Consider $\vartriangle ABC$, with $AD$ bisecting $\angle BAC$, $D$ on segment $BC$. Let $E$ be a point on $BC$, such that $BD = EC$. Through $E$ we draw the line $\ell$ parallel to $AD$ and consider a point $ P$ on it and inside the $\vartriangle ABC$. Let $G$ be the point where line $BP$ cuts side $AC$ and let F be the point where line $CP$ to side $AB$. Show that $BF = CG$.