6. As shown in Figure $2, \odot 0$ has two diameters $A B \perp$ $C D$, with its area being $S$. Point $P$ is on $\overparen{B D}$, $P A$ intersects $O D$ at point $E$, and $P C$ intersects $O B$ at point $F$. The area of quadrilateral $A C F E$ is ( ).
(A) $\frac{S}{2}$
(B) $\frac{S}{3}$
(C) $\frac{S}{\pi}$
(D) $\frac{2 S}{3 \pi}$
6.C.

Connect $P B$, then $P F$ bisects $\angle A P B$. Since $\triangle A O E \sim \triangle A P B$, we have $\frac{A O}{O E}=\frac{A P}{P B}=\frac{A F}{F B}$.
Let the radius of $\odot O$ be $r$, then $S=\pi r^{2}$.
Let $O E=x$, $O F=y$. Substituting, we get
$$
\begin{array}{l}
\frac{r}{x}=\frac{r+y}{r-y} \Rightarrow r^{2}=r x+r y+x y . \\
S_{\text {quadrilateral } A C F E}=\frac{1}{2} \cdot C E \cdot A F=\frac{1}{2}(r+x)(r+y) \\
=\frac{1}{2}\left(r^{2}+r x+r y+x y\right)=r^{2}=\frac{S}{\pi} .
\end{array}
$$