2. Given that $P$ is a point inside $\square A B C D$, and through $P$ lines $E F / / B C, G H / / A B$ are drawn, with points $E, F, G, H$ on sides $A B, C D, A D, B C$ respectively. If $S_{\square P E B H}=6, S_{\square P G D F}=4$, then the area of $\triangle P A C$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
2. A.

As shown in Figure 1, let
$$
\begin{array}{l}
S_{\square P E A G}=a, \\
S_{\square P F C H}=b .
\end{array}
$$

Then $S_{\square A B C D}$
$$
\begin{array}{l}
=a+b+6+4 \\
=a+b+10, \\
S_{\triangle A B C}=\frac{1}{2} S_{\square A B C D}=\frac{1}{2} a+\frac{1}{2} b+5, \\
S_{\text {quadrilateral } A B C}=S_{\triangle P A E}+S_{\triangle P C H}+S_{\square P E B H} \\
=\frac{1}{2} a+\frac{1}{2} b+6 .
\end{array}
$$

Therefore, $S_{\triangle P A C}=S_{\text {quadrilateral } P A B C}-S_{\triangle A B C}=1$.