SELECTION EXAMINATION

The angle ${\hat{F}}_1$ is inscribed into the circle $c_1$ with corresponding central angle $\widehat{BAE}$. Hence: Figure 4 --- $${\hat{F}}_1=\frac{BA\hat{E}}{2}=\frac{{\hat{A}}_1+{\hat{A}}_2}{2}(1)$$ From the cyclic quadrilateral AFDB we have: $${\hat{F}}_1={\hat{A}}_1(2)$$ 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 $$DK\perp BC(3)$$ 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: $$DM\perp AC(4)$$ From the inscribed quadrilateral $BKDL$ we have $B\hat{K}D=B\hat{L}D=90^{\circ }$, that is: $$DL\perp AB(5)$$ 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 $BF//LM$.

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Alternative solution.

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: $$KM//BF(6)$$ 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 $$KL//BF(7)$$ From relations (6) and (7) we get that $K,L,M$ are collinear and $BF//LM$.