SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Consider an inversion with center $B$ and radius $r>0$. The points $A^{\prime }$, $B^{\prime }$, $C^{\prime }$ are collinear and the points $K^{\prime }$, $N^{\prime }$, $M^{\prime }$ are, too. The circumcircle of quadrilateral $ACNK$ becomes the circumcircle $\omega$ of quadrilateral $A^{\prime }{C}^{\prime }{N}^{\prime }{K}^{\prime }$. In case $B$ is outside $\omega$, let $X$, $Y$ be the tangent points of 2 tangents from $B$ to $\omega$, then $O^{\prime }$ is the midpoint of $XY$. Since $M^{\prime }$, $X$, $Y$ lies on the polar of $B$ relative to $\omega$ then $\angle M^{\prime }{O}^{\prime }B=90^{\circ }$, this implies that $\angle OMB=90^{\circ }$. In the other cases, we also have $O^{\prime }$ lies on the polar of $B$ relative to $\omega$, and the argument is similar.

Remark. This problem can be solve by using the spiral similarity of center $M$ or by using the Brocard's theorem.