Unequal Sisters

For all the triangles to be similar, one of the angles of $\triangle DEC$ has to be $90^\circ$ .

This has to be at $E$ , since the other angles are less than $90^\circ$ .

The side in $\triangle DEC$ corresponding to side $EB$ in $\triangle EBC$ is $EC$ .

If $\dfrac{[DEC]}{[EBC]}=5$ Then $EC=\sqrt5 EB$ .

Since we work only with ratios of sizes, we can arbitrarily chose $EB=1$ , so that $EC=\sqrt5$ .

Pythagorean theorem results in $CB=\sqrt{(\sqrt5)^2-1}=\sqrt{5-1}=2$ .

$\frac{CB}{EB}=2\implies\frac{DE}{EC}=2\implies DE=2\sqrt5$

So the hypotenuse of $AED$ is twice the hypotenuse of $ABC$ , which means $AE=2\times CB=2\times2=4$

$AB=AE+EB=4+1=5$

$\frac{CB}{AB}=\frac25$

$2+5=\boxed7$

Since we work with ratio, let the area of ΔEBC = 1, So BC=2, and EB=1. And AE/AD=2, then AE=2AD = 2BC=4.So AB=5, then BC/AB=2/5. a+b=7

$\triangle DEC$ is $\frac{1}{2}$ of the rectangle, and $\triangle EBC$ is $\frac{1}{10}$ so $\triangle AED$ is the remaining $\frac{4}{10}$ .

If the area of $\triangle AED$ is $4$ times that of $\triangle EBC$ then its sides are $2$ times larger. Specifically $AD=2\times EB$ .

So also, $CB=2\times EB$ and we have the ratio of sides of the triangles. $AE=2\times AD=4\times EB$ .

The ratio then sought is $\frac{2}{4+1}=\frac{2}{5}$ .

$a+b=2+5=\boxed{7}$