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 point $D$, and $\overrightarrow{A D}=\frac{1}{4} \overrightarrow{A C}+$ $t \overrightarrow{A B}$. If $A B=8$, then $A D=$ $\qquad$
4. $6 \sqrt{3}$.

As shown in Figure 2, draw $D E / / A B$, intersecting $A C$ at point $E$; draw $D F / / A C$, intersecting $A B$ at point $F$. Then, from the given conditions, we have
$$
\begin{array}{l}
\overrightarrow{A D}=\frac{1}{4} \overrightarrow{A C}+t \overrightarrow{A B}=\overrightarrow{A E}+\overrightarrow{A F} \\
\Rightarrow \overrightarrow{A E}=\frac{1}{4} \overrightarrow{A C}=\frac{1}{3} \overrightarrow{E C}, \overrightarrow{A F}=t \overrightarrow{A B} \\
\Rightarrow \frac{A E}{E C}=\frac{1}{3}=\frac{B D}{C D}=\frac{B F}{F A}=\frac{A B}{A C} \\
\Rightarrow A C=24, F A=3 B F=\frac{3}{4} A B \\
\Rightarrow t=\frac{3}{4} . \\
\text { Therefore, }|\overrightarrow{A D}|^{2}=\left|\frac{1}{4} \overrightarrow{A C}+\frac{3}{4} \overrightarrow{A B}\right|^{2} \\
=\left(\frac{1}{4} \overrightarrow{A C}+\frac{3}{4} \overrightarrow{A B}\right)\left(\frac{1}{4} \overrightarrow{A C}+\frac{3}{4} \overrightarrow{A B}\right) \\
=\frac{1}{16}|\overrightarrow{A C}|^{2}+\frac{9}{16}|\overrightarrow{A B}|^{2}+\frac{6}{16} \overrightarrow{A C} \cdot \overrightarrow{A B} \\
=108 .
\end{array}
$$

Therefore, $A D=6 \sqrt{3}$.