In $\triangle{ABC}$, $\angle{BAC} = 40^\circ$ and $\angle{ABC} = 60^\circ$. Points $D$ and $E$ are on sides $AC$ and $AB$, respectively, such that $\angle{DBC}=40^\circ$ and $\angle{ECB}=70^\circ$. Let $F$ be the intersection point of $BD$ and $CE$. Show that $AF\perp BC$.