The angle $\hat{F}_1$ is inscribed into the circle $c_1$ with corresponding central angle $\hat{BAE}$. Hence:

Figure 4
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$$\hat{F}_1 = \frac{BA\hat{E}}{2} = \frac{\hat{A}_1 + \hat{A}_2}{2} \qquad (1)$$
From the cyclic quadrilateral AFDB we have:
$$\hat{F}_1 = \hat{A}_1 \qquad (2)$$
From (1) and (2) we get that $\hat{A}_1 = \hat{A}_2$, that is $AK$ is the bisector of the angle $BA\hat{E}$. Since $AB = AE$, we conclude that $AK$ and hence $DK$ is perpendicular to $BC$, that is
$$DK \perp BC \qquad (3)$$
In the circle $c$, cords $AB$ and $AF$ are equal, as radii of circle $c_1$ and so $\hat{D}_1 = \hat{C}$. Hence the quadrilateral $DKMC$ is inscribable. Therefore we have $D\hat{K}C = D\hat{M}C = 90^\circ$, that is:
$$DM \perp AC \qquad (4)$$
From the inscribed quadrilateral $BKDL$ we have $B\hat{K}D = B\hat{L}D = 90^\circ$, that is:
$$DL \perp AB \qquad (5)$$
From the relations (3), (4) and (5) we conclude that the points $K,L,M$ are on the Simson's of the triangle $ABC$ corresponding to the point $D$.
From the inscribable quadrilateral $DKMC$ we get that $\hat{M}_1 = \hat{C}_1$. Also from the inscribed quadrilateral $ABDC$ we have $\hat{C}_1 = \hat{A}_1$ and finally from the inscribed quadrilateral $ABDF$ we have $\hat{A}_1 = \hat{F}_1$. Hence $\hat{M}_1 = \hat{F}_1$ from which we get that $BF // LM$.

We work as above till relation (5) and we use the fact that $AK$ is the perpendicular bisector of $BE$ and that $AM$ is the perpendicular bisector of $EF$. Since $K$ is the midpoint of $EB$ and $M$ is the midpoint of $EF$, we conclude that:
$$KM // BF \qquad (6)$$
From the inscribed quadrilateral $BKDL$ we have $\hat{L}_1 = \hat{D}_2$. Since $\overline{AF} = \overline{AB}$ the angles $\hat{B}_1$ and $\hat{D}_2$ are equal. Therefore we conclude that $\hat{L}_1 = \hat{B}_1$ and so
$$KL // BF \qquad (7)$$
From relations (6) and (7) we get that $K,L,M$ are collinear and $BF // LM$.