Problem similar to midpoint theorem

Question

I desperately need help solving the following problem. I made a sketch here:

In a triangle ABC, let:

• D $\in$ BC \ {B,C}
• E $\in$ AC \ {A,C}
• F intercepting point of AD and BE
• G intercepting point of AB and line through C and F

To prove: AB || DE $\iff$ G is midpoint of AB

I'd really appreciate help on that one. I can't figure it out.

Thank you in advance.

Answer 1

HINT

Let use Ceva's theorem and triangle similarity.

Answer 2

If $DE$ is parallel to $AB$ then $\triangle ABD$ has equal area to $\triangle ABE$

$\triangle AEF$ has equal area to $\triangle BDF$

if we construct $PQ$ through $F$ and parallel to $AB, F$ is the midpoint of $PQ$

$\triangle ABC$ is similar to $\triangle PQC$

$G$ is the midpoint of $AB$