For, since AB is incommensurable in length with BC, therefore the squares on AB, BC are also incommensurable with twice the rectangle AB, BC; [cf. X. 36, ll. 9-20]
and, componendo, the squares on AB, BC together with twice the rectangle AB, BC, that is, the square on AC [II. 4], is incommensurable with the rectangle AB, BC. [X. 16 ]
But the rectangle AB, BC is rational,
for, by hypothesis, AB, BC are straight lines containing a rational rectangle;
therefore the square on AC is irrational;
therefore AC is irrational. [X. Def. 4 ] And let it be called a first bimedial straight line.
Q. E. D.