Length of a line

By applying the pythagorean theorem, $AB=10$ .

In an isosceles $45-45-90$ right triangle, the measure of the hypotenuse is equal to $\sqrt{2}$ times the measure of either leg so $EB=AE=5\sqrt{2}$ .

Since, $\angle AFB=\angle AEB=90^\circ$ , $AFBE$ is a cyclic.

Applying the Ptolemy's Theorem, we have

$(AF)(BE)+(AE)(FB)=(AB)(EF)$

$6(5\sqrt{2})+(5\sqrt{2})(8)=10(EF)$

$EF=7\sqrt{2}$

Finally,

$x+y=7+2=9$

Use Ptolemy's theorem for cyclic quadrilateral AEBF, given that AB=10 and AE=BE= $5\times \sqrt{2}$

In right $\bigtriangleup{AFB} \: \: (6,8,10)$ is a pythagorean triple $\implies AB = 10 \implies$ the side of the square $ABCD$ is $10 \implies$ diagonal $AC = 10 \sqrt{2} \implies AE = 5 \sqrt{2}$ and $\angle{CAB} = 45^{\circ}$ .

Let $\angle{FAB} = \theta \implies cos(\theta) = \dfrac{3}{5}$ and $sin(\theta) = \dfrac{4}{5}$ .

Using the law of cosines we obtain:

$({EF})^2 = 36 + 50 - 2(6)(5\sqrt{2})(cos(45^{\circ} + \theta)) =$

$86 - 60\sqrt{2}(\sqrt{\dfrac{1}{2}} \dfrac{3}{5} - \sqrt{\dfrac{1}{2}} \dfrac{4}{5}) =$

$86 - 60(-\dfrac{1}{5}) = 86 + 12 = 98 \implies EF = \sqrt{98} = 7 \sqrt{2} = x \sqrt{y} \implies x + y = \boxed{9}$