(Part of the proof for Hilbert's 3rd problem)
The definition of Dehn invariants seems quaint, if somehow pointless. But you'd better read on that there their definition, or this becomes an even larger load of gobbledegook than usual. Only when you see this does their relevance to equidecomposability become clear(er). The proof of this "additivity" is quite obvious; the key thing is to formulate and understand it.
Suppose we are given a decomposition of a polyhedron B⊂R^{3} into disjoint (except at the faces, etc. etc., blah blah blah) polyhedra B_{1}, ..., B_{n}. That is, B=B_{1}∪...∪B_{n}, and for any i≠j, B_{i}∩B_{j} is either empty or lies entirely within some plane. Then for any Dehn invariant D_{f} defined^{1} on the dihedral angles of B_{1},...,B_{n},
D_{f}(B) = D_{f}(B_{1})+...+D_{f}(B_{n}).
Wait for "Aha!" moment...
Idea of proof:
The edges of B, along which we shall sum up D_{f}(B), are almost precisely the edges of all the B_{i}'s. The only exceptions are such edges (or shorter segments of edges) of B_{i}'s which are completely surrounded by B, or which are halfsurrounded and become part of faces of B. For the latter 2, our insistence that f(2π)=f(π)= becomes useful: along those lengths for which the conditions are true, the linearity of f means we get a contribution of 0 to the sum.
For the remaining edges, we have merely to apply linearity, and our proof will be completed.
Notes

We need to consider all B_{i}'s simultaneously in order to have a vector space M which contains all their dihedral angles. Note, however, that we do not need explicitly to include the angles of B in M, because the construction used in the proof also shows that every dihedral angle of B is a sum of dihedral angles of the B_{i}'s.