A classical proof results by dividing the big square into two parallelograms JKDC and JKEB through the altitude from the right angle A. Then showing that each of them has area equal to the square having a common vertex with the corresponding parallelogram.

In fact, parallelogram CDKJ has area twice the area of the triangle ACD. But this triangle is equal to GCB which also has area half the area of the square GCAF. Thus the areas of CDKJ and CGFA are equal. Similarly the areas of JKEB and ABHI are equal.

p

The proof is a consequence of the theorem of Pythagoras. In fact, denote the lengths of the sides of the triangle by a=BC, b=CA, c=AB. Assume that the polygon has area (e) and its side MN has length m. Then by the similarity:

area(p

The proof results by substituting in the first equality a

Taking for shape p a half-circle and gluing it to the sides of the triangle we get as a result the well known theorem about the sum of the areas of the aqua colored lunes: the sum of their areas is equal to the area of the triangle.

There are many interesting discussions on Pythagora's theorem. Two of my favorits are: [YiuEGN, p.1] and [Bottema, p. 2].

See also the file SquaresDissection.html for another proof (of the innumerable existing) of this theorem.

[YiuEGN] Yiu, P.

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