9. Method One
$$
\begin{aligned}
& b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b c \cos \frac{B}{2} \cos \frac{C}{2} \sin \frac{A}{2} \\
= & \frac{1}{2} b^{2}(1+\cos C)+\frac{1}{2} c^{2}(1+\cos B)+2 b c \cos \frac{B}{2} \cos \frac{C}{2} \cos \frac{B+C}{2} \\
= & \frac{1}{2} b^{2}(1+\cos C)+\frac{1}{2} c^{2}(1+\cos B)+2 b c \cos ^{2} \frac{B}{2} \cos ^{2} \frac{C}{2} \\
& -2 b c \cos \frac{B}{2} \cos \frac{C}{2} \sin \frac{B}{2} \sin \frac{C}{2} \\
= & \frac{1}{2} b^{2}(1+\cos C)+\frac{1}{2} c^{2}(1+\cos B)+\frac{1}{2} b c(1+\cos B)(1+\cos C) \\
& -\frac{1}{2} b c \sin B \sin C \\
& +\frac{1}{2} b^{2}+\frac{1}{2} c^{2}+\frac{1}{2} b(b \cos C+c \cos B)+\frac{1}{2} c(b \cos C+c \cos B) \\
= & \frac{1}{2} b^{2}+\frac{1}{2} c^{2}+\frac{1}{2} a b+\frac{1}{2} a c+\frac{1}{2} b c-\frac{1}{2} b c \cos A \\
= & \frac{1}{4}\left(b^{2}+c^{2}-2 b c \cos A\right)+\frac{1}{4} b^{2}+\frac{1}{4} c^{2}+\frac{1}{2} a b+\frac{1}{2} a c+\frac{1}{2} b c \\
= & \frac{1}{4}\left(a^{2}+b^{2}+c^{2}+2 a b+2 a c+2 b c\right)^{2} \\
= & \frac{1}{4}(a+b+c)^{2}=64 .
\end{aligned}
$$

Method Two As shown in the figure, extend $BC$ to $E$ such that $CE = AC = b$, and extend $CB$ to $F$ such that $BF = AB = c$. Connect $AE$ and $AF$.

Let the midpoints of $AE$ and $AF$ be $M$ and $N$, respectively, then $AM = b \cos \frac{C}{2}$, $AN = c \cos \frac{B}{2}$,
$$
\angle MAN = \angle A + \frac{1}{2}(\angle B + \angle C) = \frac{\pi}{2} + \frac{A}{2}.
$$

Therefore,
$$
\begin{aligned}
& b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b c \cos \frac{B}{2} \cos \frac{C}{2} \sin \frac{A}{2} \\
= & AM^{2} + AN^{2} - 2 AM \cdot AN \cos \angle MAN \\
= & MN^{2} = \left(\frac{a+b+c}{2}\right)^{2} = 64.
\end{aligned}
$$
(9) (16 points) Let the sides opposite to the internal angles $A$, $B$, and $C$ of $\triangle ABC$ be $a$, $b$, and $c$ respectively. Given that $a+b+c=16$, find the value of $b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b c \cos \frac{B}{2} \cos \frac{C}{2} \sin \frac{A}{2}$.