Inspiration

We let the area of $\triangle CEF=x$ . Now what should the area of $\triangle BCF$ be in terms of $x$ ? Drop a perpendicular from $F$ to $BC$ and another from $F$ to $CE$ . Suppose these perpendicular touch $BC$ at $G$ and $CE$ at $H$ . Considering that $\angle CGF=\angle CHF=90$ degrees, $\angle GCF=\angle HCF=45$ degrees and that $CF$ is common to both $\triangle CFG$ and $\triangle CFH$ , it follows that these two triangles are congruent. Therefore $FG=FH$ .

Then considering that $ABCD$ is a square, $BC=CD$ and considering that $E$ is the midpoint of $CD$ , $CD=2CE$ . Therefore $BC=2CE$ which means $\triangle BCF$ has twice the base of $\triangle CEF$ . Considering that both triangles have the same perpendicular height ( $FG=FH)$ , $\triangle BCF$ has twice the area of $\triangle CEF$ or $2x$ .

Then the area of $\triangle BCE$ is the sum of that of $\triangle BCF$ and that of $\triangle CEF$ which gives us $2x+x=(2+1)x=3x$ . Then $\triangle ACD$ would be twice the area of $\triangle BCE$ as both share the same height of $AD=BC$ but $CD=2CE$ . Therefore the area of $\triangle ACD=2(3x)=(2\times 3)x=6x$ .

Finally, $\triangle ACD$ can be partitioned into $\triangle CEF$ and quadrilateral $ADEF$ . Therefore, the area of quadrilateral $ADEF$ can be expressed as the area of $\triangle ACD$ minus the area of $\triangle CEF$ which is $6x-x=(6-1)x=5x$ .

Therefore,

$5x=20$

$x=\boxed{4}$

Let the side length of square $ABCD$ be $a$ , the altitude of $\triangle CEF$ , $FG=h$ and $EG=x$ . Then we note that $h=2x =\dfrac a2 - x$ $\implies x = \dfrac a6$ $\implies h = \dfrac a3$ . Then the area of $\triangle CEF = \dfrac 12 \times h \times \dfrac a2 = \color{#3D99F6} \dfrac {a^2}{12}$ . We note that:

$\begin{aligned} [ACD] & = [ADEF]+[CEF] \\ \frac {a^2}2 & = 20 + \color{#3D99F6} \frac {a^2}{12} \\ \frac {5a^2}{12} & = 20 \\ \implies \color{#3D99F6} \frac {a^2}{12} & = \boxed{4} \end{aligned}$