Prove that AD : DB = GF : FB in the given figure, where DE || AC and EF || CG.

Given that $AC \parallel DE$.
According to the Basic Proportionality Theorem, if a line parallel to one side of a triangle intersects the other two sides, then it divides the two sides proportionally.
In $\triangle ABC$, $DE \parallel AC$. Therefore, we have:

$$\frac{AD}{DB} = \frac{CE}{EB} \quad (1)$$

Given that $CG \parallel EF$.
In $\triangle BCG$, $EF \parallel CG$. Therefore, we have:

$$\frac{BF}{FG} = \frac{BE}{EC}$$

Taking the reciprocal of both sides:

$$\frac{FG}{BF} = \frac{EC}{BE}$$

This can be written as:

$$\frac{GF}{FB} = \frac{CE}{EB} \quad (2)$$

From (1), we have $\frac{AD}{DB} = \frac{CE}{EB}$.
From (2), we have $\frac{GF}{FB} = \frac{CE}{EB}$.
Since both ratios are equal to $\frac{CE}{EB}$, they must be equal to each other.

$$\frac{AD}{DB} = \frac{GF}{FB}$$

This can also be written in ratio form as:

$$AD : DB = GF : FB$$

This completes the proof.

$\boxed{\text{AD : DB = GF : FB}}$