第10卷命题 47 · 两中项和之边仅可一点分割

Let AB be divided at C, so that AC, CB are incommensurable in square and make the sum of the squares on AC, CB medial, and the rectangle AC, CB medial and also incommensurable with the sum of the squares on them; I say that AB is not divided at another point so as to fulfil the given conditions. For, if possible, let it be divided at D, so that again AC is of course not the same as BD, but AC is supposed greater; let a rational straight line EF be set out, and let there be applied to EF the rectangle EG equal to the squares on AC, CB, and the rectangle HK equal to twice the rectangle AC, CB; therefore the whole EK is equal to the square on AB. [II. 4 ] Again, let EL, equal to the squares on AD, DB, be applied to EF; therefore the remainder, twice the rectangle AD, DB, is equal to the remainder MK.

设AB被C分割，使得AC、CB平方不可通约，且AC、CB上平方和为两中项，AC、CB所成矩形也为两中项，且与平方和不可通约。

And since, by hypothesis, the sum of the squares on AC, CB is medial, therefore EG is also medial. And it is applied to the rational straight line EF; therefore HE is rational and incommensurable in length with EF.

假设AB也被D分割满足同样条件，且AC大于BD。作有理线段EF，在其上作矩形EG等于AC、CB上平方和，矩形HK等于二倍AC、CB所成矩形，则EK等于AB上正方形。

[X. 22 ] For the same reason HN is also rational and incommensurable in length with EF. And, since the sum of the squares on AC, CB is incommensurable with twice the rectangle AC, CB, therefore EG is also incommensurable with GN, so that EH is also incommensurable with HN. [VI. 1 , X. 11 ] And they are rational; therefore EH, HN are rational straight lines commensurable in square only; therefore EN is a binomial straight line divided at H.

再作EL等于AD、DB上平方和，则剩余部分MK等于二倍AD、DB所成矩形。由假设，EG为两中项，故EH为有理且与EF长度不可通约；同理HN也为有理且与EF长度不可通约。

[X. 36 ] Similarly we can prove that it is also divided at M. And EH is not the same with MN; therefore a binomial has been divided at different points: which is absurd.

由于EG与GN不可通约，故EH与HN不可通约，且均为有理，因此EH、HN为仅平方可通约的有理线段，故EN为二项线被H分割。同理可证EN也被M分割，但EH与MN不同，故二项线被不同点分割，矛盾。