A contest problem in geometry to find triangle ratio $EG$ to $GF$

Question

The question is taken from IMO 1988:

Let $\Delta ABC$ be a triangle with $AB =$ $12$ and $AC =$ $16.$ Suppose $M$ is the midpoint of side $BC$ and points $E$ and $F$ are chosen on sides $AC$ and $AB$ respectively, and suppose that the lines $EF$ and $AM$ intersect at $G.$ If $AE = 2AF$ then find the ratio $EG/GF$

I have tried this question a lot but I fail to solve it .I tried construction like:

1. A line from $F$ to $X$ on $AC$ such that it divide $AE$ into equal halves.But I cant figure out anything from this approach .

2. I constructed a line parallel to $BC$ from $F$ to $AC$ but it also doesnt help.

Please guide with correct approach and solution .(I think similarity, Appolinus theorem and several constructions would help us to solve but I am not able to crack how)

Thanks in advanced.

Answer 1

Let [.] denotes triangle areas. Then,

$$\frac{FG}{GE}=\frac{[AFM]}{[AEM]} = \frac{\frac{AF}{AB}[ABM]}{\frac{AE}{AC}[ACM]} =\frac{AF}{AE}\cdot\frac{AC}{AB}= \frac12\cdot \frac{16}{12} = \frac23$$