Any Parallelogram Will Work Too?

Extend each lines $EF$ , $DC$ to meet at $G$ .
$EF \parallel AC, EF = \dfrac12 AC$ .

Since $FI \parallel ED$ , then $I$ is a midpoint of $GD$ , so $FI = \dfrac 12 ED$ .

$EK \parallel FJ , EF \parallel KJ \Rightarrow EK = FJ$ .

$\triangle FIC \sim \triangle DEA$ with ratio of similarity of $1:2$ ,
then $IC = \dfrac 12 EA$ .
$\triangle ICJ \sim \triangle EAK$ with ratio of similarity of $1:2$ .
then $IJ = \dfrac 12 EK = \dfrac12 FJ$ , therefore

$\text{ area of } \triangle ICJ = \dfrac14 \text{ area of } \triangle EAK = \dfrac12 \text{ area of } \triangle FCJ = \dfrac32 \text{ cm}^2$

Then, area of $\triangle EAK = 4 \cdot \dfrac32 = \boxed {6\text{ cm}^2 }$ .