Let the internal bisector of the angle $BAC$ meet again the circumcircle $ABC$ at point $A_1$. The lines $A_1B_1$ and $AC$ meet at point $B_2$, and the lines $AB$ and $A_1C_1$ meet at point $C_2$. Apply Pascal's theorem to the hexagon $AC_1BA_1CB_1$ to deduce that the points $B_2$, $I$ and $C_2$ are collinear; moreover, Pascal's line $B_2IC_2$ is precisely the parallel through $I$ to $BC$, for $A_1B_2$ and $A_1C_2$ are the internal bisector lines of the angles $AA_1C$ and $AA_1B$, respectively, and the segments $A_1B$ and $A_1C$ are congruent. Finally, notice that the lines $B_iB_{i+1}$ and $C_iC_{i+1}$ meet at collinear points; $B_0B_1$ and $C_0C_1$ meet at $I$, $B_1B_2$ and $C_1C_2$ meet at $A_1$, and $B_2B_0$ and $C_2C_0$ meet at $A$. Consequently, the triangles $B_0B_1B_2$ and $C_0C_1C_2$ are perspective; the lines $B_iC_i$ are concurrent (the converse of Desargues' theorem).