2. If $P$ is any point inside quadrilateral $A B C D$, and
$$
S_{\triangle P A B}+S_{\triangle P C D}=\frac{1}{2} S_{\text {quadrilateral } A B C D}
$$

always holds, then the shape of quadrilateral $A B C D$ is $\qquad$
2. Parallelogram.

As shown in Figure 4, draw $P P^{\prime} / / A B$, with point $P^{\prime}$ inside quadrilateral $A B C D$, and connect
$$
P^{\prime} A, P^{\prime} B, P^{\prime} C, P^{\prime} D \text {. }
$$

Then $S_{\triangle P R B}=S_{\triangle P A B}$.
From the given information,

Therefore, $S_{\triangle P C D}=S_{\triangle P C D}$.
Thus, $P P / / C D$. Hence, $A B / / C D$.
$$
=\frac{1}{2} S_{\text {quadrilateral } A B C D} \text {, }
$$

Similarly, $A D / / B C$.
Therefore, quadrilateral $A B C D$ is a parallelogram.