Example 1 As shown in Figure 1, given that $D$ is a point inside $\triangle A B C$, line $B D$ intersects side $A C$ and line $C D$ intersects side $A B$ at points $E$ and $F$ respectively, satisfying $A F=F B=C D, C E=D E$. Find $\angle B F C$. ${ }^{[1]}$
(2003, Japan Mathematical Olympiad (Final))
As shown in Figure 1, let $G$ be a point on segment $CF$ such that $FG = CD$. Connect $AG$.
From $CE = DE \Rightarrow \angle ACD = \angle EDC = \angle FDB$.
Also, $FG = CD$, then
$$
GC = CF - FG = CF - CD = DF \text{. }
$$

Applying Menelaus' Theorem to $\triangle ABE$ and the transversal $CF$, we get
$$
\frac{BF}{FA} \cdot \frac{AC}{CE} \cdot \frac{ED}{DB} = 1 \text{. }
$$

Given $AF = FB$ and $CE = DE \Rightarrow AC = DB$.
Thus, $\triangle ACG \cong \triangle BDF \Rightarrow AG = BF$.
Therefore, $\triangle AFG$ is an equilateral triangle
$$
\Rightarrow \angle AFG = 60^{\circ} \Rightarrow \angle BFC = 120^{\circ} \text{. }
$$