Let $S H$ be the height of the pyramid, $M$ and $N$ be the midpoints of edges $A B$ and $S A$ respectively (Fig.1). Let $A B=a, S H=a \sqrt{3}$. On the extension of edge $C D$ beyond point $D$, lay off the segment $D P=\frac{1}{4} C D=\frac{1}{4} a$. Since $E N$ is the midline of triangle $A S M$, then $N E\|A M\| D P$ and $N E=\frac{1}{2} A M=\frac{1}{4} a=D P$, hence, quadrilateral $D P N E$ is a parallelogram, so $P N=D E$ and $P N \| D E$. Therefore, the angle $\phi$ between the plane $A S C$ and the line $D E$ is equal to the angle between this plane and the line $P N$. The plane $A S C$ passes through the line $S H$, which is perpendicular to the plane $A B C D$, so the planes $A S C$ and $A B C D$ are perpendicular, meaning that the perpendicular $P Q$, dropped from point $P$ to the line $A C$, is a perpendicular to the plane $A S C$. Therefore, the distance from point $P$ to the plane $A S C$ is equal to the length of the segment $P Q$. From the isosceles right triangle $C P Q$, we find that

$$
P Q=\frac{C P}{\sqrt{2}}=\frac{\operatorname{Su}^{2}}{\sqrt{2}}=\frac{5 n}{4 \sqrt{2}}
$$

Let $F$ be the orthogonal projection of point $E$ onto the base plane of the pyramid, $K$ be the midpoint of $C D$. Then $F$ is the midpoint of $4 M$, so

$$
\begin{gathered}
F K=\frac{3}{4} a, E F=\frac{1}{2} S H=\frac{a \sqrt{3}}{2} \\
P N=D E=\sqrt{D K^{2}+K F^{2}+E F^{2}}=\sqrt{\frac{\alpha^{2}}{4}+\frac{9}{16} a^{2}+\frac{3}{4} a^{2}}=\frac{5}{4} a
\end{gathered}
$$

From the right triangle $P Q N$ (Fig.2), we find that

$$
\sin \phi=\sin \angle P N Q=\frac{P Q}{P N}=\frac{\frac{5 s}{\sqrt{2}}}{\frac{\sqrt{2}}{\sqrt{a}}}=\frac{1}{\sqrt{2}}
$$

Therefore, $\phi=45^{\circ}$.

## Answer

$45^{\circ}$.
The height of a regular quadrilateral pyramid $S A B C D$ ($S$ - the vertex) is $\sqrt{3}$ times the length of the base edge. Point $E$ is the midpoint of the apothem lying in the face $A S B$. Find the angle between the line $D E$ and the plane $A S C$.

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