Cursed angle bisectors!

Following is one method of solution; others exist.

Let $\angle BAE = x$ and $\angle BCD = y.$

$AB = 9 \times cos x = AC\times cos 2x.$
$BC = 8\sqrt2 \times cos y = AC\times cos 2y.$
Eliminating $AC:$
$\dfrac{(9 cos x)}{(cos 2x)} = \dfrac{(8\sqrt2 cos y)}{(cos 2y)}.$

$y = 45^\circ - x.$ Also, $2y = 90^\circ - 2x$ , and so $cos 2y = sin 2x.$ Therefore:
$\dfrac{(9 cos x)}{(cos 2x)} = \dfrac{(8\sqrt2 cos (45^\circ-x))}{(sin 2x)}.$
Using trigonometric identity $cos(a - b) = cos a · cos b + sin a · sin b:$
$cos (45^\circ-x) = \dfrac{(cos x + sin x)}{\sqrt2}.$

Rearranging: $tan 2x = \dfrac{8(cos x + sin x)}{9 cos x}.$

Using trigonometric identity $tan 2a = \dfrac{(2 tan a)}{(1 - tan2a)},$ and letting $t = tan x:$
$\dfrac{2t}{(1 - t^2)} = \dfrac{8(1 + t)}{9}.$
Therefore $\dfrac{9t}{4} = (1 + t)(1 - t^2) = 1 + t - t^2 - t^3.$
Hence $t^3 + t^2 + (\frac{5}{4})t - 1 = 0.$

By inspection, one root is $t = \dfrac{1}{2}.$
Therefore $(t - \dfrac{1}{2})(t^2 + \dfrac{3t}{2} + 2) = 0.$
The quadratic factor has no real roots (since $(\frac{3}{2}).2 - 4\times1\times2 < 0$ ), and so $t = \dfrac{1}{2}$ is the only real root.

$AC = \dfrac{9cos x}{cos 2x}.$

Using trigonometric identities $cos x = \dfrac{1}{\sqrt{(1 + t^2)}}, cos 2x = \dfrac{(1 - t^2)}{(1 + t^2)} :$
$AC = \dfrac{9\sqrt{(1 + t^2)}}{(1 - t^2)}.$

Therefore $AC = \dfrac{9(\frac{\sqrt5}{2})}{\dfrac{3}{4}} = 6\sqrt 5.$

So, $a=6$ and $b=5.$ Hence $a+b=6+5=\boxed{11}.$