Is this a good layman's explanation for the duality relationship of Platonic solids?

There are 5 Platonic solids These are the polyhedra (3 dimensional polygons) whose sides, edges and faces are indistinguishable from one another. The tetrahedron and cube are familiar to everyone. You should recognize the dodecahedron as being used for paperweights containing the months of a calendar on its 12 sides.

The octahedron is just a pyramid glued to an upside down pyramid along the base. To visualize the icosahedron, notice that the picture of the light colored faces forms a pyramid with a pentagonal base. Form an icosahedron by taking two such pyramids and join their bases with two rings of triangles.

The main thing that I want to try to explain is the duality relationship among the 5 shapes. Notice the line in the picture labeled dual. It shows the tetrahedron dual to itself with duality relationships between cube and octahedron and between dodecahedron and icosahedron. The dual shapes are placed alongside each other, making it easy to see that the numbers in the first few lines are the same for dual shapes, but in a different order. The line labeled vertex configuration is interpreted to mean that the cube has four sided faces and 3 lines coming out of each vertex and the reverse for the octahedron.

We get the following correspondence between dual solids:
faces <-> vertices
edges per face <-> edges at each vertex
edges <-> edges

A cube has 6 faces, 8 vertices, 12 edges, 4 edges per face and 3 edges at each vertex.

An octahedron has 8 faces, 6 vertices, 12 edges, 3 edges per face and 4 edges at each vertex.

I made a diagram to try to explain why this works out. To go from cube to octahedron, place a vertex in the center of each face of the cube. This shows why the number of vertices of the octahedron is the same as the number of faces of the cube. The centers of adjacent faces of the cube are connected.

The diagram on the left shows the view at a vertex of the octahedron. Each edge of the octahedron crosses an edge of the face of the cube, so the number of vertex edges of the octahedron is the same as the number of edges on the face of the cube.

The diagram on the right shows a vertex of the cube. Each vertex of the cube becomes a face of the octahedron and the number of edges per face of the octahedron is the number of edges at each vertex of the cube.

Does this help? Am I just stating the obvious?