Team Selection Test for IMO 2019

Let $R$ and $S$ be the reflections of $C$ in the lines $BK$ and $BN$, respectively. It is easy to see the following: $C,N,R$ are collinear, $A,B,R$ are collinear and $C,M,S$ are collinear.



By symmetry, $\angle BRK=\angle BSK=\angle KCB$. But, $\angle KCB=\angle BNK$, hence $B,K,N,R,S$ are concyclic. Let $X$ be an arbitrary point on the line which passes through $N$ and which is parallel to $DM$. Then, $\angle NBK=\angle MDN=\angle XNK$, hence $XN$ is tangent to the circle $BKNRS$. By Pascal's theorem, the following three points are collinear: $A=NK\cap BR$, $C=SK\cap NR$ and $XN\cap BS$. Thus, the three lines $XN,BS,AC$ are concurrent. Hence, $BS$ passes through $T=XN\cap AC$. Then, $\angle TBC$ is the same angle as $\angle SBC$ whose bisector is clearly $BM$.