Let $\Gamma$ be the circle with centre $E$ passing through $B$ and $C$. Since $E D \perp A C$, the point $F$ symmetric to $C$ with respect to $D$ lies on $\Gamma$. From $\angle D C I=\angle I C B=\angle C B I$, the line $D C$ is a tangent to the circumcircle of triangle $I B C$. Let $J$ be the symmetric point of $I$ with respect to $D$. Using directed lengths, from
$$D C \cdot D F=-D C^{2}=-D I \cdot D B=D J \cdot D B$$
the point $J$ also lies on $\Gamma$. Let $I^{\prime}$ be the reflection of $I$ in $A C$. Since $I J$ and $C F$ bisect each other, $C J F I$ is a parallelogram. From $\angle F I^{\prime} C=\angle C I F=\angle F J C$, we find that $I^{\prime}$ lies on $\Gamma$. This gives $E I^{\prime}=E B$. Note that $A C$ is the internal angle bisector of $\angle B D I^{\prime}$. This shows $D E$ is the external angle bisector of $\angle B D I^{\prime}$ as $D E \perp A C$. Together with $E I^{\prime}=E B$, it is well-known that $E$ lies on the circumcircle of triangle $B D I^{\prime}$.