Since $E, A, C, B$ are concyclic and $AD$ is parallel to $BC$, $\angle EAB = \angle ECB$ and $\angle BAD = -\angle ABC$. Therefore,
$$\begin{align*} \angle EAD &= \angle EAB + \angle BAD \\ &= \angle ECB - \angle ABC \\ &= \angle MCF - \angle ABC. \tag{13} \end{align*}$$

Since $M$ is the midpoint of $AB$ and $AF \perp BC$, $M$ is the circumcenter of triangle $ABF$. We then have
$$MF = MB \quad \text{and} \quad \angle ABC = -\angle MFC. \tag{14}$$

From (13) and (14),
$$\begin{align*} \angle EAD &= \angle MCF - \angle MFC \\ &= \angle MCF + \angle CFM \\ &= -\angle FMC \\ \end{align*}$$

Since $E, A, C, B$ are concyclic, we get $\triangle AEM \sim \triangle CBM$. Therefore, $\frac{AE}{EM} = \frac{CB}{BM}$.
From $CB = AD$ and $BM = MF$, we get $\frac{AE}{EM} = \frac{AD}{MF}$. So,
$$\frac{AE}{AD} = \frac{EM}{MF}. \tag{16}$$

From (15) and (16), we obtain $\triangle AED \sim \triangle MEF$. Therefore, $\angle ADE = \angle MFE$.

$$\begin{align*} \angle EDC &= \angle ADC - \angle ADE \\ &= \angle ADC - \angle MFE. \tag{17} \end{align*}$$
Applying (17) and $\angle ABC = -\angle ADC$ from the parallelogram $ABCD$ we obtain,

$$\begin{align*} \angle EDC &= \angle ADC - \angle MFE \\ &= \angle MFC - \angle MFE \\ &= \angle EFC. \end{align*}$$
Therefore, $C, D, E$, and $F$ are concyclic.

(by Wijit Yangjit)
Let $G$ be the intersection point of the lines $FM$ and $DA$. From $AM = MB$ and

$\angle AFB = 90^\circ$, we get $AM = FM = MB$. Since $AG$ is parallel to $FB$ and triangle $MBF$ is isosceles, we have $\angle AGF = \angle BFG = \angle ABF = \angle BAG$. Hence $AM = GM$. Next, since $\angle GFC = \angle GFB = -\angle ABF = \angle GDC$, it follows that $D, G, F$, and $C$ are concyclic. From $AM = GM$ and $MF = MB$, we get
$$GM \cdot MF = AM \cdot MB = EM \cdot MC.$$
By power of a point theorem, we get $G, E, F$, and $C$ are concyclic. Then, $D, E, G, F$, and $C$ are concyclic, and so $C, D, E$, and $F$ are concyclic.