<s>[INST] Mass Point Geometry Question In the triangle $ABC$, let $E$ be a point on $BC$ such that $BE : EC =
3: 2$. Pick points $D$ and $F$ on the sides $AB$ and $AC$ , correspondingly, so that
$3AD = 2AF$ . Let $G$ be the point of intersection of $AE$ and $DF$ . Given that $AB = 7$
and $AC = 9$, find the ratio $DG: GF$.
I have been working on trying to solve this problem.  I am having difficulty relating the length of $AB =7$  and the ratios given to find $AD:BD$.  Similarly I am having trouble finding ratio of $FC:AF$.  I am sure that I can solve this problem if someone can give me a hint on how to find those ratios.  Any help would be appreciated.
 [/INST] 
Let $D'\in AB$ be such that $CD'\parallel FD$ and let $G'=AE\cap CD'$.
By similarity, $\frac{DG}{GF}=\frac{DG'}{G'C}$. Let $H=BG'\cap AC$. By Ceva's theorem 
$$ 1=\frac{BE}{EC}\cdot \frac{CH}{HA}\cdot \frac{AD'}{D'B} = \frac{3}{2}\cdot\frac{CH}{HA}\cdot\frac{\frac{2}{3}b}{c-\frac{2}{3}b} $$
hence $\frac{CH}{HA}=\frac{c-\frac{2}{3}b}{b}$. By Van Obel's theorem
$$ \frac{CG'}{G'D}=\frac{CE}{EB}+\frac{CH}{HA} = \frac{2}{3}+\frac{c}{b}-\frac{2}{3}=\frac{c}{b} $$
hence if $c=AB=7$ and $b=AC=9$ it follows that $\frac{DG}{GF}=\color{red}{\large \frac{9}{7}}$.
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