Example 3 In the tetrahedron $A-B C D$, $\angle D A B+$ $\angle B A C+\angle D A C=90^{\circ}$, and $\angle A D B=\angle B D C$ $=\angle A D C=90^{\circ}$. If $D B=a, D C=b$, try to calculate the volume of the tetrahedron $A-B C D$.
Solution: Using Figure 12, the lateral surface of the triangular pyramid unfolds to $A D_{1} B C D_{2}$. Extend $D_{1} B$ and $D_{2} C$ to intersect at $P$, then $A D_{1} P D_{2}$ is a square. We have
Rt $\triangle B P C \cong \mathrm{Rt} \triangle C D B$.
Let the side length of the square be $x$, i.e., $A D=x$.
From $D_{1} B=D B=a$ we get $B P=x-a$.

From $D_{2} C=C D=b$ we get $C P=x-b$.
Applying the Pythagorean theorem in Rt $\triangle B P C$, we get $(x-a)^{2}+(x-b)^{2}=B C^{2}=a^{2}+b^{2}$.
Solving for $x$ yields $x=a+b$.
Thus, $V_{A-B C D}=\frac{1}{3} A D \cdot S_{\triangle C D}$
$$
=\frac{1}{6} a b(a+b) .
$$