Notice that it suffices to show $\angle MAD=\angle NAD$.Let $A''$ be the reflection of $A$ across $M$, then $AXA''D$ is a parallelogram, also see that $OD\|AX$ from here it follows that $O, D A''$ collinear and $\triangle OXD \sim \triangle ODN$, this is $\frac{OD}{ON}=\frac{OA}{ON}=\frac{AX}{AD}=\frac{DA''}{DA}$ but also $\angle DON=\angle ODA=\angle OAD=\angle AOF$ where $F=ON\cap (O)$(closer to $A$), then $\triangle OAN \sim \triangle DA''A$ then $\angle DAA'' =\angle ONA=\angle NAD$ and the conclusion follows