Four, (20 points) 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 $BC$.
As shown in Figure 4, the lateral faces of the tetrahedron $D-ABC$ are cut along the lateral edges and unfolded onto the plane of the base.
Since $\angle D_{1} A B + \angle B A C + \angle D_{2} A C = 180^{\circ}$,
Therefore, points $D_{1}, A, D_{2}$ are collinear.
Similarly, points $D_{1}, B, D_{3}$ are collinear.
Since $A D_{1} = A D_{2}, B D_{1} = B D_{3}$.
Therefore, $AB$ is the midline of $\triangle D_{1} D_{2} D_{3}$.
Hence $D_{2} D_{3} = 2 AB = 2c$.
And $C D_{2} = C D_{3} = c$, so $C D_{2} + C D_{3} = D_{1} D_{2}$.
Therefore, points $D_{2}, C, D_{3}$ are collinear.
Thus, $BC = D_{1} A = a, AC = D_{1} B = b$.
Therefore, the three pairs of opposite edges of the tetrahedron $D-ABC$ are equal.
As shown in Figure 5, the tetrahedron $D-ABC$ is placed in a rectangular prism. Since $A^{\prime} D^{\prime} \parallel A D$, the angle $\angle A^{\prime} O B$ (or its supplementary angle) is the angle formed by the skew lines $AD$ and $BC$. Let the length, width, and height of the rectangular prism be $x, y, z$, respectively, then we have
$$
\left\{\begin{array}{l}
x^{2} + y^{2} = a^{2} \\
x^{2} + z^{2} = b^{2} \\
y^{2} + z^{2} = c^{2} .
\end{array}\right.
$$

Solving for $x^{2} = \frac{1}{2}\left(a^{2} + b^{2} - c^{2}\right)$.
Then $\cos \angle A^{\prime} O B = \cos \left(\angle A^{\prime} C B + \angle C A^{\prime} D^{\prime}\right)$
$$
\begin{array}{l}
= \cos 2 \angle A^{\prime} C B = 2 \cos ^{2} \angle A^{\prime} C B - 1 \\
= 2\left(\frac{x}{a}\right)^{2} - 1 = \frac{a^{2} + b^{2} - c^{2}}{a^{2}} - 1
\end{array}
$$
$$
= \frac{b^{2} - c^{2}}{a^{2}} .
$$

Therefore, the angle formed by the skew lines $AD$ and $BC$ is
$$
\arccos \frac{\left|b^{2} - c^{2}\right|}{a^{2}} .
$$