$\angle DAC = \frac{A}{2} + \frac{B}{2}$ and $\angle DCA = \frac{C}{2} + \frac{B}{2}$ which give as $\angle ADC = 90 - \frac{B}{2}$ As $\angle AIC = 90 + \frac{B}{2}$ we get $A,D,C,I$ cyclic and hence $\angle ADI = \angle ACI = \frac{C}{2}$ Now from $\angle DAC = \frac{A}{2} + \frac{B}{2}$ we get $AC \perp DI$

We claim that $A,C,D,I$ are concyclic. We have $\angle DAI=\angle INC$ and $\angle DCI=\angle INA$. Thus, $\angle DAI + \angle DCI=\angle INC + \angle INA=180$. Now, we have $\angle CID=\angle CAD=\angle NAD=\angle AIN$ implying $IN$ and $ID$ are isogonal w.r.t $AIC$. On the other hand, $IN$ passes through the circumcenter of $AIC$. This means that $ID\perp AC$.