Since the sides $BC$ and $FA$ are not parallel, the lines $BC$ and $FA$ intersect. Call the point of their intersection $X$. Similarly, let $Y$ and $Z$ be the points of intersections of the lines $BC$, $DE$ and of the lines $DE$, $FA$, respectively.
Since the quadrilateral $ABCD$ is inscribed in the circle, we have $\angle YCD = \angle BAD$. As the lines $AB$, $DE$ are parallel, $\angle BAD = \angle ADE$. We also have $\angle ADE = \angle ZFE$, since the quadrilateral $ADEF$ is inscribed in the circle. Finally, we have $\angle ZFE = \angle AXB$ since the lines $BC$ and $EF$ are parallel. Putting these identities together, we get $\angle YCD = \angle AXB$, which implies that the sides $CD$ and $FA$ are parallel.

Alternate Solution: Since the lines $AB$ and $DE$ are parallel, we have $\angle ABE = \angle BED$. Let $\alpha$ be the common value of these angles. Similarly, we have $\angle CBE = \angle BEF$, whose value we call $\beta$. Using the properties of quadrilaterals inscribed in a circle, we obtain
$$\angle FCD = 180^\circ - \angle DEF = 180^\circ - \angle BED - \angle BEF = 180^\circ - \alpha - \beta,$$
$$\angle CFA = 180^\circ - \angle ABC = 180^\circ - \angle ABE - \angle CBE = 180^\circ - \alpha - \beta.$$
From this we get $\angle FCD = \angle CFA$ which implies that the sides $CD$ and $FA$ are parallel.