Example 4 In the tetrahedron $D-ABC$, $AD=a, BD=b, AB=CD=c$, and $\angle DAB + \angle BAC + \angle DAC = 180^{\circ}, \angle DBA + \angle ABC + \angle DBC = 180^{\circ}$. Find the angle formed by the skew lines $AD$ and $BC$.
Unfold the lateral faces of the tetrahedron $D-ABC$ along each edge and lay them out on the plane of the base, as shown in Figure 23-11.

From $\angle D_{1} A B + \angle B A C + \angle D_{2} A C = \angle D A B + \angle B A C + \angle D A C = 180^{\circ}$, we know that points $D_{1}, A, D_{2}$ are collinear. Similarly, points $D_{1}, B, D_{3}$ are collinear.

From $A D_{1} = A D_{2}, B D_{1} = B D_{2}$, we know that $AB$ is the midline of $\triangle D_{1} D_{2} D_{3}$, i.e., $D_{2} D_{3} = 2 AB = 2c$. And $C D_{2} = C D_{3} = c$, thus points $D_{2}, C, D_{3}$ are collinear.

Therefore, $\triangle A B D_{1} \cong \triangle B C D_{3} \cong \triangle A C D_{2}$, which means the tetrahedron is an equifacial tetrahedron. By property 1 of the equifacial tetrahedron, the angle formed by the skew lines $AD$ and $BC$ is $\arccos \frac{\left|b^{2}-c^{2}\right|}{a^{2}}$.