20. As shown in Figure 6, in square $A B C D$, $E$ is the midpoint of side $D C$, and line segment $B E$ intersects diagonal $A C$ at point $F$. If the area of quadrilateral $A F E D$ is 45, then the area of square $A B C D$ is ( ).
(A) 100
(B) 108
(C) 120
(D) 135
(E) 144
20. B.

Let the area of $\triangle C E F$ be $x$. Then the area of $\triangle A C D$ is $45+x$, and the area of square $A B C D$ is
$$
2(45+x)=90+2 x \text {. }
$$

Obviously, $\triangle C E F \backsim \triangle A B F$, with a similarity ratio of $1: 2$.
Thus, the area of $\triangle A B F$ is $4 x$.
Considering trapezoid $A B E D$, its area is $45+4 x$, and it is $\frac{3}{4}$ of the area of the square.
Therefore, $45+4 x=\frac{3}{4}(90+2 x) \Rightarrow x=9$.
Hence, the desired result is $90+2 x=90+18=108$.