5. Let $K$ be any point inside $\triangle ABC$, and the centroids of $\triangle KAB$, $\triangle KBC$, and $\triangle KCA$ be $D$, $E$, and $F$ respectively. Then the ratio $S_{\triangle DEF}: S_{\triangle ABC}$ is ( ).
(A) $\frac{1}{9}$
(B) $\frac{2}{9}$
(C) $\frac{4}{9}$
(D) $\frac{2}{3}$
5.A.

Extend $K D$, $K E$, $K F$ to intersect the sides $A B$, $B C$, and $C A$ of $\triangle A B C$ at points $M$, $N$, and $P$ respectively. Since $D$, $E$, and $F$ are the centroids of $\triangle K A B$, $\triangle K B C$, and $\triangle K C A$ respectively, it is easy to see that $M$, $N$, and $P$ are the midpoints of $A B$, $B C$, and $C A$ respectively. Therefore, $S_{\triangle M N P}=\frac{1}{4} S_{\triangle A B C}$.
It is easy to prove that $\triangle D E F \backsim \triangle M N P$, with a similarity ratio of $2: 3$. Hence, $S_{\triangle D E F}=\left(\frac{2}{3}\right)^{2} S_{\triangle M N P}=\frac{4}{9} \times \frac{1}{4} S_{\triangle A B C}=\frac{1}{9} S_{\triangle A B C}$.