Let $X$, $Y$ be the feet from $B$ onto $\overline{IC}$, $\overline{IA}$ and let $BX$, $BY$ cut $AC$ at $D'$, $E'$. $\angle AIC = 90 + \tfrac{\angle B}{2} = 135^\circ \implies \angle XID = \angle YIE = 45^\circ$. Noticing that $E$ lies on the perpendicular bisector of $AD'$ and similarly for $D$ we get $\angle XED' = \angle IED' = 45^\circ = \angle IDE' = \angle YDE'$ and rest is easy by chasing to yield $IDD'EE'$ is cyclic.