For let a rational straight line DE be set out, and let the parallelogram DF equal to the square on AC be applied to DE, producing DG as breadth. [I. 44 ]
Then, since the square on AC is equal to the squares on AB, BC and twice the rectangle AB, BC, [II. 4 ] let EH, equal to the squares on AB, BC, be applied to DE;
therefore the remainder HF is equal to twice the rectangle AB, BC.
And, since each of the straight lines AB, BC is medial, therefore the squares on AB, BC are also medial.
But, by hypothesis, twice the rectangle AB, BC is also medial.
And EH is equal to the squares on AB, BC, while FH is equal to twice the rectangle AB, BC;
therefore each of the rectangle EH, HF is medial.
And they are applied to the rational straight line DE;
therefore each of the straight lines DH, HG is rational and incommensurable in length with DE. [X. 22 ]
Since then AB is incommensurable in length with BC, and, as AB is to BC, so is the square on AB to the rectangle AB, BC,
therefore the square on AB is incommensurable with the rectangle AB, BC. [X. 11 ]
But the sum of the squares on AB, BC is commensurable with the square on AB, [X. 15 ] and twice the rectangle AB, BC is commensurable with the rectangle AB, BC. [X. 6 ]
Therefore the sum of the squares on AB, BC is incommensurable with twice the rectangle AB, BC. [X. 13 ]
But EH is equal to the squares on AB, BC, and HF is equal to twice the rectangle AB, BC.
Therefore EH is incommensurable with HF, so that DH is also incommensurable in length with HG. [VI. 1 , X. 11 ]
Therefore DH, HG are rational straight lines commensurable in square only;
so that DG is irrational. [X. 36 ]
But DE is rational; and the rectangle contained by an irrational and a rational straight line is irrational; [cf. X. 20 ]
therefore the area DF is irrational, and the side of the square equal to it is irrational. [X. Def. 4 ]
But AC is the side of the square equal to DF; therefore AC is irrational.
And let it be called a second bimedial straight line.
Q. E. D.
Complete text
Book I