Given a triangle $ABC $, $A {{A} _ {1}} $, $B {{B} _ {1}} $, $C {{C} _ {1}}$ - its chevians intersecting at one point. ${{A} _ {0}}, {{C} _ {0}} $ - the midpoint of the sides $BC $ and $AB$ respectively. Lines ${{B} _ {1}} {{C} _ {1}} $, ${{B} _ {1}} {{A} _ {1}} $and ${ {B} _ {1}} B$ intersect the line ${{A} _ {0}} {{C} _ {0}} $ at points ${{C} _ {2}} $ , ${{A} _ {2}} $ and ${{B} _ {2}} $, respectively. Prove that the point ${{B} _ {2}} $ is the midpoint of the segment ${{A} _ {2}} {{C} _ {2}} $. (Eugene Bilokopitov)