4. As shown in the figure, in $\triangle A B C$, $\angle A=60^{\circ}$, points $D, E$ are on side $A B$, and $F$, $G$ are on side $C A$. Connecting $B F, F E, E G, G D$ divides $\triangle A B C$ into five smaller triangles of equal area, i.e., $S_{\triangle C B F}=S_{\triangle F B E}=S_{\triangle F E C}=S_{\triangle G E D}=S_{\triangle G D A \text { V }}$ and $E B=2 C F$. Find the value of $\frac{B C}{E F}$.
4. $\frac{\sqrt{7}}{2}$.

It is known that $S_{\triangle F E A}=3 S_{\triangle F B E}$, so $A E=3 B E$.
Similarly, we get $A F=4 C F$.
And $B E=2 C$, so we can set $C F=x$, then $A F=4 x, A C=$ $5 x, A E=6 x, A B=8 x$.
In $\triangle F E A$, using the cosine rule we get
$$
E F^{2}=36 x^{2}+4^{2} x^{2}-2 \cdot 6 x \cdot 4 x \cdot \cos 60^{\circ}=28 x^{2} \text {. }
$$

In $\triangle A B C$, using the cosine rule we get
$$
\begin{aligned}
B C^{2} & =64 x^{2}+25 x^{2}-2 \cdot 8 x \cdot 5 x \cdot \cos 60^{\circ} \\
& =49 x^{2} .
\end{aligned}
$$
$$
\text { Then } \frac{B C}{E F}=\frac{7 x}{2 \sqrt{7} x}=\frac{\sqrt{7}}{2} \text {. }
$$