#### PROPOSITION 44.

A second bimedial straight line is divided at one point only.

Let AB be a second bimedial straight line divided at C, so that AC, CB are medial straight lines commensurable in square only and containing a medial rectangle; [X. 38 ] it is then manifest that C is not at the point of bisection, because the segments are not commensurable in length.

I say that AB is not so divided at another point.

For, if possible, let it be divided at D also, so that AC is not the same with DB, but AC is supposed greater; it is then clear that the squares on AD, DB are also, as we proved above [Lemma], less than the squares on AC, CB; and suppose that AD, DB are medial straight lines commensurable in square only and containing a medial rectangle.

Now let a rational straight line EF be set out, let there be applied to EF the rectangular parallelogram EK equal to the square on AB, and let EG equal to the squares on AC, CB be subtracted; therefore the remainder HK is equal to twice the rectangle AC, CB. [II. 4 ]

Again, let there be subtracted EL, equal to the squares on AD, DB, which were proved less than the squares on AC, CB [Lemma ]; therefore the remainder MK is also equal to twice the rectangle AD, DB.

Now, since the squares on AC, CB are medial, therefore EG is medial.

And it is applied to the rational straight line EF; therefore EH is rational and incommensurable in length with EF. [X. 22 ]

For the same reason HN is also rational and incommensurable in length with EF.

And, since AC, CB are medial straight lines commensurable in square only, therefore AC is incommensurable in length with CB.

But, as AC is to CB, so is the square on AC to the rectangle AC, CB; therefore the square on AC is incommensurable with the rectangle AC, CB. [X. 11 ]

But the squares on AC, CB are commensurable with the square on AC; for AC, CB are commensurable in square. [x. 15 ]

And twice the rectangle AC, CB is commensurable with the rectangle AC, CB. [X. 6 ]

Therefore the squares on AC, CB are also incommensurable with twice the rectangle AC, CB. [X. 13 ]

But EG is equal to the squares on AC, CB, and HK is equal to twice the rectangle AC, CB; therefore EG is incommensurable with HK, so that EH is also incommensurable in length with HN. [VI. 1 , X. 11 ]

And they are rational; therefore EH, HN are rational straight lines commensurable in square only.

But, if two rational straight lines commensurable in square only be added together, the whole is the irrational which is called binomial. [X. 36 ]

Therefore EN is a binomial straight line divided at H.

In the same way EM, MN will also be proved to be rational straight lines commensurable in square only; and EN will be a binomial straight line divided at different points, H and M.

And EH is not the same with MN.

For the squares on AC, CB are greater than the squares on AD, DB.

But the squares on AD, DB are greater than twice the rectangle AD, DB; therefore also the squares on AC, CB, that is, EG, are much greater than twice the rectangle AD, DB, that is, MK, so that EH is also greater than MN.

Therefore EH is not the same with MN. Q. E. D.

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