A first bimedial straight line is divided at one point only.
第一双中项线只能被分割于一点。
本页以“第一双中项线仅一点可分”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 a first bimedial straight line divided at C, so that AC, CB are medial straight lines commensurable in square only and containing a rational rectangle; I say that AB is not so divided at another point.
设AB为第一双中项线，被分割于C，使得AC、CB是仅平方可通约的中项线，且它们所成矩形为有理。
For, if possible, let it be divided at D also, so that AD, DB are also medial straight lines commensurable in square only and containing a rational rectangle.
假设AB也被分割于另一点D，使得AD、DB也是仅平方可通约的中项线，且它们所成矩形为有理。
Since, then, that by which twice the rectangle AD, DB differs from twice the rectangle AC, CB is that by which the squares on AC, CB differ from the squares on AD, DB, while twice the rectangle AD, DB differs from twice the rectangle AC, CB by a rational area—for both are rational— therefore the squares on AC, CB also differ from the squares on AD, DB by a rational area, though they are medial: which is absurd.
由于两倍矩形AD、DB与两倍矩形AC、CB的差等于AC、CB上的正方形和与AD、DB上的正方形和的差，而两倍矩形AD、DB与两倍矩形AC、CB的差是有理面积（因为两者均为有理），因此AC、CB上的正方形和与AD、DB上的正方形和的差也是有理面积。
但AC、CB上的正方形和与AD、DB上的正方形和均为中项面积，其差为有理面积是不可能的（根据X.26）。