Quadrilateral $ABCD$ is inscribed in a circle with $\angle BAC=70^{\circ}, \angle ADB=40^{\circ}, AD=4,$ and $BC=6$. What is $AC$?
$\textbf{(A)}\; 3+\sqrt{5} \qquad\textbf{(B)}\; 6 \qquad\textbf{(C)}\; \dfrac{9}{2}\sqrt{2} \qquad\textbf{(D)}\; 8-\sqrt{2} \qquad\textbf{(E)}\; 7$
$\angle ADB$ and $\angle ACB$ are both subtended by segment $AB$, hence $\angle ACB = \angle ADB = 40^\circ$. By considering $\triangle ABC$, it follows that $\angle ABC = 180^\circ - (70^\circ + 40^\circ) = 70^\circ$. Hence $\triangle ABC$ is isosceles, and $AC = BC = \boxed{\textbf{(B)}\; 6}.$