The side of a rational plus a medial area is divided at one point only.
有理中矩面边（即两线段平方和为中矩面、二倍矩形为有理面的线段）仅能被一点分割。
本页以“有理中矩面边仅一点可分”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 the side of a rational plus a medial area divided at C, so that AC, CB are incommensurable in square and make the sum of the squares on AC, CB medial, but twice the rectangle AC, CB rational; [X. 40 ] I say that AB is not so divided at another point.
设AB为有理中矩面边，被C分割，使AC、CB平方不可通约，且AC、CB平方和为中矩面，二倍矩形AC·CB为有理面。
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 medial, but twice the rectangle AD, DB rational.
假设AB也被D分割，使AD、DB平方不可通约，且AD、DB平方和为中矩面，二倍矩形AD·DB为有理面。
Since then that by which twice the rectangle AC, CB differs from twice the rectangle AD, DB is also that by which the squares on AD, DB differ from the squares on AC, CB, while twice the rectangle AC, CB exceeds twice the rectangle AD, DB by a rational area, therefore the squares on AD, DB also exceed the squares on AC, CB by a rational area, though they are medial: which is impossible.
由于二倍矩形AC·CB与二倍矩形AD·DB的差等于AD、DB平方和与AC、CB平方和的差，且前者差为有理面，故后者差也为有理面。
但AD、DB平方和与AC、CB平方和均为中矩面，其差为有理面不可能，故假设不成立，AB仅能被一点分割。