4. Given that $A D, B E, C F$ are the altitudes of $\triangle A B C$ (with $D, E, F$ being the feet of the altitudes), $\angle B=45^{\circ}, \angle C=60^{\circ}$. Then the value of $\frac{D F}{D F}$ is
(A) $\frac{2}{\sqrt{3}}$
(B) $\frac{1}{2}$
(C) $\frac{1}{\sqrt{2}}$
(D) $\frac{\sqrt{3}}{2}$
4. D.
$\because A, B, D, E$ are concyclic,
$$
\therefore \angle C E D=\angle C B A \text {. }
$$

Therefore, $\triangle C E D \sim \triangle C B A$.

Thus, $\frac{D E}{A B}=\frac{D C}{A C}=$ $\cos 60^{\circ}=\frac{1}{2}, D E=\frac{1}{2} A B$.
Similarly, $D F=\frac{\sqrt{2}}{2} A C$.
Therefore, $\frac{D E}{D F}=\frac{1}{\sqrt{2}} \cdot \frac{A B}{A C}$ $=\frac{1}{\sqrt{2}} \cdot \frac{\sin 60^{\circ}}{\sin 45^{\circ}}=\frac{\sqrt{3}}{2}$.