Then, since the straight line BF has been cut into equal segments at G, and into unequal segments at E, the rectangle contained by BE, EF together with the square on EG is equal to the square on GF. [II. 5]
But GF is equal to GH;
therefore the rectangle BE, EF together with the square on GE is equal to the square on GH.
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.
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.
But the rectangle BE, EF is BD,
for EF is equal to ED;
therefore the parallelogram BD is equal to the square on HE.
And BD is equal to the rectilineal figure A.
Therefore the rectilineal figure A is also equal to the square which can be described on EH.