The Garden of Archimedes
 A Museum for Mathematics

Pythagoras and his theorem

A non right-angled triangle

|    Pappo´s construction    |    The Pythagorean Theorem as a specific case    |   

scheda4_1.gif In the Mathematical Collection by Pappo, a Greek Mathematician of the V century B.C., we find the following construction, which is also valid when the triangle ABC is not right angled. On the two sides we draw two parallelograms BCDE and ABFG having the sides AM and FG that meet in the point H when prolonged. We draw now a line HB, and taking the segment IL equal to HB , we draw a parallelogram ACMN, having the sides AM and CN parallel to IL. This parallelogram is equal to the sum of BCDE and AFGB.
To demonstrate that, we prolong the sides CN and AM. The parallelogram BCDE is equal to BCOH since they have the same base and stand between the two parallels BC and HD. For the same reason BCOH is equal to ICNL since the base HB is equal to IL and stands between the two parallels HL and OC. Consequently, the parallelograms BCDE and ICNL are equal. Similarly AFGB and AILM are equal and therefore the sum of BCDE and AFGB is equal to ACML.

scheda4_2.gif This result contains the Pythagorean Theorem as a specific case. In fact if the angle in B is a right angle and BCDE and AFGB are two squares, the segment HB, and therefore IL, is equal to the hypothenuse AC (since GBEH is a rectangle with the sides equal to the cathets), and the line HL is perpendicular to AC, so the parallelogram ACNM is the square of the hypothenuse.



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