Let $E$ be the intersection of $A M$ with the circumcircle of $A B C$. We note that, by equal angles $A D C \sim A B M$, so that $$A D=A C\left(\frac{A B}{A M}\right)=\frac{20}{A M}$$ Using the law of cosines on $A B C$, 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 $A B M$, we get that $$A M=\sqrt{4^{2}+4^{2}-2(4)(4) \cos B}=\frac{3}{\sqrt{2}} \Rightarrow A D=\frac{20 \sqrt{2}}{3}$$ Applying Power of a Point on $M$, $$(A M)(M E)=(B M)(M C) \Rightarrow M E=\frac{16 \sqrt{2}}{3} \Rightarrow A E=\frac{41 \sqrt{2}}{6}$$ Then, we note that $A X B \sim A C E$, so that $$A X=A B\left(\frac{A C}{A E}\right)=\frac{60 \sqrt{2}}{41} \Rightarrow \frac{A X}{A D}=\frac{9}{41}$$