smc

Let $AD=y$. Since $AD$ bisects $\angle BAC$, the Angle Bisector Theorem gives $\frac{DB}{CD}=\frac{AB}{AC}=2$, so let $CD=x$ and $DB=2x$. Applying the Law of Cosines to $△CAD$ gives $x^2=3^2+y^2-3y$, and to $△DAB$ gives $(2x{)}^2=6^2+y^2-6y$. Subtracting $4$ times the first equation from the second equation therefore yields $0=6y-3y^2\Rightarrow y(y-2)=0$, so $y$ is $0$ or $2$. But since $y\ne 0$ ($y$ is the length of a side of a triangle), $y$ must be $2$, so the answer is $\boxed{2}$.