4. As shown in Figure 2, in $\triangle A B C$, it is known that $A B=9, B C$ $=8, C A=7, A D$ is the angle bisector, and a circle is drawn with $A D$ as a chord, 
touching $B C$ and intersecting $A B$ and $A C$ at points $M$ and $N$, respectively. Then $M N=$
$\qquad$
4. 6 .

As shown in Figure 4, connect $D M$.
$$
\begin{array}{l}
\text { By } \angle B D M \\
=\angle B A D \\
=\angle C A D \\
=\angle D M N,
\end{array}
$$

we get $M N \parallel B C$.
Thus, $\triangle A M N \sim \triangle A B C$.
It is easy to know $B D=\frac{9}{9+7} \times 8=\frac{9}{2}, B M \cdot B A=B D^{2}$
$\Rightarrow B M \cdot 9=\left(\frac{9}{2}\right)^{2} \Rightarrow B M=\frac{9}{4}$
$\Rightarrow A M=9-\frac{9}{4}=\frac{3}{4} \times 9$.
Also, $\frac{M N}{B C}=\frac{A M}{A B}=\frac{\frac{3}{4} \times 9}{9}=\frac{3}{4}$, therefore,
$$
M N=\frac{3}{4} \times 8=6
$$