18.41 As shown in the figure, $P$ is a point inside the square $ABCD$, and $S_{\triangle PAB}=5, S_{\triangle PAD}=$ 2, then $S_{\triangle PAC}$ equals
(A) 2.
(B) 3.
(C) $3 \frac{1}{2}$.
(D) 4.
(China Hubei Province Huanggang Junior High School Mathematics Competition, 1993)
[Solution] Consider the extreme case where $P$ is on the diagonal $BD$ of $\square ABCD$.
According to the problem, $S_{\triangle PAB}=5, S_{\triangle PAD}=2$. Let $S_{\triangle APO}=S$,
then $S_{\triangle AOH}+S=5, S_{\triangle AOD}-S=2$,
and $S_{\triangle AOH}=S_{\triangle AOH}$,
solving this gives $S=1.5$, thus $S_{\triangle APC}=2S=3$.
Therefore, the answer is $(B)$.