4. In $\triangle A B C$, $\angle B A C=45^{\circ}, \angle A B C=$ $60^{\circ}$, the altitudes $A D$ and $B E$ intersect at point $H$. If $A B=1$, then the area of quadrilateral $C D H E$ is $\qquad$ .
4. $\frac{2-\sqrt{3}}{8}$.

Given $\angle B A C=45^{\circ}, \angle A B C=60^{\circ}$
$\Rightarrow \angle A C B=75^{\circ}$.
Thus, $\triangle A B C$ is an acute triangle, and the altitudes $A D$ and $B E$ intersect the sides $B C$ and $A C$ at their feet, respectively.
Then $\angle A B E=\angle B A E=45^{\circ}$
$\Rightarrow A E=B E=\frac{\sqrt{2}}{2}$
$\Rightarrow S_{\triangle A B E}=\frac{1}{2} \times \frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{2}=\frac{1}{4}$.
It is easy to see that $\angle E A H=\angle E B C=15^{\circ}$.
Therefore, $\triangle A E H \cong \triangle B E C$.
Also, $\angle B A D=30^{\circ} \Rightarrow B D=\frac{1}{2}, A D=\frac{\sqrt{3}}{2}$
$\Rightarrow S_{\triangle A B D}=\frac{1}{2} \times \frac{1}{2} \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{8}$.
Let the area of quadrilateral $C D H E$ be $S$. Then
$$
\begin{array}{l}
S_{\triangle A B E}-S_{\triangle A B D}=S_{\triangle A E H}-S_{\triangle B D H} \\
=S_{\triangle A E H}-\left(S_{\triangle A E H}-S\right)=S \\
\Rightarrow S=\frac{1}{4}-\frac{\sqrt{3}}{8}=\frac{2-\sqrt{3}}{8} .
\end{array}
$$