CMO 1998 Problem 4

Let $ABC$ be a triangle with $\angle{BAC} = 40^{\circ}$ and $\angle{ABC}=60^{\circ}$ . Let $D$ and $E$ be the points lying on the sides $AC$ and $AB$ , respectively, such that $\angle{CBD} = 40^{\circ}$ and $\angle{BCE} = 70^{\circ}$ . Let $F$ be the point of intersection of the lines $BD$ and $CE$ . Show that the line $AF$ is perpendicular to the line $BC$ .