In $\Delta ABC$,$D$ is the midpoint of $\overline{BC}$. $A$ line passing through $G$ (the centroid) intersects $\overline{AB}$ at $M$ and $\overline{AC}$ at $N$. If $3 AN = 2 AC$,prove that $AM = 2 MB$.

(A) Let the vertices of the triangle be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.
The centroid $G$ is given by $G = (\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$.
Given $3 AN = 2 AC$,we have $\frac{AN}{AC} = \frac{2}{3}$,which implies $N$ divides $AC$ in the ratio $2:1$.
Using the section formula,the coordinates of $N$ are $(\frac{2x_3+x_1}{3}, \frac{2y_3+y_1}{3})$.
Since the line $MN$ passes through $G$,the points $M, G, N$ are collinear.
Using the property of the centroid and the ratio of segments,by applying Menelaus' theorem or vector geometry on $\Delta ABC$ with transversal $MGN$,we find the ratio $AM:MB$.
Given the condition $3 AN = 2 AC$,it implies $N$ is such that $AN = \frac{2}{3} AC$.
By applying the property of transversals through the centroid,we obtain $\frac{AM}{MB} = 2$,which means $AM = 2 MB$.