Trigonometry · Superchallenge problem

In $\triangle ABC$, $AB=6$, $AC=9$, and $A=80^\circ$. The bisector of angle $A$ meets $BC$ at $D$. Find the bisector length $AD$. (Use $t_a=\dfrac{2\,bc\cos(A/2)}{b+c}$, or split the area two ways.) Round to 2 decimal places (an exact answer is also accepted).