Dear Mathlinkers, First step of this cleaver problem : A0A1 is parallel to the A-bissector of ABC. Sincerely Jean-Louis

A solution, please.

Well after jayme posted a useful hint I have solved the problem.WLOG we can assume that $AC>AB$.So $A_1$ lies on $AC$ and $AA_1=\frac {b-c}{2}$ and $A_1C=\frac {b+c}{2}$.Let the angle bisector of $A$ cuts the line $BC$ at $D$.Now $CD$ can be easily found and $A_0C=\frac {a}{2}$.So it's easy to show that $\frac {CA_1}{AA_1}=\frac {CA_0}{A_0D}$. So $A_0A_1||AD$.Now $A_0B_0C_0$ is the medial triangle of $\triangle ABC$ and the cevian $A_0A_1$ is also the angle bisector of $\angle B_0A_0C_0$. Hence the result follows.

External bisectors of $\angle A, \angle B, \angle C$ cut circumcircle $(O)$ of $\triangle ABC$ again at $X, Y, Z.$ $A_0$ is feet of perpendicular from $X$ to $BC.$ Let $A_2, A_3$ be feet of perpendiculars from $X$ to $AB, AC$ $\Longrightarrow$ $A_0A_2A_3$ is Simson line with pole $X$. $XA_2AA_3$ is kite $\Longrightarrow$ $A_0A_2A_3 \perp AX$. By Archimedes theorem, $A_1 \equiv A_2$ for $AB > AC$ and $A_1 \equiv A_3$ for $AB < AC$ $\Longrightarrow$ $A_0A_1 \perp AX$ is internal bisector of $\angle B_0A_0C_0$. Likewise, $B_0B_1 \perp BY, C_0C_1\perp CZ$ are internal bisectors of $\angle C_0B_0A_0, \angle A_0C_0B_0,$ concurrent at incenter of $\triangle A_0B_0C_0$.