The construction is this: squares are formed outwardly on sides AC and BC of ΔABC and the diagram os complemented by three parallelograms - CTPR, STPM, and QRPN. The resulting heptagon ABQNPMS can be cut in two ways ...
The proof follows from an observation that CP is equal and perpendicular to AB. Indeed, triangles ABC and CTR are friendly, which implies that the median of the latter through the vertex C is altitude of the former. But, since CTPR is a parallelogram, its diagonals divide each other into equal halves, CP is indeed the median of ΔCTR.
As Thâbit ibn Qurra's proof of the Pythagorean proposition that has an unfolded variant, so does its generalization above. The unfolded variant of the proof, appears on a separate page.
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