8. In Rt $\triangle A B C$, the incircle $\odot O$ touches the hypotenuse $A B$ at point $D$, and touches $B C, C A$ at points $E, F$ respectively. Draw $D K \perp A C$ at point $K$, and $D P \perp B C$ at point $P$. Given $A D$ $=m, B D=n$. Express the area $S$ of rectangle $C K D P$ in terms of $m, n$ as $\qquad$
8. $\frac{2 m^{2} n^{2}}{(m+n)^{2}}$.

Let the inradius be $r$. As shown in Figure 7, connect $O D$, $O E$, and $O F$.
Then $O D=O E=O F=r$.
By the tangent segment theorem,
$$
A D=A F=m, B D=B E=n, C E=C F=r \text {. }
$$

Let the area of $\triangle A B C$ be $S_{1}$. Then
$$
\begin{aligned}
S_{1}= & \frac{(r+m)(r+n)}{2} \\
\Rightarrow & 2 S_{1}=r^{2}+r m+r n+m n \\
& =r(r+m+n)+m n=S+m n \\
\Rightarrow & S_{1}=m n .
\end{aligned}
$$

Since $D K \parallel B C$, 
$\triangle A D K \sim \triangle A B C$.
Then $S_{\triangle M D K}=S_{1} \cdot \frac{m^{2}}{(m+n)^{2}}=\frac{m^{3} n}{(m+n)^{2}}$.
Similarly, $S_{\triangle R D P}=\frac{m n^{3}}{(m+n)^{2}}$.
Thus $S=m n-\frac{m^{3} n}{(m+n)^{2}}-\frac{m n^{3}}{(m+n)^{2}}$ $=\frac{2 m^{2} n^{2}}{(m+n)^{2}}$.
(Li Yanlin provided)