Let. AB be rational, and CD medial; I say that the side

of the area AD is a binomial or a first bimedial or a major or a side of a rational plus a medial area.

Then, since AB is rational and is equal to EG, therefore EG is also rational.

And it has been applied to EF, producing EH as breadth; therefore EH is rational and commensurable in length with EF. [[X. 20](elem.10.20)]

Again, since CD is medial and is equal to HI, therefore HI is also medial.

And it is applied to the rational straight line EF, producing HK as breadth; therefore HK is rational and incommensurable in length with EF [[X. 22](elem.10.22)]

And, since CD is medial, while AB is rational, therefore AB is incommensurable with CD, so that 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 also incommensurable in length with HK. [[X. 11](elem.10.11)]

And both are rational; therefore EH, HK are rational straight lines commensurable in square only; therefore EK is a binomial straight line, divided at H. [[X. 36](elem.10.36)]

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](elem.10.def.2.1)]

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](elem.10.54)]

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](elem.10.def.2.4)]

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](elem.10.57)]

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.

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

Now the lesser straight line EH is commensurable in length with the rational straight line EF set out; therefore EK is a second binomial. [[X. Deff. II. 2](elem.10.def.2.2)]

But EF is rational, and, if an area be contained by a rational straight line and the second binomial, the side of the square equal to it is a first bimedial; [[X. 55](elem.10.55)] therefore the side

of the area EI is a first bimedial, so that the side

of AD is also a first bimedial.

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

Now the lesser straight line EH is commensurable with the rational straight line EF set out; therefore EK is a fifth binomial. [[X. Deff. II. 5](elem.10.def.2.5)]

But EF is rational; and, if an area be contained by a rational straight line and the fifth binomial, the side of the square equal to the area is a side of a rational plus a medial area. [[X. 58](elem.10.58)]

Therefore the side

of the area EI is a side of a rational plus a medial area, so that the side

of the area AD is also a side of a rational plus a medial area.

Therefore etc. Q. E. D.