Few Congruent Dimensions in a Trapezoid

From the above figure we notice that:

$a+2y = x$ $\dots \boxed{1}$

$(a+y)^2+ a^2 = x^2$ $\dots\boxed{ 2}$

Plugging the value of $x$ in $\boxed{2}$ , we obtain:

$(a+y)^2+a^2=(a+2y)^2\implies a^2+y^2+2ay+a^2=a^2+4ay+4y^2\implies 3y^2+2ay-a^2=0\implies y=\frac{a}{3}$

But, we know that $x=2y+a\implies x= \frac{2a}{3}+a= \frac{5a}{3}\implies \frac{x}{y}= \frac{3}{5}\implies A=3, B=5, A+B= \boxed{8}$