3. (10 points) As shown in the figure, in the right triangle $ABC$, point $F$ is on $AB$ and $AF=2FB$, quadrilateral $EBCD$ is a parallelogram, then $FD: EF$ is $\qquad$ .
【Analysis】Since $A F=2 F B$, we have $A F: F B=2: 1$. Since quadrilateral $E B C D$ is a parallelogram, $B E / / A C$, so $\triangle A D F \sim \triangle B E F$. Therefore, $F D: E F=A F: F B=2: 1$. The answer can be derived from this.

【Solution】Solution: Since $A F=2 F B$,
we have $A F: F B=2: 1$,
Since quadrilateral $E B C D$ is a parallelogram,
we have $B E / / A C$,
thus $\angle A D F=\angle B E F, \angle E F B=\angle D F A$,
so $\triangle A D F \sim \triangle B E F$,
therefore $F D: E F=A F: F B=2: 1$,
hence the answer is: $2: 1$.