5. As shown in Figure $9, D$ 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 $\angle B F C$.
Prompt: On $F C$, intercept $F G=A F$, and connect $A G$. Then $G C=D F, \angle E C D=\angle E D C=\angle F D B$.

For $\triangle A B E$ and the transversal $F D C$, applying Menelaus' theorem yields $A C=D B$.

Therefore, $\triangle A C G \cong \triangle B D F, A G=B F, \triangle A F G$ is an equilateral triangle, $\angle A F G=60^{\circ}$.
Thus, $\angle B F C=120^{\circ}$.