8. As shown in Figure 3, given that $M$ and $N$ are points on the sides $BC$ and $CD$ of the unit square $ABCD$ respectively, $\angle MAN=45^{\circ}, BE \perp$ $AN$ at point $E, DF \perp AM$ at point $F, BG \perp AM$ at point $G$, $DH \perp AN$ at point $H$. Then the area of quadrilateral $EFGH$ is
8. $\frac{1}{4}$.

Let $\angle M A B=\alpha, \angle M A N=\beta=45^{\circ}, \angle N A D=\gamma$. Then
$$
\begin{array}{l}
S_{\triangle A E F}=\frac{1}{2} A E \cdot A F \sin \beta \\
=\frac{1}{2} A B \sin \gamma \cdot A D \sin \alpha \cdot \sin \beta \\
=\frac{1}{2} \sin \alpha \cdot \sin \beta \cdot \sin \gamma .
\end{array}
$$

Similarly, $S_{\triangle A C H}=\frac{1}{2} \cos \alpha \cdot \sin \beta \cdot \cos \gamma$.
$$
\begin{array}{l}
\text { Therefore, } S_{\text {quadrilateral EFGH }}=S_{\triangle A G H}-S_{\triangle A E F} \\
=\frac{1}{2} \sin \beta(\cos \alpha \cdot \cos \gamma-\sin \alpha \cdot \sin \gamma) \\
=\frac{1}{2} \sin \beta \cdot \cos (\alpha+\gamma)=\frac{1}{2} \sin 45^{\circ} \cdot \cos 45^{\circ} \\
=\frac{1}{4} .
\end{array}
$$