Note that $IA_{1} = IB_{1} = IC_{1} = 2r$, where $r$ is the in-radius of the triangle $ABC$. Hence $I$ is the circum-centre of the triangle $A_{1}B_{1}C_{1}$.

Let $K$ be the point of intersection of $IB_{1}$ and $AC$. Then $IK = r$, $IA = 2r$ and $\angle IKA = 90^{\circ}$. It follows that $\angle IAK = 30^{\circ}$ and hence $\angle IAB_{1} = 60^{\circ}$. Thus $AIB_{1}$ is an equilateral triangle. Similarly triangle $AIC_{1}$ is also equilateral. We hence obtain $AB_{1} = AC_{1} = AI = IB_{1} = IC_{1} = 2r$.

We also observe that $\angle B_{1}IC_{1} = 120^{\circ}$ and $IB_{1}AC_{1}$ is a rhombus. Thus $\angle B_{1}AC_{1} = 120^{\circ}$ and by concyclicity $\angle A_{1} = 60^{\circ}$. Since $AB_{1} = AC_{1}$, $A$ is the midpoint of the arc $B_{1}AC_{1}$. It follows that $A_{1}A$ bisects $\angle A_{1}$ and $I_{1}$ lies on the line $A_{1}A$. This implies that
$$\angle B_{1}I_{1}C_{1} = 90^{\circ} + \angle A_{1}/2 = 90^{\circ} + 30^{\circ} = 120^{\circ}$$
Since $\angle B_{1}IC_{1} = 120^{\circ}$, we conclude that $B_{1}, I, I_{1}, C_{1}$ are concyclic. (Further $A$ is the centre.)