Romanian Master of Mathematics

The line $C_1{A}_1$ is parallel to the external angle bisector of $\angle B$, so the reflection in $C_1{A}_1$ maps the segment $A_{1}C$ to the segment $A_{1}E$ parallel to $AB$.

Similarly, $B_{1}F\parallel AB$. Notice also that $A_{1}E=A_{1}C=B_{1}C=B_{1}F=r$, where $r$ is the inradius of $△ABC$. Let $M$ be the midpoint of $AB$. Let $X$ be the point of $\omega$ such that $\overrightarrow{IX}\nearrow \overrightarrow{CM}$. Notice that $\angle E{A}_{1}I=90^{\circ }+\angle E{A}_{1}B=90^{\circ }+\angle CBM=90^{\circ }+\angle BCM=\angle A_{1}IX$, and $A_{1}E=I{A}_1=IX$; thus, $XI{A}_{1}E$ is an isosceles trapezoid. Hence $X$ lies on the circle $I{A}_{1}E$, and $EX\parallel A_{1}I$. Similarly, $X$ lies on the circle $I{B}_{1}F$, and $FX\parallel B_{1}I$. It remains to show that $X$ lies on the circle $C_{1}KL$. Under the symmetry in $C_1{A}_1$, the line $CH$ (perpendicular to $AB$) maps to the line through $E$ perpendicular to $BC$ — i.e., $CH$ maps to $EX$. Therefore, the projection $H$ of $C_1$ onto $CH$ maps to the projection $K$ of $C_1$ onto $EX$. Similarly, $L$ is the projection of $C_1$ onto $FX$. So the quadrilateral $C_{1}KXL$ is cyclic, due to right angles at $K$ and $L$.