4. In $\triangle A B C$, $\angle B A C=60^{\circ}$, the angle bisector $A D$ of $\angle B A C$ intersects $B C$ at $D$, and $\overrightarrow{A D}=\frac{1}{4} \overrightarrow{A C}+t \overrightarrow{A B}$. If $A B=8$, then $A D=$ . $\qquad$
Answer: $6 \sqrt{3}$.
[Analysis] It is easy to know that $t=\frac{3}{4}$, thus: $A C=24, A D^{2}=\frac{1}{16} \times 24^{2}+\frac{9}{16} \times 8^{2}+\frac{3}{16} \times 8 \times 24=108$, hence: $A D=6 \sqrt{3}$.