4. In right $\triangle ABC$, the area is 120, and $\angle BAC=90^{\circ}$. $AD$ is the median of the hypotenuse. A line $DE \perp AB$ is drawn from point $D$ to point $E$. Connect $CE$ to intersect $AD$ at point $F$. Then the area of $\triangle AFE$ is ( ).
(A) 18
(B) 20
(C) 22
(D) 24
4. (B).

As shown in Figure 6, draw $F G \perp A B$ at $G$, then $F G / / D E / / A C$. Therefore, we have
$$
\begin{array}{l}
\frac{F G}{\frac{1}{2} A C}=\frac{F G}{D E}=\frac{A G}{A E}, \\
\frac{F G}{A C}=\frac{E G}{A E} .
\end{array}
$$
(1) + (2) gives $\frac{3 F G}{A C}=\frac{A E}{A E}=1$.

Thus, $F G=\frac{1}{3} A C$.
Hence, $S_{\text {LAFE }}=\frac{1}{2} A E \times F G$
$$
=\frac{1}{2}\left(\frac{1}{2} A B \times \frac{1}{3} A C\right)=\frac{1}{6} S_{\triangle A B C}=20 \text {. }
$$