Let $AI \cap \circ ABC =H, C_1H \cap AB=F, B_1H \cap AC=G$Tahmid Hasan wrote:Problem $8$:Let $ABC$ be a triangle such that $AB \neq AC$. The
internal bisector lines of the angles $ABC$ and $ACB$ meet the opposite
sides of the triangle at points $B_0$ and $C_0$, respectively, and the
circumcircle $ABC$ at points $B_1$ and $C_1$, respectively. Further, let
$I$ be the incentre of the triangle $ABC$. Prove that the lines
$B_0C_0$ and $B_1C_1$ meet at some point lying on the parallel through
$I$ to the line $BC$.
Now $F,G,I$ are colliner using pascal's theorem .
$AIGB_1$ is cyclic.
So $\angle B_1IG=\angle CAB_1=\angle B_1BC$.
that imply $FG || BC$ .
As $A.I,H$ are colliner ,we can say $B_1C_1,B_0C_0,FG$ concur using
(wow !! i used Desargues for the first time .. yey)