Let $k$ be the incircle of the triangle $ABC$ with the center of the incircle $I$. The circle $k$ touches the sides $BC, CA$ and $AB$ in points $D, E$ and $F$. Let $G$ be the intersection of the straight line $AI$ and the circle $k$, which lies between $A$ and $I$. Assume $BE$ and $FG$ are parallel. Show that $BD = EF$.