Problem

Source:

Tags: concurrent, geometry, circles



Three circles are constructed for the triangle $ABC $: the circle ${{w} _ {A}} $ passes through the vertices $B $ and $C $ and intersects the sides $AB $ and $ AC $ at points ${{A} _ {1}} $ and ${{A} _ {2}} $ respectively, the circle ${{w} _ {B}} $ passes through the vertices $A $ and $C $ and intersects the sides $BA $ and $BC $ at the points ${{B} _ {1}} $ and ${{B} _ {2}} $, ${{w} _ {C}} $ passes through the vertices $A $ and $B $ and intersects the sides $CA $ and $CB $ at the points ${{C} _ {1}} $ and ${{C} _ {2}} $. Let ${{A} _ {1}} {{A} _ {2}} \cap {{B} _ {1}} {{B} _ {2}} = {C} '$, ${{A} _ {1}} {{A} _ {2}} \cap {{C} _ {1}} {{C} _ {2}} = {B} '$ ta ${ {B} _ {1}} {{B} _ {2}} \cap {{C} _ {1}} {{C} _ {2}} = {A} '$ is Prove that the perpendiculars, which are omitted from the points ${A} ', \, \, {B}', \, \, {C} '$ to the lines $BC $, $CA $ and $AB $ respectively intersect at one point. (Rudenko Alexander)