Problem

Source: Moldova TST 2018

Tags: geometry



Let the triangle $ABC $ with $m (\angle ABC)=60^{\circ} $ and $m (\angle BAC)=40^{\circ}$ . Points $D $ and $E $ are on the sides $(AB) $ and $(AC) $ such that $m (\angle DCB )=70^{\circ}$ and $m (\angle EBC)=40^{\circ}$ . $BE$ and $CD$ intersect in $F $ . Prove that $BC $ and $AF $ are perpendicular.