The Garden of Archimedes  A Museum for Mathematics Pythagoras and his Theorem

Another very simple demonstration.

|    visual demonstration    |    and Airy's explanation    |    demonstration of Thabit Ibn Qurra    |

 The white area with the two yellow triangles forms the square of the hypothenuse, while with the two green triangles, equal to the previous ones, gives the squares of the cathets. Naturally, even here the visual evidence must be supported by a demonstration, which can be performed by anybody. It seems that the previous demonstration was found in 1855 by G. B. Airy , the Greenwich observatory astronomer from 1836 to 1881. In the white part of the figure, Airy wrote the poem that follows: I am, as you may see, a2 + b2 – ab. When two triangles on me stand, Square of hypothenuse is plann’d; But if I stand on them instead, The squares of both sides are read. The demonstration of Thabit Ibn Qurra. The following demonstration is attributed to the Arab mathematician Thabit Ibn Qurra (826-901). Starting from the right angled triangle ABC we make an irregular polygon ABDGLA by adding to the triangle the squares on the cathets ALHC and CBDE and the rectangle HCEG. This last one is divided by the diagonal GC in two right angled triangles, equal to the triangle ABC. Let now LI equal to BC and FD equal to AC; also the triangles ALI and BFD are equal to ABC. The same is true for the triangle IGF, because we have GI=AC and GF=BC. . Lastly, the quadrilateral AIFB has all sides equal and the angle IAB is a right angle, since it is equal to the angle LAC (the angles LAI and CAB are equal and the angle IAC is held in common) so AIFB is the square on the hypothenuse AB. At this point the demonstration is immediate. In fact, the irregular polygon ABDGLA can be divided both in the two squares on the cathets and the three triangles ABD, HCG and GCE; and in the square on the hypothenuse and the three triangles (equal to the first ones) FBD, IFG and ILA.