In the circle $c$, the chords $AB$ and $BE$ are equal and hence
$$\hat{A}_1 = \hat{F}_1 = \hat{F}_2 = \hat{E}_1 = \hat{C},$$

Figure 1
because in the circle $c$ the above angles go to the equal arches $AB$ and $BE$. $OB$ is the line of the centers of the circles $c$ and $c_1$, and so it is perpendicular bisector of the common chord $AE$, i.e. $AD \perp BM$. We will prove that $BF \perp AC$, whence point $D$ will be the orthocenter of the triangle $ABM$.

Moreover in the triangle $BKF$ we have: $\hat{K}_1 = \hat{F}_2 + \hat{B}_2$. (1)

The triangle $BKE$ is isosceles ($BE = BK$, as radii of the circle $c_1$), whence: $\hat{K}_1 = \hat{E}_3 = \hat{E}_1 + \hat{E}_2$. (2)

From the cyclic quadrilateral $ABEF$ we have:
$\hat{B}_1 = \hat{E}_2$ (3)

From relations (1), (2) and (3) (taking in mind equalities $\hat{F}_2 = \hat{E}_1 = \hat{C}$) we find: $\hat{B}_1 = \hat{B}_2$ (4).

From the equalities $\hat{B}_1 = \hat{B}_2$ and $\hat{F}_1 = \hat{F}_2 = \hat{C}$, we conclude that $ABF$ and $KBF$ are equal. Hence $BF$ is the perpendicular bisector of $AK$. Since $AD \perp BM$, we conclude that $D$ is the orthocenter of the triangle $ABM$. Therefore we have $\hat{M}_1 = \hat{A}_1 = 90^\circ - MBA$ and in combination with $\hat{A}_1 = \hat{F}_2 = \hat{C}$ we have that $\hat{M}_1 = \hat{F}_2$, whence the quadrilateral $DLMF$ is cyclic.

Since $\hat{B}_2 = \hat{E}_2$, the quadrilateral $DKEB$ is cyclic and since $\hat{E}_1 = \hat{M}_1$, the quadrilateral $DBEM$ is cyclic. Therefore the polygon $BDKME$ is cyclic.