Croatia_2018

Denote by $P$ and $Q$ the midpoints of $\overline{BC}$ and $\overline{AD}$, respectively. We will show that the quadrilateral $ABNM$ is cyclic. From $\angle MQN=\angle NPM=90^{\circ }$, we get that the quadrilateral $MNPQ$ is cyclic. This implies $\angle PQM+\angle MNP=180^{\circ }$. The segment $\overline{QP}$ is the midsegment of the trapezium $ABCD$, which means that it is parallel to $AB$. From here we conclude that $\angle MAB=\angle DQP$. Now we have $$\angle MAB=\angle DQP=180^{\circ }-\angle PQM=\angle MNP=180^{\circ }-\angle BNM,$$ which is enough to conclude that the quadrilateral $ABNM$ is cyclic. Analogously, we show that the quadrilateral $MNCD$ is cyclic. This shows that the segment $\overline{MN}$ is simultaneously a chord for both circumscribed circles of triangles $ABN$ and $CDM$. We conclude that the line $O_1{O}_2$, connecting the centres of these circles, must bisect the segment $\overline{MN}$.