In quadrilateral $ABCD$ with diagonals $AC$ and $BD$, intersecting at $O$, $BO=4$, $OD = 6$, $AO=8$, $OC=3$, and $AB=6$. The length of $AD$ is:
$\textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 6\sqrt{3}\qquad \textbf{(D)}\ 8\sqrt{2}\qquad \textbf{(E)}\ \sqrt{166}$
After drawing the diagram, we see that we actually have a lot of lengths to work with. Considering triangle ABD, we know values of $AB, BD(BD = BO + OD)$, but we want to find the value of $AD$. We can apply stewart's theorem now, letting $m = 4, n = 6, AD = X, AB = 6$, and we have $10 \cdot 6 \cdot 4 + 8 \cdot 8 \cdot 10 = x^2 + 36 \cdot 6$, and we see that $x = \boxed{\textbf{(E)}~\sqrt{166}}$