第10卷命题 42 · 二项线段的唯一分割点

Let AB be a binomial straight line divided into its terms at C; therefore AC, CB are rational straight lines commensurable in square only. [X. 36 ] I say that AB is not divided at another point into two rational straight lines commensurable in square only. For, if possible, let it be divided at D also, so that AD, DB are also rational straight lines commensurable in square only.

设AB为二项线段，在C点分割为AC和CB，则AC和CB是仅平方可公度的有理线段。

It is then manifest that AC is not the same with DB. For, if possible, let it be so.

假设AB在另一点D也被分割为AD和DB，且AD和DB也是仅平方可公度的有理线段。

Then AD will also be the same as CB, and, as AC is to CB, so will BD be to DA; thus AB will be divided at D also in the same way as by the division at C: which is contrary to the hypothesis. Therefore AC is not the same with DB. For this reason also the points C, D are not equidistant from the point of bisection.

若AC等于DB，则AD等于CB，导致分割方式相同，与假设矛盾；故AC不等于DB，且C、D不关于中点对称。

Therefore that by which the squares on AC, CB differ from the squares on AD, DB is also that by which twice the rectangle AD, DB differs from twice the rectangle AC, CB, because both the squares on AC, CB together with twice the rectangle AC, CB, and the squares on AD, DB together with twice the rectangle AD, DB, are equal to the square on AB. [II. 4 ] But the squares on AC, CB differ from the squares on AD, DB by a rational area, for both are rational; therefore twice the rectangle AD, DB also differs from twice the rectangle AC, CB by a rational area, though they are medial [X. 21 ]: which is absurd, for a medial area does not exceed a medial by a rational area.

由平方差关系，AC、CB的平方和与AD、DB的平方和之差为有理面积，但两倍矩形AD、DB与两倍矩形AC、CB之差也为有理面积，而两者均为中项面积，矛盾（中项面积之差不能为有理面积）。