2. As shown in Figure 1, in $\triangle A B C$, a semicircle with diameter $B C$ intersects sides $A B$ and $A C$ at points $D$ and $E$, respectively. If $D E=E C=4, B C-B D$ $=\frac{16}{5}$, then the value of $\sqrt{\frac{B D-A D}{B C}}$ is ( ).
(A) $\frac{3}{5}$
(B) $\frac{2}{3}$
(C) $\frac{4}{5}$
(D) $\frac{5}{6}$
2. A.

Since $B C$ is the diameter of the semicircle, therefore,
$$
\begin{array}{l}
\angle B E C=90^{\circ} . \\
\text { By } D E=E C \Rightarrow \angle A B E=\angle C B E \\
\Rightarrow A B=B C, A E=E C=D E \\
\Rightarrow A D=A B-B D=B C-B D=\frac{16}{5} .
\end{array}
$$
$$
\begin{array}{l}
\text { By } A E=D E=E C=4 \Rightarrow A C=8 \text {. } \\
\text { By } \triangle A E D \backsim \triangle A B C \\
\Rightarrow \frac{A E}{A B}=\frac{A D}{A C}=\frac{2}{5} \Rightarrow A B=B C=10 \text {. }
\end{array}
$$

Thus $B D=\frac{34}{5} \Rightarrow \sqrt{\frac{B D-A D}{B C}}=\frac{3}{5}$.