In the tetrahedron $A B C D$, $\angle A C B = \angle C A D = 90^{\circ}$ and $C A = C B = A D / 2$ and $C D \perp A B$. What is the angle between the faces $A C B$ and $A C D$?
Based on the conditions, the $ABC$ plane is an isosceles right triangle. Consider the plane $S$ of this triangle, and denote by $D^{\prime}$ the orthogonal projection of point $D$ onto $S$ (Figure 1). Since edge $CD$ is perpendicular to $AB$, the orthogonal projection of line $CD$ onto the plane $S$ containing $AB$ is also perpendicular to $AB$, hence $CD^{\prime} \perp AB$. Similarly, from the condition $AD \perp AC$, we get that $AD^{\prime} \perp AC$. Thus, $D^{\prime}$ is at the intersection of these two perpendicular lines, making $AD^{\prime}BC$ a square (Figure 2).

The problem asks for the dihedral angle between the planes $ACB$ and $ACD$. By the dihedral angle of adjacent planes, we mean the angle between two half-lines that start from a point on the common edge, are perpendicular to this edge, and lie in the half-planes containing the respective faces. In our case, the common edge is $AC$, and the two half-lines are $AD$ and $AD^{\prime}$, respectively, since the first lies in the plane of $ACD$ and the second in the plane of $ABC$, and both are perpendicular to the line $AC$. Thus, the desired angle is the angle at $A$ in the right triangle $ADD^{\prime}$, which, based on $AD^{\prime} = AC = AD / 2$, is $60^{\circ}$.