If a straight line be cut in extreme and mean ratio, the square on the whole and the square on the lesser segment together are triple of the square on the greater segment.
若一线段被分成中外比，则整段上的正方形与较小段上的正方形之和等于较大段上的正方形的三倍。
线段 AB 在 C 处被分为中末比（AC 为较大段）。AB 上立正方形 ADEB（A、B 为底边，D、E 为顶边）；过 C 的竖线与过 H 的横线把方形分成正方形 HG（边为 AC）、矩形 AK（即 AB·BC）以及拐尺形 LMN。F、G、H、K、L、M、N 为相应交点／标注。
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 straight line, let it be cut in extreme and mean ratio at C, and let AC be the greater segment; I say that the squares on AB, BC are triple of the square on CA. For let the square ADEB be described on AB, and let the figure be drawn.
设线段AB被点C分成中外比，且AC为较大段。在AB上作正方形ADEB，并完成图形。
Since then AB has been cut in extreme and mean ratio at C, and AC is the greater segment, therefore the rectangle AB, BC is equal to the square on AC. [VI. Def. 3, VI. 17] And AK is the rectangle AB, BC, and HG the square on AC; therefore AK is equal to HG.
由于AB被C分成中外比，AC为较大段，故矩形AB·BC等于AC上的正方形。又AK为矩形AB·BC，HG为AC上的正方形，因此AK等于HG。
And, since AF is equal to FE, let CK be added to each; therefore the whole AK is equal to the whole CE; therefore AK, CE are double of AK. But AK, CE are the gnomon LMN and the square CK; therefore the gnomon LMN and the square CK are double of AK.
因AF等于FE，各加CK，则整体AK等于整体CE，故AK与CE之和为AK的二倍。但AK与CE构成拐尺形LMN加正方形CK，故拐尺形LMN与正方形CK之和为AK的二倍。
But, further, AK was also proved equal to HG; therefore the gnomon LMN and the squares CK, HG are triple of the square HG. And the gnomon LMN and the squares CK, HG are the whole square AE and CK, which are the squares on AB, BC, while HG is the square on AC.
又已证AK等于HG，故拐尺形LMN与正方形CK、HG之和为HG的三倍。而拐尺形LMN与正方形CK、HG构成整个正方形AE加CK，即AB与BC上的正方形，HG为AC上的正方形。