A major straight line is divided at one and the same point only.
大线段只能被分割于同一点。
本页以“大线段仅能被唯一分割”整体图解辅助阅读；点、线、角、圆索引已按命题文字和证明步骤校订，可与证明和问答联动。
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 a major straight line divided at C, so that AC, CB are incommensurable in square and make the sum of the squares on AC, CB rational, but the rectangle AC, CB medial; [X. 39 ] I say that AB is not so divided at another point.
设AB为大线段，被分割于C，使得AC、CB平方不可通约，且AC、CB上的平方和为有理，但AC、CB所成矩形为中项线。
For, if possible, let it be divided at D also, so that AD, DB are also incommensurable in square and make the sum of the squares on AD, DB rational, but the rectangle contained by them medial.
假设AB也被分割于另一点D，使得AD、DB平方不可通约，且AD、DB上的平方和为有理，但AD、DB所成矩形为中项线。
Then, since 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, while the squares on AC, CB exceed the squares on AD, DB by a rational area—for both are rational— therefore twice the rectangle AD, DB also exceeds twice the rectangle AC, CB by a rational area, though they are medial: which is impossible.
由于AC、CB上的平方和与AD、DB上的平方和之差等于两倍AD、DB所成矩形与两倍AC、CB所成矩形之差，且前者之差为有理面积（因两者均为有理），故两倍AD、DB所成矩形也以有理面积超过两倍AC、CB所成矩形。
但两倍AD、DB所成矩形与两倍AC、CB所成矩形均为中项线，中项线之差不能为有理面积，矛盾。因此AB不能被分割于另一点。