If we orient positively the line $AB$ from $A$ to $B$, the line $BC$ from $B$ to $C$ and the line $CA$ from $C$ to $A$, we get
$$\overline{B_{1}A} - \overline{AC_{1}} = \overline{A_{1}C} - \overline{CB_{1}} = \overline{C_{1}B} - \overline{BA_{1}}.$$
Let $A_{0}, B_{0}, C_{0}$ be the intouch points of the incircle of triangle $ABC$ and the sides $BC$, $CA$, $AB$, respectively. We have
$$\overline{B_{1}A} - \overline{AC_{1}} = \overline{B_{1}B_{0}} + \overline{B_{0}A} - \overline{AC_{0}} - \overline{C_{0}C_{1}} = \overline{B_{1}B_{0}} + \overline{C_{1}C_{0}},$$
and similarly $\overline{A_{1}C} - \overline{CB_{1}} = \overline{A_{1}A_{0}} + \overline{B_{1}B_{0}}$, and $\overline{C_{1}B} - \overline{BA_{1}} = \overline{C_{1}C_{0}} + \overline{A_{1}A_{0}}$. We deduce that
$$\overline{B_{1}B_{0}} + \overline{C_{1}C_{0}} = \overline{A_{1}A_{0}} + \overline{B_{1}B_{0}} = \overline{C_{1}C_{0}} + \overline{A_{1}A_{0}},$$
which is equivalent to
$$\overline{A_{0}A_{1}} = \overline{B_{0}B_{1}} = \overline{C_{0}C_{1}}.$$
Because $IA_{0} = IB_{0} = IC_{0}$ and $\angle A_{1}A_{0}I = \angle B_{1}B_{0}I = \angle C_{1}C_{0}I$, the triangles $IA_{0}A_{1}$, $IB_{0}B_{1}$ and $IC_{0}C_{1}$ are congruent and $I$ is the circumcenter of triangle $A_{1}B_{1}C_{1}$.

Because $\angle AB_{0}I = \angle BC_{0}I = 90^{\circ}$, quadrilateral $AC_{0}IB_{0}$ is cyclic. Because $\angle B_{0}IB_{1} = \angle C_{0}IC_{1}$, we have $\angle B_{1}IC_{1} = \angle B_{0}IC_{0} = 180^{\circ} - \angle B_{1}AC_{1}$. We deduce that quadrilateral $AC_{1}IB_{1}$ is cyclic.
Because $I$ is the midpoint of the minor $\operatorname{arc} \widehat{C_{1}B_{1}}$ of the circumcircle of triangle $AC_{1}B_{1}$ not containing $A$, we deduce that $IB_{1} = IC_{1} = II_{1}$. Therefore $I_{A}$ is on the circumcircle of triangle $A_{1}B_{1}C_{1}$ of center $I$.
Similarly, one can prove that $I_{B}, I_{C}$ are on the circumcircle of triangle $A_{1}B_{1}C_{1}$. Hence the circumcenter of triangle $I_{A}I_{B}I_{C}$ is the incenter of triangle $ABC$.