If a straight line be cut in extreme and mean ratio, the square on the greater segment added to the half of the whole is five times the square on the half.
若一线段被分为中外比，则较大段与整段一半之和上的正方形是整段一半上的正方形的五倍。
线段 AB 在 C 处被分为中末比（AC 为较大段），DA 沿 CA 延长且 AD = AB/2。AB 上立正方形 AE（顶点 A、B、E、K），DC 上立正方形 DF（顶点 D、C、F、G）；H、P 为通过 C、A 的水平/竖直辅助线落在边上的分点；L、M、N、O 标注拐尺形 MNO 各转角。
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, and let AC be the greater segment; let the straight line AD be produced in a straight line with CA, and let AD be made half of AB; I say that the square on CD is five times the square on AD. For let the squares AE, DF be described on AB, DC, and let the figure in DF be drawn; let FC be carried through to G. Now, since AB has been cut in extreme and mean ratio at C, therefore the rectangle AB, BC is equal to the square on AC.
设线段AB被点C分为中外比，AC为较大段；延长CA至D，使AD等于AB的一半。
[VI. Def. 3, VI. 17] And CE is the rectangle AB, BC, and FH the square on AC; therefore CE is equal to FH. And, since BA is double of AD, while BA is equal to KA, and AD to AH, therefore KA is also double of AH. But, as KA is to AH, so is CK to CH; [VI. 1] therefore CK is double of CH.
在AB和DC上分别作正方形AE和DF，并完成DF内的图形；延长FC至G。
But LH, HC are also double of CH. Therefore KC is equal to LH, HC. But CE was also proved equal to HF; therefore the whole square AE is equal to the gnomon MNO.
由于AB被C分为中外比，故矩形AB·BC等于AC上的正方形；又BA是AD的两倍，故CK是CH的两倍，从而KC等于LH与HC之和。
And, since BA is double of AD, the square on BA is quadruple of the square on AD, that is, AE is quadruple of DH. But AE is equal to the gnomon MNO; therefore the gnomon MNO is also quadruple of AP; therefore the whole DF is five times AP. And DF is the square on DC, and AP the square on DA; therefore the square on CD is five times the square on DA.
因此正方形AE等于拐尺形MNO；而BA是AD的两倍，故AE是DH的四倍，从而拐尺形MNO是AP的四倍，故整个DF是AP的五倍，即CD上的正方形是DA上的正方形的五倍。