC^{2}=0 \text {. } \\
\end{array}
$$

Since $AC=BC$, $PA \cdot PB$ are the two roots of the quadratic equation $t^{2}-PC \cdot t+PC^{2}-BC^{2}=0$ in terms of $t$.
By $\Delta \geqslant 0$, we know $PC^{2}-4\left(PC^{2}-BC^{2}\right) \geqslant 0$.
Simplifying, we get $4 BC^{2} \geqslant 3 PC^{2}$. Therefore, $BC \geqslant \frac{\sqrt{3}}{2} PC$.