6. As shown in the figure, the diagonals of isosceles trapezoid $ABCD$ intersect at point $O$, and the areas of $\triangle OBC$, $\triangle OCD$, $\triangle ODA$, and trapezoid $ABCD$ are $S_{1}$, $S_{2}$, $S_{3}$, and $S$, respectively. For the following three conclusions:
(1) Knowing $S_{1}: S$, the ratio $AD: BC$ can be determined;
(2) Knowing $\left(S_{1}+S_{2}\right): S$, the ratio $AD: BC$ can be determined;
(3) Knowing $S_{2}: S$, the ratio $AD: BC$ can be determined;
The correct understanding is
(A) Only (1) is correct;
(B) Only (2) is correct;
(C) Only (1) and (2) are correct;
(D) (1), (2), and (3) are all correct.
Answer $(\quad)$
6. (D)

Solution Extend $B C$ to $E$, such that $C E=A D$,
then
$$
\begin{array}{l} 
S_{\triangle B D E}=S^{\prime} . \\
\because \quad A D: B C=\frac{B E}{B C}-1 \\
=\sqrt{\frac{S}{S_{1}}}-1,
\end{array}
$$

and $A D: B C=\frac{S_{2}+S_{3}}{S_{1}+S_{2}}$
$$
=\frac{S}{S_{1}+S_{2}}-1,
$$
$\therefore$ (1) and (2) are both correct.
Also, since $S_{1} S_{3}=S_{2}^{2}$, we have $\frac{S_{1}}{S} \cdot \frac{S_{3}}{S}=\left(\frac{S_{2}}{S}\right)^{2}$,
thus $\frac{B C \cdot A D}{(B C+A D)^{2}}=\frac{S_{2}}{S}$.
That is, $\frac{\frac{A D}{B C}}{\left(1+\frac{A D}{B C}\right)^{2}}=\frac{S_{2}}{S}$.
Solving it, we get $\frac{A D}{B C}=\frac{1-\frac{2 S_{2}}{S}-\sqrt{1-\frac{4 S_{2}}{S}}}{\frac{2 S_{2}}{S}}$.
$\therefore$ (3) is also correct, that is, (1), (2), and (3) are all correct, choose (D).