smc

$\angle ADB$ and $\angle ACB$ are both subtended by segment $AB$, hence $\angle ACB=\angle ADB=40^{\circ }$. By considering $△ABC$, it follows that $\angle ABC=180^{\circ }-(70^{\circ }+40^{\circ })=70^{\circ }$. Hence $△ABC$ is isosceles, and $AC=BC=\boxed{6}.$ Note: The statement $AD=4$ is not needed.