Solution:

Extend lines $BC$ and $AE$ to intersect at $H$. Then the circle inside quadrilateral $ABCE$, tangent to $AB$, $BC$, and $EA$, is really the inscribed circle of $\triangle HBA$. (Actually, this is only true if the circle lies inside $\triangle HBA$ rather than outside it. However, the fact that $CE$ is parallel to $AB$ with $CE < CD = AB$ readily implies that $E$ lies between $H$ and $A$, and $C$ lies between $H$ and $B$, so that the whole quadrilateral $ABCE$ lies within $\triangle HBA$, so the circle drawn inside it does too.)

Now let $P$ be the intersection point of lines $BD$ and $AE$. Consider the homothety (scaling) about $P$ that sends point $D$ to point $B$. Since homotheties preserve directions of lines, this map takes line $AD$ to the line through $B$ and parallel to $AD$, namely line $HB$. Similarly, it takes line $DE$ to line $BA$. And line $EA$ passes through $P$, the center of the homothety, so it goes to itself.

Thus, our homothety takes lines $AD$, $DE$, $EA$ to lines $HB$, $BA$, $AH(=AE)$, respectively, so it takes $\triangle ADE$ to $\triangle HBA$. Consequently, the incircle of $\triangle ADE$ is mapped to the incircle of $\triangle HBA$, and the map also matches corresponding tangency points: $F$ goes to $G$. But if a homothety about $P$ takes $F$ to $G$, then $P$, $F$, $G$ must be collinear. We now know that $P$ lies on lines $AE$, $BD$, and $FG$, which is what we need.