4. A. Given points $D$ and $E$ are on the sides $AB$ and $AC$ of $\triangle ABC$, respectively, and $BE$ intersects $CD$ at point $F$. Let
$S_{\text{quadrilateral EADF}}=S_{1}, S_{\triangle BDF}=S_{2}$,
$S_{\triangle BCF}=S_{3}, S_{\triangle CEF}=S_{4}$.
Then the relationship between $S_{1} S_{3}$ and $S_{2} S_{4}$ is ().
(A) $S_{1} S_{3} < S_{2} S_{4}$
(B) $S_{1} S_{3} > S_{2} S_{4}$
(C) $S_{1} S_{3} = S_{2} S_{4}$
(D) Cannot be determined
4. A. C.

As shown in Figure 8, connect $D E$.
Let $S_{\triangle D E F}=S_{1}^{\prime}$. Then,
$$
\frac{S_{1}^{\prime}}{S_{2}}=\frac{E F}{B F}=\frac{S_{4}}{S_{3}} \text{. }
$$

Thus, $S_{1}^{\prime} S_{3}=S_{2} S_{4}$.
Since $S_{1}>S_{1}^{\prime}$, we have,
$$
S_{1} S_{3}>S_{2} S_{4} .
$$