$18 \cdot 117$ As shown in the figure, in trapezoid $ABCD$, $CD // AB, AC, BD$ intersect at point $P$. Let the areas of $\triangle PCD, \triangle PDA, \triangle PAB, \triangle PBC$ be $S_{1}, S_{2}, S_{3}, S_{4}$, respectively. Then the relationship that holds is
(A) $S_{1}+S_{3}=S_{2}+S_{4}$.
(B) $S_{1}+S_{3} \geqslant S_{2}+S_{4}$.
(C) $S_{1}+S_{3}>S_{2}+S_{4}$.
(D) $S_{1}+S_{3} \leqslant S_{2}+S_{4}$.
(Anhui Province, Anqing and other counties and cities Junior High School Mathematics Competition, 1991)
[Solution] Without loss of generality, let $DC \leqslant AB$, and set $DC=a, AB=b$. Also, draw the common perpendicular line $EF$ through $P$, and let $EF=h, \quad EP=h_{1}, \quad PF=h_{2}$.
From $\triangle PDC \backsim \triangle PBA$, and $DC \leqslant AB$, we know $h_{1} \leqslant h_{2}$.
$$
\begin{aligned}
S_{ABCD} & =\frac{1}{2}(a+b) h \\
& =\frac{1}{2}(a+b)\left(h_{1}+h_{2}\right) \\
& =\frac{1}{2}\left(a h_{1}+b h_{2}\right)+\frac{1}{2}\left(a h_{2}+b h_{1}\right), \\
& S_{1}+S_{3}=\frac{1}{2}\left(a h_{1}+b h_{2}\right),
\end{aligned}
$$

Also, $S_{1}+S_{3}=\frac{1}{2}\left(a h_{1}+b h_{2}\right)$,
then $S_{2}+S_{4}=S_{ABCD}-\left(S_{1}+S_{3}\right)=\frac{1}{2}\left(a h_{2}+b h_{1}\right)$.
Notice that if $x \geqslant y, \quad u \geqslant v$, then $x(u-v)-y(u-v) \geqslant 0$.
That is, $x u+y v \geqslant x v+y u$,

Similarly, by $a \leqslant b, h_{1} \leqslant h_{2}$, then $a h_{1}+b h_{2} \geqslant a h_{2}+b h_{1}$.
$$
\therefore \quad S_{1}+S_{3} \geqslant S_{2}+S_{4} \text {. }
$$

Therefore, the answer is $(B)$.