$A,B,C$ and $D$ are concyclic.$AC$ is the diameter of the circle and $AD=DC$.The area of quadrilateral $ABCD$ is $20cm^2$. $A,B,C$ and $D$ are concyclic . $AC$ is the diameter of the circle and $AD=DC$ . The area of quadrilateral $ABCD$ is $20c$m$^2$.
Draw a line $DE$ such that $E$ is a point on $AB$, and $DE$ $\bot$ $AB$.
Find the length of $DE$.
The answer keys says that it is ${2}\sqrt{5}$ $cm$.  How was the length found?
I imagine that the use of similar triangles and using the fact that angle in a semicircle is $90^\circ $may be involved, but I cannot go beyond that.

Let $AB=x$ and $BC=y$.
Thus, $$AC=\sqrt{x^2+y^2}$$ and by the given:
$$\frac{xy}{2}+\frac{x^2+y^2}{4}=20,$$ which gives $$x+y=4\sqrt5.$$
Id est,  $$DE=AD\sin\left(45^{\circ}+\measuredangle BAC\right)=$$
$$=\sqrt{\frac{x^2+y^2}{2}}\cdot\frac{1}{\sqrt2}\left(\sin\arctan\frac{y}{x}+\cos\arctan\frac{y}{x}\right)=$$
$$=\sqrt{\frac{x^2+y^2}{2}}\cdot\frac{1}{\sqrt2}\left(\frac{\frac{y}{x}}{\sqrt{1+\left(\frac{y}{x}\right)^2}}+\frac{1}{\sqrt{1+\left(\frac{y}{x}\right)^2}}\right)=\frac{1}{2}(x+y)=2\sqrt5.$$
