Question 182
In the convex quadrilateral $ABCD$ the angles $ABC$ and $BCD$ are equal to $120^\circ$, $O$ is the intersection of the diagonals, $M$ is the midpoint of $BC$, and $K$ is the point of intersection of $MO$ and $AD$. It happens that $\angle BKC = 60^\circ$. Prove that $\angle BKA = \angle CKD = 60^\circ$.
(Serduk Nazar)
(Serduk Nazar)