85 In $\triangle A B C$, $P$ is a moving point on the line segment $B C$, $R, Q$ are on sides $A B, A C$ respectively, such that $P Q / / A B, P R / / A C$. Let $A C=b, A B=c, \angle B A C=\alpha$, try to find the shortest length of $R Q$ (expressed in the simplest form using $b, c, \alpha$).
85 As shown in Figure 33(a), let $B P=\lambda B C, P C=\mu B C$, then
$$
\lambda+\mu=1 .
$$

And $P R=\lambda b, P Q=\mu c$, so
$$
\begin{aligned}
R Q^{2} & =(\lambda b)^{2}+(\mu c)^{2}-2 \lambda \mu b c \cos \alpha \\
& =\lambda^{2}\left(b^{2}+c^{2}+2 b c \cos \alpha\right)-2 \lambda\left(c^{2}+b c \cos \alpha\right)+c^{2}
\end{aligned}
$$

When $\lambda=\frac{c^{2}+b c \cos \alpha}{b^{2}+c^{2}+2 b c \cos \alpha}$, $R Q$ reaches its minimum value
$$
\sqrt{\frac{\left(b^{2}+c^{2}+2 b c \cos \alpha\right) c^{2}-\left(c^{2}+b c \cos \alpha\right)^{2}}{b^{2}+c^{2}+2 b c \cos \alpha}}=\frac{b c \sin \alpha}{\sqrt{b^{2}+c^{2}+2 b c \cos \alpha}}
$$

Another solution As shown in Figure 33(b), extend $B A$ to $B^{\prime}$, such that $A B^{\prime}=B A=c$.
Connect $B^{\prime} C$. Let $P Q$ intersect $B^{\prime} C$ at $S$. It is easy to see that
$$
Q S=P Q, Q S \parallel A R
$$

Thus, $A S \perp R Q$. Therefore, the minimum value of $R Q$ is the distance from $A$ to $B^{\prime} C$. This distance
$$
=\frac{2 S_{\triangle A B^{\prime} C}}{B^{\prime} C}=\frac{b c \sin \alpha}{\sqrt{b^{2}+c^{2}+2 b c \cos \alpha}} .
$$

When $A S$ takes its minimum value, as shown in Figure 33(c), draw the altitude $C H$ of $B A$, then
$$
\frac{B P}{B C}=\frac{B^{\prime} S}{B^{\prime} C}=\frac{1}{B^{\prime} C} \times \frac{B^{\prime} A \times B^{\prime} H}{B^{\prime} C}=\frac{c(c+b \cos \alpha)}{b^{2}\left(c^{2}+2 b c \cos \alpha\right)}
$$