Problem: We have the quadrilateral $ABCD$. The middles of $AB$, $BC$, $CD$ and $DA$, are respectively $M$, $N$, $P$ and $Q$. The centroid of $BNP$ is $F$, and the centroid of $NPD$ is $G$. $MG$ intersects $FQ$ at $K$. We know that $FK = 6$ and we have to show $KQ = 9$.
I made а diagram, but I need some help.
Since $$\frac{GT}{DT}=\frac{1}{3}=\frac{FT}{TB},$$ we obtain $$GF||DB$$ and $$GF=\frac{1}{3}DB.$$ On the other hand, $$QM||DB$$ and $$QM=\frac{1}{2}DB.$$
Thus, $$GF||QM$$ and $$GF=\frac{2}{3}QM.$$ Id est, $$\frac{FK}{QK}=\frac{GF}{QM}=\frac{2}{3} $$ or $$\frac{6}{QK}=\frac{2}{3},$$ which gives $$QK=9.$$