Both angle conditions are equivalent to $ABCD$ being harmonic quadrilateral.

Pretty much @above, but with a bit more detail. Interesect $AC$ and $BD$ at $E$. Let $OM \cap BD = T$. Note that since $OM \perp AC \implies TM \perp ME$ and since $\angle AMB = \angle AMD$ this tells us that\[(TE; BD) = -1\]which then implies that $ABCD$ is harmonic. We can reverse this on $N$ to conclude that $\angle ANB = \angle BNC$ and we are done.

Above wait, why must $CT$ or $AT$ strictly a tangent? If it is clear, please add it

I just think the best solution is with harmonic quadrilateral