2015 Danube Mathematical Competition

We show that $\angle OMN\equiv \angle ONM$. To this end, let the line $IJ$ cross the segments $AD$ and $BC$ at $P$ and $Q$, respectively, and consider the position of $J$ relative to the line $AC$, to write $\angle OMN\equiv \angle AJP\pm \angle JAM\equiv \angle AJP\pm (\angle JAD\mp \angle CAD)\equiv \angle AJP+\angle JAD-\angle CAD\equiv \angle AJP+\angle BAJ-\angle CAD$. Similarly, $\angle ONM\equiv \angle BIQ+\angle ABI-\angle CBD$. Since the quadrangle $ABCD$ is cyclic, $\angle CAD\equiv \angle CBD$ and $\angle ACB\equiv \angle ADB$. The latter implies that $\angle AIB\equiv \angle AJB$, so the quadrangle $ABIJ$ is cyclic. Consequently, $\angle AJP\equiv \angle ABI$ and $\angle BAJ\equiv \angle BIQ$, and the conclusion follows.