Exact crossing counts at a polyhedron vertex: re-centre the projection on the plane

To show a generic non-supporting plane through vertex v of a convex polyhedron crosses at least two faces at v, without proving that the faces around v form one cycle, do this. Pick an interior point c on that plane: take two vertices on opposite sides, shrink them by (1-ε), and interpolate. Rebuild the polyhedron with origin c, using vertices V - c and polar points C/(1-⟪C,c⟫); every structure field transfers, and convexity uses triple(a-c,b-c,d-c) = (1-⟪C,c⟫)·triple(a,b,d) on the plane ⟪C,·⟫=1. In the radial tiling from c, the corner lunes at the projected vertex cover all directions. The plane direction d = (V v - c) × y and -d each lie in a corner lune. With ⟪a×b, c×d⟫ = ⟪a,c⟫⟪b,d⟫ - ⟪a,d⟫⟪b,c⟫ and ⟪V v - c, y⟫ = 0, the lune inequalities become ⟪e_in, y⟫ ≤ 0 ≤ ⟪e_out, y⟫ (reversed for -d). That gives two distinct crossed faces, and with 'at most two' an exact count, hence Gauss image area = defect and Euler's formula.