Euclid, Elements
Thomas L. Heath, Sir Thomas Little Heath, Ed.

And, since AB is greater than CD, while AB is equal to EG and CD to HI, therefore EG is also greater than HI; therefore EH is also greater than HK.

The square, then, on EH is greater than the square on HK either by the square on a straight line commensurable in length with EH or by the square on a straight line incommensurable with it.

First, let the square on it be greater by the square on a straight line commensurable with itself.

Now the greater straight line HE is commensurable in length with the rational straight line EF set out; therefore EK is a first binomial. [X. Deff. II. 1]

But EF is rational; and, if an area be contained by a rational straight line and the first binomial, the side of the square equal to the area is binomial. [X. 54]

Therefore the “side” of EI is binomial; so that the “side” of AD is also binomial.

Next, let the square on EH be greater than the square on HK by the square on a straight line incommensurable with EH.

Now the greater straight line EH is commensurable in length with the rational straight line EF set out; therefore EK is a fourth binomial. [X. Deff. II. 4]

But EF is rational; and, if an area be contained by a rational straight line and the fourth binomial, the “side” of the area is the irrational straight line called major. [X. 57]

Therefore the “side” of the area EI is major; so that the “side” of the area AD is also major.

Next, let AB be less than CD; therefore EG is also less than HI, so that EH is also less than HK.

Now the square on HK is greater than the square on EH either by the square on a straight line commensurable with HK or by the square on a straight line incommensurable with it.