Let $ABC$ be an isosceles triangle with $AB=AC$. Points $D$ and $E$ lie on the sides $AB$ and $AC$, respectively. The line passing through $B$ and parallel to $AC$ meets the line $DE$ at $F$. The line passing through $C$ and parallel to $AB$ meets the line $DE$ at $G$. Prove that \[\frac{[DBCG]}{[FBCE]}=\frac{AD}{DE} \]