8. As shown in the figure, in $\triangle A B C$, $A B=c, \angle A=\alpha\left(30^{\circ}<\alpha<45^{\circ}\right), \angle C=90^{\circ}$, the midpoint of side $A C$ is $M$, and point $P$ on side $A B$ (not the midpoint of $A B$) satisfies $P C=\frac{c}{2}, P C$ intersects $B M$ at point $D$, then the length of $C D$ is $\qquad$ . (express in terms of $c, \alpha$)
Given, let the midpoint of $AB$ be $Q, \angle CPQ = \angle CQP = 2\alpha$
$\Rightarrow PQ = 2 \cdot \frac{c}{2} \cos 2\alpha = c \cos 2\alpha$,
thus $\frac{BP}{PA} = \frac{\frac{c}{2} - c \cos 2\alpha}{\frac{c}{2} + c \cos 2\alpha} = \frac{1 - 2 \cos 2\alpha}{1 + 2 \cos 2\alpha}$.
In $\triangle APC$, $B, D, M$ are collinear, by Menelaus' theorem we get
$$
\begin{array}{l}
\frac{AB}{BP} \cdot \frac{PD}{DC} \cdot \frac{CM}{MA} = 1 \Rightarrow \frac{PD}{DC} = \frac{1 - 2 \cos 2\alpha}{2} \Rightarrow \frac{PC}{DC} = \frac{3 - 2 \cos 2\alpha}{2} \\
\Rightarrow CD = \frac{2}{3 - 2 \cos 2\alpha} \cdot \frac{c}{2} = \frac{c}{3 - 2 \cos 2\alpha}.
\end{array}
$$