In the pyramid $A B C D$, the dihedral angle at edge $A C$ is $90^{\circ}, A B=B C=C D, B D=A C$. Find the dihedral angle at edge $A D$.

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Since the planes $A B C$ and $A C D$ are perpendicular (the dihedral angle between them is $90^{\circ}$), the height $B M$ of triangle $A B C$ is perpendicular to the plane $A C D$, meaning that triangle $A M D$ is the orthogonal projection of triangle $A B D$ onto the plane $A C D$. Since triangle $A B C$ is isosceles ($A B = B C$), $M$ is the midpoint of $A C$. Triangles $A C D$ and $D B A$ are equal by three sides, and $D M$ is the median of triangle $A D C$, therefore

$$
S_{\triangle A M D}=\frac{1}{2} S_{\triangle A C D}=\frac{1}{2} S_{\triangle A B D}
$$

Let $\alpha$ be the angle between the faces $B A D$ and $C A D$. Then

$$
\cos \alpha=\frac{S_{S A M A D}}{S_{\triangle A B D}}=\frac{\frac{1}{S_{\triangle A A B D}}}{S_{\triangle A B D}}=\frac{1}{2}
$$

Therefore, $\alpha=60^{\circ}$.

## Answer

$60^{\circ}$.

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