在下图中，ABGF 和 BCDG 是菱形。$\angle ABF = 31^\circ$，$\angle CBG = 40^\circ$ 且 EF 平行于 DG。

1. Find $\angle FGB$ using the properties of rhombus ABGF

In rhombus $ABGF$, the diagonal $BF$ bisects $\angle ABG$.
$\angle ABG = 2 \times 31^\circ = 62^\circ$
Adjacent angles in a rhombus add up to $180^\circ$, so:
$\angle FGB = 180^\circ - 62^\circ = 118^\circ$

2. Find $\angle BGD$ using the properties of rhombus BCDG

In rhombus $BCDG$, adjacent angles add up to $180^\circ$.
$\angle BGD = 180^\circ - 40^\circ = 140^\circ$

3. Calculate $\angle FGD$ using angles at a point

Angles around point $G$ sum to $360^\circ$.
$\angle FGD = 360^\circ - 118^\circ - 140^\circ = 102^\circ$

4. Find $\angle GDF$ using the isosceles triangle formed by the equal sides of the rhombuses

Since $ABGF$ and $BCDG$ are rhombuses sharing side $BG$, all their sides are equal. Thus, $GF = GB = GD$.
Triangle $FGD$ is an isosceles triangle ($GF = GD$).
$\angle GDF = \frac{180^\circ - 102^\circ}{2} = 39^\circ$

5. Find $\angle DFE$ using parallel lines

Given $EF \parallel DG$, $\angle DFE$ and $\angle GDF$ are alternate interior angles.
$\angle DFE = \angle GDF = 39^\circ$