Kitap 2
14
But the squares on HE, EG are equal to the square on GH; [I. 47] therefore the rectangle BE, EF together with the square on GE is equal to the squares on HE, EG.
14
Let the square on GE be subtracted from each; therefore the rectangle contained by BE, EF which remains is equal to the square on EH.
14
But the rectangle BE, EF is BD, for EF is equal to ED; therefore the parallelogram BD is equal to the square on HE.
14
And BD is equal to the rectilineal figure A.
14
Therefore the rectilineal figure A is also equal to the square which can be described on EH.
14
Therefore a square, namely that which can be described on EH, has been constructed equal to the given rectilineal figure A. Q. E. F.
