Silk Road Mathematics Competition

W.l.o.g. assume that $AC\le AB$. Then $A_1$ belongs to the segment $AB$, moreover, it lies between $A$ and $C_0$. Then notice that $AC+A{A}_1=B{A}_1⟺AC+A{C}_0-A_1{C}_0=C_{0}B+A_1{C}_0⟺A_1{C}_0=\frac{1}{2}AC\Rightarrow A_1{C}_0=A_0{C}_0$ i.e. $\angle C_0{A}_1{A}_0=\angle A_1{A}_0{C}_0\Rightarrow \angle A_1{A}_0{C}_0=\angle A_1{A}_0{B}_0$ since $AB\parallel A_0{B}_0$. So, $A_0{A}_1$ is bisector of the angle $\angle C_0{A}_0{B}_0$.

Similarly, $B_0{B}_1$ is bisector of $\angle C_0{B}_0{A}_0$ and $C_0{C}_1$ is bisector of $\angle A_0{C}_0{B}_0$. So the lines $A_0{A}_1,B_0{B}_1$ and $C_0{C}_1$ intersect in the incenter of the triangle $A_0{B}_0{C}_0$.