Selection Examination A

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.