18.39 As shown in the figure: on the four sides of $\square P Q R S$ there are points $A$, $B, C, D$, such that $S_{\triangle P Q R S}=2 S_{\text {quadrilateral } A B C D}$, then it must be true that
(A) $A C / / Q R$ or $B D / / P Q$.
(B) $A C / / Q R$ and $B D / / P Q$.
(C) $A C \perp P Q$ or $B D \perp Q R$.
(D) $A C \perp P Q$ and $B D \perp Q R$.
(1st "Five Sheep Cup" Junior High School Mathematics Competition, 1989)
[Solution] If $BD$ is not parallel to $PQ$, then take another point $D'$ on $SP$ such that $BD' \parallel PQ$, as shown in the figure. This gives us
$$
\begin{array}{l}
S_{\triangle ABC} + S_{\triangle ACD'} \\
= S_{\text{quadrilateral } ABCD'} = S_{\triangle ABD'} + S_{\triangle BCD'} \\
= \frac{1}{2} S_{\square PQBD'} + \frac{1}{2} S_{\square D'BR S} \\
= \frac{1}{2} S_{\square PQRS} \\
= S_{\text{quadrilateral } ABCD} = S_{\triangle ABC} + S_{\triangle ACD}, \\
\therefore S_{\triangle ACD'} = S_{\triangle ACD}, \\
\end{array}
$$

Thus, $AC \parallel DD'$, i.e., $AC \parallel RQ$.
Similarly, if $AC$ is not parallel to $QR$, we can prove that $BD \parallel PQ$. Therefore, at least one of $AC \parallel RQ$ and $BD \parallel PQ$ must hold. Hence, the answer is $(A)$.