18-119 As shown in the figure, the diagonals of trapezoid $ABCD$ intersect at point $O$. The areas of $\triangle OBC$, $\triangle OCD$, $\triangle ODA$, $\triangle AOB$, and trapezoid $ABCD$ are $S_{1}$, $S_{2}$, $S_{3}$, $S_{4}$, 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.
(China Junior High School Mathematics League, 1986)
[Solution 1] Extend $BC$ to $E$ such that $CE = AD$, then $S_{\triangle BDE} = S$.
$$
\begin{aligned}
\because \quad AD: \dot{B}C & =\left(\frac{BE}{BC}\right)-1 \\
& =\sqrt{\frac{S}{S_{1}}}-1,
\end{aligned}
$$

and $AD: BC = \frac{S_{3}+S_{4}}{S_{1}+S_{2}} = \frac{S}{S_{1}+S_{2}}-1$,
$\therefore$ (1), (2) are both correct.
Also, since $S_{1} S_{3} = S_{2}^{2}$, we have $\frac{S_{1}}{S} \cdot \frac{S_{3}}{S} = \left(\frac{S_{2}}{S}\right)^{2}$, hence $\frac{BC \cdot AD}{(BC+AD)^{2}} = \frac{S_{2}}{S}$, which means $\frac{AD}{BC} / \left(1+\frac{AD}{BC}\right)^{2} = \frac{S_{2}}{S}$. Solving this, we get $\frac{AD}{BC} = \left(1-\frac{2S_{2}}{S} - \sqrt{1-\frac{4S_{2}}{S}}\right) / \frac{2S_{2}}{S}$.
$\therefore$ (3) is also correct, so the answer is $(D)$.
[Solution 2] It is easy to prove that $S_{2} = S_{4} = \sqrt{S_{1} S_{3}}$,

then $S = S_{1} + S_{3} + 2 \sqrt{S_{1} S_{3}}$;
$\therefore \quad \frac{S}{S_{1}} = 1 + \frac{S_{3}}{S_{1}} + 2 \sqrt{\frac{S_{3}}{S_{1}}}$,
then given $S_{1}: S$, we can find $S_{3}: S_{1}$.
By $\triangle ADO \sim \triangle CBO$, we can find $\frac{AD}{BC} = \sqrt{\frac{S_{3}}{S_{1}}}$, so (1) is correct.
Also, $1 = \frac{S_{1} + S_{3} + 2S_{2}}{S} = \frac{S_{1} + S_{2}}{S} + \frac{S_{2} + S_{3}}{S}$,
$\therefore \quad \frac{S_{1} + S_{2}}{S} = 1 - \frac{S_{2} + S_{3}}{S}$,
then if $\frac{S_{1} + S_{2}}{S}$ is known, we can find $\frac{S_{2} + S_{3}}{S}$, and thus $\frac{S_{2} + S_{3}}{S_{1} + S_{2}}$,
$$
\therefore \frac{S_{\triangle ACD}}{S_{\triangle BCD}} = \frac{AD}{BC} = \frac{S_{2} + S_{3}}{S_{1} + S_{2}}
$$

can be found, so (2) is correct.
Furthermore, $\frac{S}{S_{2}} = \frac{S_{1} + S_{3} + 2 \sqrt{S_{1} S_{3}}}{S_{2}} = \frac{S_{1} + S_{3}}{\sqrt{S_{1} S_{3}}} + 2$,
if $S_{2}: S$ is known, we can find $\left(S_{1} + S_{3}\right): \sqrt{S_{1} S_{3}}$,
let $\sqrt{\frac{S_{1}}{S_{3}}} + \sqrt{\frac{S_{3}}{S_{1}}} = \frac{S_{1} + S_{3}}{\sqrt{S_{1} S_{3}}} = k$,

then
$$
\frac{S_{3}}{S_{1}} - k \sqrt{\frac{S_{3}}{S_{1}}} + 1 = 0,
$$

from which we can solve for $\frac{S_{3}}{S_{1}}$, and thus find $\frac{AD}{BC}$, so (3) is correct. Hence, the answer is $(D)$.