1. In Rt $\triangle A B C$, the area is 120, and $\angle B A C=$ $90^{\circ}, A D$ is the median of the hypotenuse, a line $D E \perp$ $A B$ is drawn through point $D$, and $C E$ is connected to intersect $A D$ at point $F$. Then the area of $\triangle A F E$ is ( ).
(A) 18
(B) 20
(C) 22
(D) 24
Draw $F G \perp A B$ at point $G$. Then $F G / / D E / / A C$,
we have $\frac{F G}{\frac{1}{2} A C}=\frac{F G}{D E}=\frac{A G}{A E}, \frac{F G}{A C}=\frac{E G}{A E}$.
Adding the two equations, we get
$$
\frac{3 F G}{A C}=\frac{A E}{A E}=1 \Rightarrow F G=\frac{1}{3} A C \text {. }
$$

Then $S_{\triangle A F E}=\frac{1}{2} A E \cdot F G=20$.