NMO Selection Tests for the Balkan and International Mathematical Olympiads

Let the internal bisector of the angle $BAC$ meet again the circumcircle $ABC$ at point $A_1$. The lines $A_1{B}_1$ and $AC$ meet at point $B_2$, and the lines $AB$ and $A_1{C}_1$ meet at point $C_2$. Apply Pascal's theorem to the hexagon $A{C}_{1}B{A}_{1}C{B}_1$ to deduce that the points $B_2$, $I$ and $C_2$ are collinear; moreover, Pascal's line $B_{2}I{C}_2$ is precisely the parallel through $I$ to $BC$, for $A_1{B}_2$ and $A_1{C}_2$ are the internal bisector lines of the angles $A{A}_{1}C$ and $A{A}_{1}B$, respectively, and the segments $A_{1}B$ and $A_{1}C$ are congruent. Finally, notice that the lines $B_i{B}_{i+1}$ and $C_i{C}_{i+1}$ meet at collinear points; $B_0{B}_1$ and $C_0{C}_1$ meet at $I$, $B_1{B}_2$ and $C_1{C}_2$ meet at $A_1$, and $B_2{B}_0$ and $C_2{C}_0$ meet at $A$. Consequently, the triangles $B_0{B}_1{B}_2$ and $C_0{C}_1{C}_2$ are perspective; the lines $B_i{C}_i$ are concurrent (the converse of Desargues' theorem).