6. Given that $D$ is the intersection of the tangents to the circumcircle of $\triangle A B C$ at $A$ and $B$, the circumcircle of $\triangle A B D$ intersects the line $A C$ and the segment $B C$ at another point $E$ and $F$ respectively, and $C D$ intersects $B E$ at point $G$. If $\frac{B C}{B F}=2$, find $\frac{B G}{G E}$.
(2010-2011, Hungarian Mathematical Olympiad)
Using Property 1, let the circumcenter of $\triangle ABC$ be $O$. Then $O, A, D, B$ are concyclic. If point $O$ is not on side $BC$, then $FO \perp BC \Rightarrow \angle OAB=90^{\circ}$, which is a contradiction. Therefore, point $O$ is on $BC$.
Let $CD$ intersect $AB$ at point $M$. Then
$$
\begin{array}{l}
\frac{AM}{MB}=\frac{S_{\triangle ACD}}{S_{\triangle BCD}}=\frac{AC \cdot AD \sin \angle CAD}{BC \cdot BD \sin \angle CBD} \\
=\frac{AC}{BC} \cos \angle BCA=\left(\frac{AC}{BC}\right)^{2} .
\end{array}
$$

Applying Menelaus' Theorem to $\triangle BAE$ and the transversal $GMC$ yields
$$
\frac{BG}{GE}=\frac{CA}{CE} \cdot \frac{BM}{MA}=\frac{CA}{CE} \cdot \frac{BC^2}{AC^2}=2 \cdot \frac{CB}{CA} \cdot \frac{CO}{CE}=2 .
$$