6. As shown in Figure $2, \odot O$ is the circumcircle of square $ABCD$, with $O$ as the center. Point $P$ is on the minor arc $AB$, and $DP$ intersects $AO$ at point $Q$. If $PQ=QO$, then $\frac{QC}{AQ}$ equals ( ).
(A) $2 \sqrt{3}-1$
(B) $\sqrt{3}+2$
(C) $\sqrt{3}+\sqrt{2}$
(D) $2 \sqrt{3}$
6. B.

As shown in Figure 7, let's assume $A O=$ $O C=1$, then $A D=C D=\sqrt{2}$.
Let $Q O=Q P=x$, then
$$
\begin{array}{l}
A Q=1-x, \\
Q C=1+x .
\end{array}
$$

By the intersecting chords theorem, we get
$$
\begin{array}{l}
D Q=\frac{A Q \cdot Q C}{Q P} \\
=\frac{1-x^{2}}{x} .
\end{array}
$$

Construct $Q H \perp A D, Q K \perp C D$, with $H$ and $K$ being the feet of the perpendiculars, then $Q H=A H=\frac{1-x}{\sqrt{2}}, D H=Q K=C K=\frac{1+x}{\sqrt{2}}$. By the Pythagorean theorem, $Q H^{2}+H D^{2}=D Q^{2}$, i.e., $\left(\frac{1-x}{\sqrt{2}}\right)^{2}+\left(\frac{1+x}{\sqrt{2}}\right)^{2}=\left(\frac{1-x^{2}}{x}\right)^{2}$.

Therefore, $x^{2}=\frac{1}{3} \Rightarrow x=\frac{1}{\sqrt{3}}$.
Thus, $\frac{Q C}{A Q}=\frac{1+x}{1-x}=\frac{\sqrt{3}+1}{\sqrt{3}-1}=2+\sqrt{3}$.