Solution:

Let $E$ be the intersection of $AM$ with the circumcircle of $ABC$. We note that, by equal angles, $ADC \sim ABM$, so that
$$AD = AC \left(\frac{AB}{AM}\right) = \frac{20}{AM}$$
Using the law of cosines on $ABC$, we get that
$$\cos B = \frac{4^2 + 8^2 - 5^2}{2(4)(8)} = \frac{55}{64}$$
Then, using the law of cosines on $ABM$, we get that
$$AM = \sqrt{4^2 + 4^2 - 2(4)(4) \cos B} = \frac{3}{\sqrt{2}} \Rightarrow AD = \frac{20 \sqrt{2}}{3}.$$
Applying Power of a Point on $M$,
$$(AM)(ME) = (BM)(MC) \Rightarrow ME = \frac{16 \sqrt{2}}{3} \Rightarrow AE = \frac{41 \sqrt{2}}{6}$$
Then, we note that $AXB \sim ACE$, so that
$$AX = AB \left(\frac{AC}{AE}\right) = \frac{60 \sqrt{2}}{41} \Rightarrow \frac{AX}{AD} = \frac{9}{41}$$