To a straight line which produces with a rational area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of the squares on them medial, but twice the rectangle contained by them rational.
对于一条与有理面构成中项全体的线段，只能附加一条线段，该线段与整条线段平方不可通约，且与整条线段所成两平方和为中项面，而所成矩形之二倍为有理面。
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正文图形由校订坐标生成；点、线、角、圆可与证明和问答联动。
Let AB be the straight line which produces with a rational area a medial whole, and let BC be an annex to AB; therefore AC, CB are straight lines incommensurable in square which fulfil the given conditions.
设AB为与有理面构成中项全体的线段，BC为其附加线，则AC、CB为平方不可通约且满足给定条件的线段。
[X. 77] I say that no other straight line can be annexed to AB which fulfils the same conditions.
假设可附加另一线段BD，则AD、DB也为平方不可通约且满足给定条件的线段。
For, if possible, let BD be so annexed; therefore AD, DB are also straight lines incommensurable in square which fulfil the given conditions.
与前例相同，AD、DB上平方和超过AC、CB上平方和的部分等于AD、DB所成矩形二倍超过AC、CB所成矩形二倍的部分，而AD、DB所成矩形二倍超过AC、CB所成矩形二倍的部分为有理面，因两者均为有理面。
[X. 77] Since then, as in the preceding cases, the excess of the squares on AD, DB over the squares on AC, CB is also the excess of twice the rectangle AD, DB over twice the rectangle AC, CB, while twice the rectangle AD, DB exceeds twice the rectangle AC, CB by a rational area, for both are rational, therefore the squares on AD, DB also exceed the squares on AC, CB by a rational area: which is impossible, for both are medial.
因此AD、DB上平方和超过AC、CB上平方和的部分为有理面，但两者均为中项面，此为不可能。