Answer: E

Small Hint:

Use triangle $AOB$ to find $\cos\angle AOB$

Big Hint:

Angles $AOB$ and $AOD$ are supplementary

Solution:

In triangle $AOB,$ $$\begin{aligned} \cos\angle AOB &=\frac{8^2+4^2-6^2}{2\cdot8\cdot4}\\ &=\frac{11}{16}. \end{aligned}$$ Therefore $\cos\angle AOD=-\frac{11}{16}.$ Applying the law of cosines in triangle $AOD$ gives $$\begin{aligned} AD^2 &=8^2+6^2\\ &\quad+2(8)(6)\frac{11}{16}\\ &=166. \end{aligned}$$ Hence $AD=\sqrt{166}.$

Therefore, the correct answer is E.