### Equidecomposeable polygons

We suppose that the two polygons below have the same area:  The theorem that we want to prove states that we can decompose the two polygons in lesser polygons, with the parts of first congruen to the parts of the second.
How do we do this? It can be laborious, but the idea is very simple!

We transform, as seen previously, both the polygons into squares of same area. This means that the squares are composites from the parts that compose the polygons.

It is clear that the parts of the squares are not necessarily in congruent. To get the congruence between the parts, it is enough to overlap the two squares and to make new cuts so that all the cuts in the first square appear in the second, and vice versa.

The final dissection, to dissect both the polygons with the same parts, can be quite complicated, but we conclude that it is possible! The figures below give an idea of how difficult can be:    To get the equidecomposition of the pentagon and hexagon it still remains to overlap the two squares and to make the necessary cuts, as explained previously, to have the same parts in both.

We reasoned using regular polygons. The assumption of regularity at no time is basic, it was only for the ease of drawing them; thus we can think in the same way for any two polygons that have the same area.

Final conclusion: With the construction of the puzzles, step by the step, we finished by showing that it is always possible to make equidecomposition of polygons that have the same area.
Mathematically it is a pretty and interesting result, involving perfectly adjusted knowledge and reasonings to high school level Geometry . Return to the Presentation