Proportional segments between parallelogram and triangle

Question

In the figure, ABCD is a parallelogram, $F$ is in the prolongation of $AD$. If $EF=32$, $GF=24$ and $AD=DF$, $BE=?$

I made the drawing

My try: I tried to do some proportions between the triangles $FDG$ and $BCG$ but got nothing. I'm 100% sure that this problem is about propotionality. Any hints?

Answer 1

By intercept theorem, $BG=GF=24$

$BE=BG+GF-EF=24+24-32=16$

Answer 2

Hint: Triangles DGF and ABF are similar.