Solution 1. Construct point $F_1$ such that $EF_1 = BF_1$ and ray $DF_1$ is perpendicular to line $ED$. It suffices to show that $F = F_1$ or $ABDF_1$ is cyclic; that is,
$$\angle BAD = \angle BF_1 D. \qquad \textcircled{1}$$
Set $\angle BAD = \angle CAD = x$. Because $EC \perp AB$ and $AD \perp BC$,
$$\angle ECB = 90^\circ - \angle ABD = \angle BAD = x.$$
Note that $MD$ is a midline of triangle $BCE$. In particular, $MD \parallel EC$ and
$$\angle MDB = \angle ECD = x. \qquad \textcircled{2}$$
In isosceles triangle $EF_1M$, we may set $\angle EF_1M = \angle BF_1M = y$. Because $EM \perp MF_1$ and $MD \perp DF_1$,
$$\angle EMF_1 = \angle EDF_1 = 90^\circ,$$
implying that $EMDF_1$ is cyclic. Consequently, we have
$$\angle EDM = \angle EF_1 M = y. \qquad \textcircled{3}$$
Combining ② and ③, we obtain

Fig. 2. 2
$$\angle BDE = \angle EDM + \angle MDB = x + y.$$
Because $BE = BD$, we conclude that triangle $BED$ is isosceles with $\angle MED = \angle BED = \angle BDE = x + y$. Because $EMDF_1$ is cyclic, we have $\angle MF_1 D = \angle MED = x + y$. It is then clear that
$$\angle BF_1 D = \angle MF_1 D - \angle MF_1 B = x = \angle BAD,$$
which is ①.

Solution 2. (Based on work by Sherry Gong and Inna Zakharevich) We maintain the notations used in Solution 1. Let $\omega$ and $O$ denote the circumcircle and the circumcenter of triangle $ABD$, and let $T$ be second intersection (other than $B$) between line $BE$ and $\omega$. Extend segment $DE$ through $E$ to meet $\omega$ at $S$. Point $F_2$ lies on $\omega$ such that $DF_2 \perp DE$. We will show that $F_2 = F$ or $F_2B = F_2E$. Let $M_2$ denote the foot of the perpendicular from $F_2$ to line $BE$. It suffices to show that $M_2$ is the midpoint of segment $BE$, that is, $EM_2 = M_2B$.
Because $\angle SDF_2 = \angle EDF_2 = 90^\circ$, $O$ is the midpoint of $SF_2$. Because $BTAD$ is cyclic, $\angle BTD = \angle BAD = \angle BCE = x$. Note also that $BD = BE$ and $\angle EBC = \angle DBT$. We conclude that triangle $BDT$ and $BEC$ are congruent to each other, implying that $BE = ET$. Hence, $OE \perp BT$. Let $N$ be the foot of the perpendicular from $S$ to line $BE$.

Fig. 2. 3
Note that segments $NE$ and $EM_2$ are the respective perpendicular projections of segments $SO$ and $OF_2$ onto line $BE$. Because $SO = OF_2$, $NE = EM_2$. Because $TE = EB$, it suffices to show that $TN = NE$, which is evident since
$$\angle STE = \angle SEB = \angle SDB = \angle EDB = \angle BED = \angle SET$$
and so triangle $SET$ is isosceles with $SE = ST$.