A Square Problem .

$\angle EDC= \angle ECA$ Hence $\angle ECD= \angle EDK$ and $\angle DEC = 90^\circ$

Let $|CD|=x$ and $|ED|=y$ ; $\frac{|CD|}{|KD|}=\frac{|ED|}{|KE|}=\frac{|CE|}{|ED|}$

$\frac{x}{\frac{x}{2}}=\frac{y}{|KE|}=\frac{|CE|}{y}=2$ hence $|KE|=\frac{y}{2}$ and $|CE|=2y$

Let $F$ be point on $[CK]$ such that $[AF]⟂[CK]$

$|AC|=|CD|=x$ Hence $△AFC$ and $△ECD$ are congruent and $|AF|=2y$

Area of $△AKE$ is $\frac{2y \cdot \frac{y}{2}}{2}=\frac{y^{2}}{2}$

Area of $ABCD$ is $x^{2}$ and $5y^{2}=x^{2}\Rightarrow \frac{y^{2}}{x^{2}}=\frac{1}{5}$ (Pythagoras Theorem)

Required ratio is $\frac{ \frac{y^{2}}{2}}{x^{2}}=\frac{y^{2}}{2x^{2}}=\frac{1}{10}=\boxed{0.1}$

Note that $\triangle DEK$ and $\triangle CDE$ are similar. Since $\dfrac {DK}{CD} = \dfrac 12$ , the ratio of the areas of the two triangles $\dfrac {[DEK]}{[CDE]} = \dfrac 14$ . Since $\triangle DEK$ and $\triangle CDE$ are sharing the same height, this means that $\dfrac {EK}{CE} = \dfrac 14$ $\implies EK = \dfrac {CK}5$ . Again $\triangle AKE$ and $\triangle ACK$ are sharing the same height, $\implies \dfrac {[AKE]}{[ACK]} = \dfrac 15$ . But $\dfrac {[ACK]}{[ABCD]} = \dfrac 12$ . Therefore $\dfrac {[AKE]}{[ABCD]} = \dfrac {[AKE]}{[ACK]} \times \dfrac {[ACK]}{[ABCD]} = \dfrac 15 \times \dfrac 12 = \dfrac 1{10} = \boxed{0.1}$ .