First solution: Note $M$ lie on perpendicular bisector of the segment $AD$(so $AM = MD$), on angle bisector and the quadrilateral $AMDB$ is not symmetrical with respect to the diagonal $BM$(This is impossible because the triangle then degenerates because $\angle BID = 90^{\circ}$) then well known that $A, M, D, B$ are concircle(fake deltoid I call it). Similary, $A, N, D, C$ are concircle. Then $\angle ANI + \angle AMI = \angle ADC + \angle ADB = 180^{\circ}$ $\square$

Second solution: Let tangle to $(ABC)$ at point $A$ intersect $BC$ at point $K$. Well known that(easy angles): $AK = KD \Rightarrow K$ lie on the perpendicular bisector $AD \Rightarrow MN$ - bisector $\angle AKD \Rightarrow AM$ - bicetor internal $\angle BAK$ и $AN$ - external bicetor $\angle CAK \Rightarrow \angle NMI = \angle MBK + \angle MKB = \frac{\angle B + \angle C - \angle B}{2} = \frac{\angle C}{2} = \angle NAC - \angle IAC = \angle NAI \Rightarrow A, M, N, I$ are concircle. $\square$