But if they are not equal, produce one of them BE to F make EF equal to ED, [I. 3.
and bisect BF at G; [I. 10.
from the centre G, at the distance GB, or GF, describe the semicircle BHF, and produce DE to H.
The square described on EH shall be equal to the given rectilineal figure A.
Join GH. Then, because the straight line BF is divided into two equal parts at the point G, and into two unequal parts at the point E, the rectangle BE, EF, together with the square on GE, 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 square on GH is equal to the squares on GE, EH;[I.47.
therefore the rectangle BE, EF, together with the square on GE, is equal to the squares on GE, EH.
Take away the square on GE, which is common to both;
therefore the rectangle BE, EF is equal to the square on EH. [Axiom 3.
But the rectangle contained by BE, EF is the parallelogram BD,
because EF is equal to ED. [Construction.
Therefore BD is equal to the square on EH.
But BD is equal to the rectilineal figure A. [Construction.
Therefore the square on EH is equal to the rectilineal figure A.
Wherefore a square has been made equal to the given rectilineal figure A, namely, the square described on EH.q.e.f
