In $\triangle ABC$ shown in the adjoining figure, $M$ is the midpoint of side $BC, AB=12$ and $AC=16$. Points $E$ and $F$ are taken on $AC$ 
and $AB$, respectively, and lines $EF$ and $AM$ intersect at $G$. If $AE=2AF$ then $\frac{EG}{GF}$ equals

$\textbf{(A)}\ \frac{3}{2} \qquad \textbf{(B)}\ \frac{4}{3} \qquad \textbf{(C)}\ \frac{5}{4} \qquad \textbf{(D)}\ \frac{6}{5}\\ \qquad \textbf{(E)}\ \text{not enough information to solve the problem}$
Here, we use Mass Points.
Let $AF = x$. We then have $AE = 2x$, $EC = 16-2x$, and $FB = 12 - x$
Let $B$ have a mass of $2$. Since $M$ is the midpoint, $C$ also has a mass of $2$. 
Looking at segment $AB$, we have
\[2 \cdot (12-x) = \text{m}A_{AB} \cdot x\]
So
\[\text{m}A_{AB} = \frac{24-2x}{x}\]
Looking at segment $AC$,we have
\[2 \cdot (16-2x) = \text{m}A_{AC} \cdot 2x\]
So
\[\text{m}A_{AC} = \frac{16-2x}{x}\]
From this, we get
\[\text{m}E = \frac{16-2x}{x} + 2 \Rrightarrow \text{m}E = \frac{16}{x}\]
and
\[\text{m}F = \frac{24-2x}{x} + 2 \Rrightarrow \text{m}F = \frac{24}{x}\]
We want the value of $\frac{EG}{GF}$. This can be written as
\[\frac{EG}{GF} = \frac{\text{m}F}{\text{m}E}\]
Thus
\[\frac{\text{m}F}{\text{m}E} = \frac {\frac{24}{x}}{\frac{16}{x}} = \frac{3}{2}\]
$\boxed{A}$
~JustinLee2017