6. As shown in Figure 2, in isosceles $\triangle ABC$, $O$ is the midpoint of the base $BC$. A semicircle is drawn with $O$ as the center, tangent to $AB$ and $AC$ at points $D$ and $E$, respectively. A tangent line is drawn from a point $F$ on the semicircle, intersecting $AB$ and $AC$ at points $M$ and $N$, respectively. Then, the value of $\frac{BM \cdot CN}{BC^{2}}$ is ( ).
(A) $\frac{1}{8}$
(B) $\frac{1}{4}$
(C) $\frac{1}{2}$
(D) 1
6. B.

As shown in Figure 7, connect $OD$, $OM$, $OF$, $ON$, $OE$, then $\angle 1=\angle 6$, $\angle 2=\angle 3$, $\angle 4=\angle 5$.

Therefore, $\angle 1+\angle 2=90^{\circ}-\angle 5$.
Also, $\angle B O M$
$=\angle 1+\angle 2$
$=90^{\circ}-\angle 5$
$=\angle O N E$,
$\angle M B O=\angle N C O$,

Thus,
$$
\begin{array}{l}
\triangle M B O \backsim \triangle O C N, \\
\frac{B M}{C O}=\frac{O B}{C N} .
\end{array}
$$

Hence, $B M \cdot C N=C O \cdot O B=\frac{1}{4} B C^{2}$.
Therefore, $\frac{B M \cdot C N}{B C^{2}}=\frac{1}{4}$.