5. $P$ is a point inside $\triangle A B C$, connecting $P A, P B, P C$, dividing the triangle's area into three equal parts. Then $P$ is the ( ) of $\triangle A B C$.
(A) Incenter
(B) Circumcenter
(C) Orthocenter
(D) Centroid
5. (D).

As shown in the figure, extend $A P$ to intersect $B C$ at $D$. By the ratio of the areas of triangles with the same height, which is equal to the ratio of their bases, we have
$$
\begin{array}{l}
\frac{S_{\triangle A P B}}{S_{\triangle A B C}}=\frac{A P}{A D}=\frac{S_{\triangle A P C}}{S_{\triangle A C D}} . \\
\therefore \frac{S_{\triangle A P B}+S_{\triangle A P C}}{S_{\triangle A B C}}=\frac{A P}{A D}, \\
\frac{1+1}{3}=\frac{A P}{A D} .
\end{array}
$$

Thus, $A P=-\frac{2}{3} A D, P D=\frac{1}{2} P A$.
From this, we know that $S_{\triangle B P D}=\frac{1}{2} S_{\triangle A P B}=\frac{1}{2} S_{\triangle B P C}$.
Therefore, $B D=D C$, which means $P$ is the centroid of $\triangle A B C$.