see here

By Pascal's Theorem on $AACC_1B_1B$, $AA, B_1C_1, B_0C_0$ are collinear. Let these three lines concur at $X$. $AC_1=IC_1, AB_1=IB_1, C_1B_1=C_1B_1 \implies AB_1C_1, IB_1C_1$ congruent by SSS. Thus $\angle AC_1X=\angle IC_1X$. Also $XC_1$ common and $AC_1=IC_1$ so $AC_1X, IC_1X$ are congruent by SAS. Hence $\angle C_1IX=\angle C_1AX=\angle ACC_1=\angle C_1CB \implies XI$ is parallel to $BC$. So we are done.

WLOG define $A_0$ and $A_1$ and let the parallel line through $I$ intersect $AB$ and $AC$ in $C_2$ respectively $B_2$.Now it is easily to observe that $\angle AB_1I=\angle ACB=\angle AB_2I$.Thus,$B_1,B_2$ and $A_1$ are collinear.Now,by Desargues for triangles $C_1C_0C_2$ and $B_1B_0B_2$ we get that lines $ B_2C_2,B_0C_0$ and $B_1C_1$ are concurent. Q.E.D

Dear Mathlinkers, see also http://www.artofproblemsolving.com/Forum/viewtopic.php?t=293685 Sincerely Jean-Louis

Here's a solution with projective geometry/inversion: We begin with a lemma. Lemma. Let $\triangle ABC$ be a triangle with orthocenter $H$, circumcenter $O$, nine-point center $N$, and nine-point circle $\gamma.$ Let $M_a, H_a, T_a, X_a$ be the midpoint of $\overline{BC}$, the projection of $H$ onto $BC$, the midpoint of $\overline{AH}$, the second intersection of $OM_a$ with $\gamma$, respectively. Define $M_b, M_c, H_b, H_c, T_b, T_c, X_b, X_c$ similarly, and denote $R \equiv T_bX_b \cap T_cX_c.$ Let $A'$ be the reflection of $A$ in $M_a.$ Then $RA' \parallel OA.$ Proof of Lemma. Note that $M_aT_a$ and $H_aX_a$ both pass through $N$, since they are diameters of $\gamma.$ Then since $R \equiv T_bX_b \cap T_cX_c$, a homothety with center $N$ and ratio $-1$ sends $R$ to $M_bH_b \cap M_cH_c \equiv A.$ Hence, $R$ is just the reflection of $A$ in $N.$ Now, note that $T_aHM_aO$ is a parallelogram because $N$ is the midpoint of both of its diagonals. Then if $R'$ is the reflection of $O$ in $BC$, it follows from $T_aH = M_aO$ that $AH = R'O.$ Furthermore, $AH \parallel R'O$, which implies that $AHR'O$ is a parallelogram. Therefore, $N$ is the midpoint of $\overline{AR'}$, and hence $R' \equiv R.$ It follows that a reflection in $M_a$ sends $R \mapsto O$ and $A' \mapsto A.$ Therefore, $RA' \parallel OA$ as desired. $\blacksquare$ Back to the main proof, let $D, E, F$ be the points of tangency of the incircle $\omega$ with sides $\overline{BC}, \overline{CA}, \overline{AB}$, respectively. Let $A', B', C'$ be the midpoints of $\overline{EF}, \overline{FD}, \overline{DE}$, respectively. It is sufficient to show that the polar of $B_0C_0 \cap B_1C_1$ (i.e. the line passing through the poles of $B_0C_0$ and $B_1C_1$) is parallel to $ID.$ Note that $F$ lies on the polar of $C_0$ because $C_0F$ is tangent to $\omega.$ Furthermore, the polar of $C_0$ is parallel to $DE$ since $IC_0 \perp DE.$ The polar of $B_0$ is analyzed similarly. It follows that if $Z$ is the intersection of the polars of $B_0, C_0$ (i.e. the pole of $B_0C_01$), then $DEZF$ is a parallelogram. Meanwhile, the inversion in the incircle $\omega$ sends $A \mapsto A', B \mapsto B', C \mapsto C'.$ Hence the image of $\odot (ABC)$ is just the nine-point circle of $\triangle DEF.$ The image of $C_1$ is then the second intersection of $C'I$ with the nine-point circle. Keeping in mind that the polar of $C_1$ is perpendicular to $IC'$, we see that it is analagous to the line $T_cX_c$ from the lemma. A similar statement holds for the polar of $B_1$, and it follows that the pole of $B_1C_1$ (i.e. the intersection of the polars of $B_1$ and $C_1$) is analagous to the point $R.$ Employing the lemma, we're done. $\square$

Let $B_1C_1$ and $B_0C_0$ intersect at $P$. By Pascal on $AABB_1C_1C$, $P$ lies on the tangent to the circumcircle at $A$. Now we use barycentric coordinates on $\triangle ABC$. Remark that line $B_0C_0$ has equation $bcx=acy+abz$ and the tangent to $A$ is given by $\frac{y}{z}=\frac{-b^2}{c^2}$. Intersecting these two lines gives $P=(ac-ab:-b^2:c^2)$. Normalizing, we find that $P$ and $I$ lie on the line $(a+b+c)x=a$, which is clearly parallel to $BC$, so we're done.

The feet of angle bisectors are $A_{0},B_{0},C_{0}$ $A_{1}$ is the midpoint if the arc$BC$ $A_{1}B_{1}$ intersect $AC$ in $B_{2}$ $C_{1}A_{1}$ intersect $AB$ in $C_{2}$ By Pascals on $BC_{1}AB_{1}CA_{1}$ $I,B_{2},C_{2}$ are collinear. Now assume for a moment that $C_{1}B_{1}$,$C_{0}B_{0}$,$C_{2}B_{2}$ are concurrent.It easily follow that $B_{1}A_{1}$ is the perpendicular bisector of $IC$ so $IB_{2}=B_{2}C$ so $IB_{2}||AB$ so we are done. Now we prove the above claim Claim: $C_{1}B_{1}$,$C_{0}B_{0}$,$C_{2}B_{2}$ are concurrent. Proof: We have by some trig manipulations the following $$\frac{A_{1}B_{2}}{B_{1}B_{2}}=\frac{\sin(\gamma+\frac{\alpha}{2})\sin\frac{\alpha}{2}}{\sin^2 \frac{\beta}{2}}$$$$\frac{A_{1}C_{2}}{C_{1}C_{2}}=\frac{\sin(\beta+\frac{\alpha}{2})\sin\frac{\alpha}{2}}{\sin^2\frac{\gamma}{2}}$$$$\frac{IB_{0}}{B_{0}B_{1}}=\frac{\sin\alpha\sin\frac{\gamma}{2}}{\sin\frac{\beta}{2}\cos\frac{\alpha}{2}}$$$$\frac{IC_{0}}{C_{1}C_{0}}=\frac{\sin\alpha\sin\frac{\gamma}{2}}{\cos\frac{\alpha}{2}\sin\frac{\beta}{2}}$$substituting this into the Menaluas and using $(\gamma+\frac{\alpha}{2})+(\beta+\frac{\alpha}{2})=\pi$ everyhting vanishes and we are done.$\square$