Let $T$ be the midpoint of $\operatorname{arc} \overparen{BAC}$ and let lines $BA$ and $CD$ intersect $EF$ at $K$ and $L$, respectively. Note that $T$ lies on the perpendicular bisector of segment $BC$.

Since $ABCD$ is cyclic, $\frac{BD}{\sin \angle BAD} = \frac{AC}{\sin \angle ADC}$. From parallel lines we have $\angle DAF = \angle ADC$ and $\angle BAD = \angle EDA$. Hence,
$$AF \cdot \sin \angle DAF = BD \cdot \sin \angle ADC = AC \cdot \sin \angle BAD = DE \cdot \sin \angle EDA.$$
So $F$ and $E$ are equidistant from the line $AD$, meaning that $EF$ is parallel to $AD$.

We have that $KADE$ and $FADL$ are parallelograms, hence we get $KA = DE = AC$ and $DL = AF = BD$. Also, $KE = AD = FL$ so it suffices to prove the perpendicular bisector of $KL$ passes through $T$.

Triangle $AKC$ is isosceles so $\angle BTC = \angle BAC = 2 \angle BKC$. Likewise, $\angle BTC = 2 \angle BLC$. Since $T, K$, and $L$ all lie on the same side of $BC$ and $T$ lies on the perpendicular bisector of $BC$, $T$ is the centre of circle $BKLC$. The result follows.

Let $AF$ and $DE$ meet $\omega$ at $X$ and $Y$, respectively, and let $T$ be as in Solution 1.

As $BD < AD$, $DY \parallel AB$ and $\angle BAY = \angle DBA < 90^{\circ}$, we have $DY < AB$ and $Y$ lies on the opposite side of line $AD$ to $C$. Also from $BD < AD$, we have $B, C$, and $D$ all lie on the same side of the perpendicular bisector of $AB$ which shows $AC > AB$. Combining these, we get $DY < AB < AC = DE$ and, as $Y$ and $E$ both lie on the same side of line $AD$, $Y$ lies in the interior of segment $DE$. Similarly, $X$ lies in the interior of segment $DF$.

Since $AB$ is parallel to $DY$, we have $YA = BD = FA$. Likewise $XD = AC = ED$.

Claim 1. $T$ is the midpoint of $\operatorname{arc} \overparen{XY}$.

Proof. From $AX \parallel CD$ and $AB \parallel DY$ we have
$$\angle CAX = \angle AXD = \angle AYD = \angle YDB.$$
Since $T$ is the midpoint of $\operatorname{arc} \overparen{BAC}$, we have $\angle BAT = \angle TDC$, so
$$\angle TAX = \angle CAX + \angle BAC - \angle BAT = \angle YDB + \angle BDC - \angle TDC = \angle YDT.$$
Recall from above we have $AB < AC$ and analogously, $DC < DB$, which shows that $X, Y$ and $T$ all lie on the same side of line $AD$. In particular, $T$ and $A$ lie on opposite sides of $XY$ so $T$ lies on the internal angle bisector of $\angle XAY$. Since $AF = AY$, we have $\triangle ATF \cong \triangle ATY$, giving $TF = TY$.

Likewise, $TE = TX$, so $TE = TF$, meaning that $T$ lies on the perpendicular bisector of segment $EF$ as required.

From $AF = DB$, $AC = DE$ and
$$\angle(AC, AF) = \angle(AC, CD) = \angle(AB, BD) = \angle(DE, DB),$$
triangles $ACF$ and $DEB$ are congruent, so $CF = BE$.

Let $P = BE \cap CF$. Since
$$\angle(CP, BP) = \angle(CF, BE) = \angle(AF, DB) = \angle(DC, DB),$$
we have that $P$ lies on circle $ABCD$.

Finally, let $T$ be the Miquel point of the quadrilateral $BCFE$ so $T$ lies on circles $EFP$ and $ABCD$. Note that $T$ is the centre of spiral similarity taking segments $BE$ to $CF$ and since $BE = CF$, this is in fact just a rotation, so $TB = TC$ and $TE = TF$; that is, the perpendicular bisectors of $BC$ and $EF$ meet at $T$, on circle $ABCD$.