If a straight line be cut in extreme and mean ratio, and there be added to it a straight line equal to the greater segment, the whole straight line has been cut in extreme and mean ratio, and the original straight line is the greater segment.
若一线段被分成中外比，且在其上加上等于较大分段的线段，则整条线段被分成中外比，且原线段为较大分段。
线段 AB 在 C 处被分为中末比（AC 为较大段）；在 A 左侧延长 AD 使 AD = AC，故 D-A-C-B 四点共线。AB 上立正方形 AE（顶点 A、B、E、K）；正方形 CH 落在 AC 上（顶点 A、C、L、H），矩形 DK 横跨 D-A-C 上方。
正文图形由校订坐标生成；点、线、角、圆可与证明和问答联动。
For let the straight line AB be cut in extreme and mean ratio at the point C, let AC be the greater segment, and let AD be equal to AC. I say that the straight line DB has been cut in extreme and mean ratio at A, and the original straight line AB is the greater segment. For let the square AE be described on AB, and let the figure be drawn.
设线段AB被点C分成中外比，AC为较大分段，作AD等于AC。
Since AB has been cut in extreme and mean ratio at C, therefore the rectangle AB, BC is equal to the square on AC. [VI. Def. 3, VI. 17] And CE is the rectangle AB, BC, and CH the square on AC; therefore CE is equal to HC.
在AB上作正方形AE，并完成图形。因AB被C分成中外比，故矩形AB, BC等于AC上的正方形。
But HE is equal to CE, and DH is equal to HC; therefore DH is also equal to HE. Therefore the whole DK is equal to the whole AE. And DK is the rectangle BD, DA, for AD is equal to DL; and AE is the square on AB; therefore the rectangle BD, DA is equal to the square on AB.
矩形CE为AB, BC，正方形CH为AC上的正方形，故CE等于HC。又HE等于CE，DH等于HC，故DH等于HE。
Therefore, as DB is to BA, so is BA to AD. [VI. 17] And DB is greater than BA; therefore BA is also greater than AD.
因此整个DK等于整个AE。而DK为矩形BD, DA（因AD等于DL），AE为AB上的正方形，故矩形BD, DA等于AB上的正方形。所以DB比BA等于BA比AD，且DB大于BA，故BA大于AD。