For AB is either greater or less than CD.

First, if it so chance, let AB be greater than CD.

Let the rational straight line EF be set out, and to EF let there be applied the rectangle EG equal to AB and producing EH as breadth, and the rectangle HI equal to CD and producing HK as breadth.

Now, since each of the areas AB, CD is medial, therefore each of the areas EG, HI is also medial.

And they are applied to the rational straight line FE, producing EH, HK as breadth; therefore each of the straight lines EH, HK is rational and incommensurable in length with EF. [[X. 22](elem.10.22)]

And, since AB is incommensurable with CD, and AB is equal to EG, and CD to HI, therefore EG is also incommensurable with HI.

But, as EG is to HI, so is EH to HK; [[VI. 1](elem.6.1)] therefore EH is incommensurable in length with HK. [[X. 11](elem.10.11)]

Therefore EH, HK are rational straight lines commensurable in square only; therefore EK is binomial. [[X. 36](elem.10.36)]

But the square on EH is greater than the square on HK either by the square on a straight line commensurable 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 in length with itself.

Now neither of the straight lines EH, HK is commensurable in length with the rational straight line EF set out; therefore EK is a third binomial. [[X. Deff. II. 3](elem.10.def.2.3)]

But EF is rational; and, if an area be contained by a rational straight line and the third binomial, the side

of the area is a second bimedial; [[X. 56](elem.10.56)] therefore the side

of EI, that is, of AD, is a second bimedial.

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

Now each of the straight lines EH, HK is incommensurable in length with EF; therefore EK is a sixth binomial. [[X. Deff. II. 6](elem.10.def.2.6)]

But, if an area be contained by a rational straight line and the sixth binomial, the side

of the area is the side of the sum of two medial areas; [[X. 59](elem.10.59)] so that the side

of the area AD is also the side of the sum of two medial areas.

Therefore etc. Q. E. D.