Really easy problem Redefine $C_0=C_1A_1 \cap AB$ and $A_0=C_1A_1 \cap BC$. $$\measuredangle{IAC_0}=\measuredangle{A_1AC}=\measuredangle{A_1C_1C}=\measuredangle{C_0C_1I}$$hence $C_1,C_0,I,A$ cyclic. $$\measuredangle{ACC_1}=\measuredangle{AA_1C_1}=\measuredangle{A_0A_1I}$$ hence $A_1,A_0,I,C$ cyclic.

Let $A_1C_1$ meet $AB$ at $C_0'$. Define $A_0'$ similarly. Then we have $$\measuredangle IC_1C_0' = \measuredangle CC_1A_1 = \measuredangle CAA_1 = \measuredangle A_1AB = \measuredangle IAC_0'$$$$\implies A, I, C_0', C_1 \text{are cyclic}$$$$\implies C_0' = C_0$$Similarly, $A_0' = A_0$. Therefore, $A_0, C_0$ lie on $A_1C_1$ and we are done. $\square$

Claim: $C_0$ lies on $C_1A_1$ $$\angle C_0C_1I=\angle C_0AI=\angle CAI=\angle IC_1A_1$$Claim: $A_0$ lies on $A_1C_1$ $$\angle A_0A_1I=\angle A_0CI=\angle ACI=\angle IA_1C_1$$

An elemantary solution for an easy problem: Let $\angle ACB=2\theta$ and $\angle BAC=2\alpha$. By simple angle chasing, one can get $\angle A_1BC=\angle A_1CB=\alpha$ and $\angle C_1AB=\angle C_1BA=\theta$. Since $(AC_1I)\cap AB=C_0$ and $\angle (A_1CI)\cap BC=A_0$, $AC_1C_0I$ and $ICA_1A_0$ cyclic and this implies $\angle IA_1A_0=\theta=\angle A_0A_1B$ and $\angle IC_1C_0=\alpha=\angle C_0C_1B$. Thus $C_1C_0$ and $A_1A_0$ bisects $\angle BC_1C$ and $\angle BA_1C$ respectively. On the other hand, by Incenter Lemma, $BC_1=IC_1$ and $BA_1=IA_1$ which implies $\boxed{C_1BA_1I \quad \text{is a kite}}$. Because of both $C_1C_0$ and $A_1A_0$ lies on kite $C_1B_1I$ 's $\angle BC_1C$ bisector, we have $A_0,A_1,C_0,C_1$ collinear, as desired.

We have $\angle{BA_0I} = 180^{\circ} - \angle{IA_0C} = 180^{\circ} - \angle{IA_1C} = 180^{\circ} - \angle{ABC}$. Then $\triangle IA_0B$ is $A_0$ - isosceles. Note that $A_1C_1$ is perpendicular bisector of $IB,$ we have $A_0 \in A_1C_1$. Similarly, we have $C_0 \in A_1C_1$

very cute

easy just show that $C_1$ $ C_0$ and $A_1$ are allcollinear just by strict angle chasing on both circles $AC_1 C_0 I$ and $(ABC)$ and thus by symetry the result follows