61-A property of incircle X(1)

Problem: Let $ABC$ be a triangle, $A_1B_1C_1$ be cevian triangle of the point $D$ (on the plane). I is the in center of the triangle ABC. Reflection of BC on IA1 meets $B_1C_1$ at $A_2$. Define $B_2,C_2$ cycllicaly. Show that $A_2,B_2,C_2$ are collinear.

                             

Solution by Telv Cohl https://www.facebook.com/telv.cohl

Let Incircle of triangle $ABC$ touch $BC, CA, AB$ at $G, E, F$. $L_g$ is the line through $G$ which is perpendicular to $IA_1$.  $L_e$ is the line through $E$ which is perpendicular to $IB_1$. $L_f$ is the line through $F$ which is perpendicular to $IC_1$. $L_g, L_e, L_f$ form triangle $A_3B_3C_3$
Let $G'$ be the reflection point of $G$ wrt $IA_1$, similarly define $E'$ and $F'$ Let $A_4$ be the intersection of $B_1C_1$ and $BC$, similarly define $B_4$ and $C_4$. ($A_4, B_4, C_4$ not appear in my picture)

Since $EE', FF'$ are the polar of $(B_1)$, $(C_1)$ wrt Incircle, So $B_1C_1$ is the polar of the intersection of $EE'$ and $FF'$ ie. $A_3$ similarly discussion we get $C_1A_1, A_1B_1$ is the polar of $(B_3), (C_3)$.

Since $AA_1, BB_1, CC_1$ concur at $D$ so triangle ABC and triangle A1B1C1 are perspective. By Desargue theorem we deduce that$A_4, B_4 C_4$ are collinear ... (1). Since $BC, CA, AB$ is the polar of $G, E, F$ wrt Incircle respectively. By Pole and Polar theorem and (1) we get $A_3G, B_3E, C_3F$ are concur. By Terquem theorem for triangle $A_3B_3C_3$ and Incircle we get $A_3G', B_3E', C_3F'$ are concur ... (2). Since $A_2, B_2, C_2$ is the pole of $A_3G', B_3E', C_3F'$ respectively. By Pole and Polar theorem and (2) we get $A_2, B_2,C_2$ are collinear. Q.E.D