The line passing through the center of the equilateral triangle $ ABC $ intersects the lines $ AB $, $ BC $ and $ CA $ at the points $ {{C} _ {1}} $, $ {{A} _ {1}} $ and $ {{B} _ {1}} $, respectively. Let $ {{A} _ {2}} $ be a point that is symmetric $ {{A} _ {1}} $ with respect to the midpoint of $ BC $; the points $ {{B} _ {2}} $ and $ {{C} _ {2}} $ are defined similarly. Prove that the points $ {{A} _ {2}} $, $ {{B} _ {2}} $ and $ {{C} _ {2}} $ lie on the same line tangent to the inscribed circle of the triangle $ ABC $. (Serdyuk Nazar)