6. Given that the sides of $\square A B C D$ are $A B=m, A D=n$ $(m>n)$. If $P$ is a moving point on side $C D$, and let $D P=x$, the line $A P$ intersects the extension of $B C$ at point $Q$, such that $D P+C Q$ is minimized, then the relationship that $m$, $n$, and $x$ should satisfy is ( ).
(A) $x=\frac{m+n}{2}$
(B) $x=\sqrt{m n}$
(C) $m^{2}-n^{2}=x^{2}$
(D) $\frac{1}{x}=\frac{1}{m}+\frac{1}{n}$
6. B.

As shown in Figure 5, we have
$$
\begin{array}{l}
\triangle C P Q \backsim \triangle D P A \\
\Rightarrow \frac{C Q}{D A}=\frac{C P}{D P} \\
\Rightarrow \frac{C Q}{n}=\frac{m-x}{x} \\
\Rightarrow C Q=\frac{m-x}{x} \cdot n .
\end{array}
$$

Let's assume $y=D P+C Q$. Then
$$
y=x+\frac{n(m-x)}{x}=x+\frac{m n}{x}-n \geqslant 2 \sqrt{m n}-n \text {. }
$$

Therefore, $D P+C Q$ is shortest when and only when $x=\sqrt{m n}$.