4. As shown in Figure 3, in $\triangle A B C$, a semicircle is drawn with side $B C$ as the diameter, intersecting sides $A B$ and $A C$ at points $D$ and $E$ respectively. If $D E = E C = 4$, and $B C - B D = \frac{16}{5}$, then $\sqrt{\frac{B D - A D}{B C}}=$ $\qquad$ .
4. $\frac{3}{5}$.

As shown in Figure 7, connect $B E$. Since $B C$ is the diameter, we have
$$
\begin{array}{l}
\angle B E C=\angle B E A \\
=90^{\circ} .
\end{array}
$$

Also, $D E=E C$, thus
$$
\overparen{D E}=\overparen{E C}, \angle 1=\angle 2 \text {. }
$$

Therefore, Rt $\triangle A E B \cong \mathrm{Rt} \triangle C E B$.
Hence, $A E=E C=4$.
Since $\angle A E D=\angle A B C, \angle A=\angle A$, we have $\triangle A D E \backsim \triangle A C B$.

Thus, $\frac{A D}{A C}=\frac{D E}{B C}$.
Substituting the values, we get $\frac{A D}{8}=\frac{4}{B C}$, which means $B C=\frac{32}{A D}$.
Also, $B C=\frac{16}{5}+B D$, so,
$$
A D \cdot B D+\frac{16}{5} A D=32 \text {. }
$$

By the secant theorem,
$$
\begin{array}{l}
A D \cdot A B=A E \cdot A C=4 \times 8=32 . \\
\text { Therefore, } \frac{16}{5} A D+A D \cdot B D=A D \cdot A B \\
=A D(A D+B D)=A D^{2}+A D \cdot B D .
\end{array}
$$

Thus, $A D=\frac{16}{5}, B C=10, B D=\frac{34}{5}$.
Hence, $\sqrt{\frac{B D-A D}{B C}}=\sqrt{\frac{\frac{34}{5}-\frac{16}{5}}{10}}=\frac{3}{5}$.