6. As shown in Figure 3, in quadrilateral $A B C D$, the angle bisectors of $\angle A$, $\angle B$, $\angle C$, and $\angle D$ intersect at a single point $P$. Let the areas of $\triangle A P D$, $\triangle A P B$, $\triangle B P C$, and $\triangle D P C$ be $S_{1}$, $S_{2}$, $S_{3}$, and $S_{4}$, respectively. Then ( . ).
(A) $S_{1}+S_{3}=S_{2}+S_{4}$
(B) $S_{1}+S_{2}=S_{3}+S_{4}$
(C) $S_{1}+S_{4}=S_{2}+S_{3}$
(D) $S_{1}+S_{3} \neq S_{2}+S_{4}$
6. A.

From the point on the angle bisector to the two sides of the angle, the distances are equal, so the distances from point $P$ to each side of the quadrilateral are equal, denoted as $h$ (as shown in Figure 11). It is easy to see that
$\triangle A P E \cong \triangle A P F$,
$\triangle B P E \cong \triangle B P H$,
$\triangle C P G \cong \triangle C P H$,
$\triangle D P G \cong \triangle D P F$.

Adding them up, we get $S_{\triangle A B P}+S_{\triangle C D P}=S_{\triangle B C P}+S_{\triangle D P}$.