ABCD is a parallelogram and G is a point on BD. The line BF produced meets DC at F and AD produced at E. GF is parallel to AE.

1. Identify one pair of equal corresponding angles using parallel lines.

In parallelogram $ABCD$, $AD \parallel BC$. Since $E$ lies on $AD$ produced, $DE$ is along $AD$.

Also, $F$ lies on $DC$, so $CF$ and $DF$ are on the same straight line.

Hence,
$\angle BCF = \angle EDF$
because $BC \parallel DE$ and $CF$ is collinear with $DF$.

2. Identify a second pair of equal angles and conclude similarity.

Since $B,F,E$ are collinear and $C,F,D$ are collinear,
$\angle BFC = \angle EFD$
as vertically opposite angles.

Therefore, two pairs of corresponding angles are equal, so by AA similarity,
$\triangle BCF \sim \triangle EDF.$

1. Name the similar triangle using corresponding angles from parallel lines.

Since $GF \parallel DE$ and $G$ lies on $BD$, corresponding angles are equal:
$\angle BGF = \angle BDE.$
Also, $B$, $F$, $E$ are collinear and $B$, $G$, $D$ are collinear, so
$\angle GBF = \angle DBE.$
Hence, by AA similarity,
$\triangle BFG \sim \triangle BDE.$
Therefore, a triangle similar to $\triangle BDE$ is $\triangle BFG$.

1. Use corresponding sides of similar triangles to find the ratio $CF:DF$, then relate it to $AB$.

From part (a), $\triangle BCF \sim \triangle EDF$.

Corresponding sides give
$\frac{CF}{DF}=\frac{BC}{DE}.$
Since $ABCD$ is a parallelogram, $BC=AD$ and $AB=DC$.
Given $AD=2DE$, we have $BC=2DE$.
Hence
$\frac{CF}{DF}=\frac{2DE}{DE}=2,$
so $CF=2DF$.

Since $C,F,D$ are collinear,
$DC=CF+DF=2DF+DF=3DF.$
But $AB=DC$, so
$AB=3DF.$
Therefore
$\frac{DF}{AB}=\frac{DF}{3DF}=\frac{1}{3}.$

1. Find the scale factor from $\triangle BDE$ to $\triangle BGF$.

From part (a), $\triangle BCF \sim \triangle EDF$.

Since $AD=2DE$ and $AD=BC$ in a parallelogram,
$BC=2DE.$
Hence
$\frac{BF}{EF}=\frac{BC}{DE}=2.$
So
$BF:EF=2:1,$
and therefore
$\frac{BF}{BE}=\frac{2}{2+1}=\frac{2}{3}.$

2. Use $GF \parallel AE$ to relate $GF$ to $DE$, then convert to $AE$.

Since $GF \parallel AE$ and $AE$ is on the same straight line as $DE$,
$GF \parallel DE.$
Thus
$\triangle BGF \sim \triangle BDE.$
So
$\frac{GF}{DE}=\frac{BF}{BE}=\frac{2}{3}.$

Also,
$AE=AD+DE=2DE+DE=3DE.$
Therefore
$\frac{GF}{AE}=\frac{\frac{2}{3}DE}{3DE}=\frac{2}{9}.$

1. Express the base of the parallelogram in terms of $DE$.

Since $AD=2DE$, and $ABCD$ is a parallelogram,
$CD=AB=AD=2DE.$

2. Use the earlier result for $DF$ to express $DF$ in terms of $DE$.

From part (c)(i),
$\frac{DF}{AB}=\frac{1}{3}.$
Since $AB=2DE$,
$DF=\frac{1}{3}AB=\frac{1}{3}(2DE)=\frac{2}{3}DE.$

3. Compare the areas using common perpendicular height.

The perpendicular height of $\Delta EDF$ from $F$ to $DE$ is the same as the perpendicular height of parallelogram $ABCD$ to base $AD$, since $F$ lies on $DC$ and $DC\parallel AD$.

Hence,

But this uses base $DE$ directly. Equivalently, using the height to $DF$ compared with the parallelogram height requires the angle between $DE$ and $DF$; the valid area comparison is instead obtained from similarity with $\triangle BCF$ and the whole parallelogram decomposition. Since $AD=2DE$ and $DF=\frac{2}{3}DE$, let $DE=x$. Then $AD=AB=2x$ and $DF=\frac{2x}{3}$.

Taking $DE$ as horizontal and the height of the parallelogram as $h$, $F$ lies on $DC$, so the vertical height of $\Delta EDF$ is $h$. Thus
$\text{Area of }\Delta EDF=\frac12(x)(h),$
while
$\text{Area of }ABCD=(2x)(h).$
Therefore,
$\frac{\text{Area of }\Delta EDF}{\text{Area of }ABCD}=\frac{\frac12xh}{2xh}=\frac14.$