Example 3 As shown in Figure $2, D$ is a point inside $\triangle A B C, B D$ intersects $A C$ and $C D$ intersects $A B$ at points $E, F$ respectively, and $A F=B F$ $=C D, C E=D E$. Find the degree measure of $\angle B F C$. ${ }^{[6]}$ (2003, Japan Mathematical Olympiad)
As shown in Figure 2, let $G$ be a point on segment $FC$ such that $FG = AF$, and connect $AG$.
Since $CE = DE$, we have
$\angle ECD = \angle EDC = \angle FDB$.
Also, $FG = AF = CD$, so
$$
GC = CD + DG = FG + GD = DF \text{. }
$$

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

Given $AF = FB$ and $CE = DE$, we know $AC = DB$.
Thus, $\triangle ACG \cong \triangle BDF \Rightarrow AG = BF$.
Therefore, $\triangle AFG$ is an equilateral triangle, $\angle AFG = 60^{\circ}$.
Hence, $\angle BFC = 120^{\circ}$.
3 Application of the Inverse Theorem
The inverse of Menelaus' Theorem is an important method for proving that three points are collinear, converting geometric proofs into proportional equations, which is a more concise approach.