5. As shown in Figure 1, if $P$

is a point inside the square
$A B C D$, and $S_{\triangle P A B}=5$,
$S_{\triangle P A D}=2$, then $S_{\triangle P A C}$ $=$ ( ).
(A) 1
(B) 2
(C) 3
(D) 4
5. C.

Let $S_{\triangle P A H}=m, S_{\triangle P C H}=n$.
Then $2+S_{\triangle B H C}+n=S_{\triangle P A D}+S_{\triangle B P C}$ $=\frac{1}{2} S_{\square A B C D}=S_{\triangle A B C}=S_{\triangle A B H}+S_{\triangle B H C}$ $=5-m+S_{\triangle B H C}$.
Therefore, $S_{\triangle P A C}=m+n=3$.