Incircle and intersecting lines

[ins-]

Let $A_0, B_0, C_0$ are the tangency points of the incircle of the triangle $ABC$ with the sides $BC, CA, AB$ respectively. $A', B', C'$ are the symmetric points of the $A, B, C$ with respect to $A_0, B_0, C_0$ respectively. $A'', B'', C''$ are feets of the perpendiculars from $A', B', C'$ to the sides $BC, CA, AB$ respectively. Prove that $A'A'', B'B'', C'C''$ intersects at a common point.

[Ilya Nek]

Hi!
For any point P let XYZ is triangle with feets of the perpendiculars from P to sides ABC(X - BC, Y - AC, Z - AB). If X' , Y' , Z' are the symmetric points of the A,B,C with respect to X,Y,Z, and X'', Y'' ,Z'' are feets of the perpendiculars from X' , Y' , Z' to the sides BC,AC,AB, then X'X'', Y'Y'', Z'Z'' intersects at a common point which is image of homotety with center P and k=-1 of H - orthocenter of ABC .
Ok?