If a straight line be cut in extreme and mean ratio, the square on the lesser segment added to the half of the greater segment is five times the square on the half of the greater segment.
若一线段被分成中外比，则较小线段上的正方形加上较大线段一半上的正方形等于较大线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），D 为 AC 的中点。AB 上立正方形 AE 并“将图加倍”，得到下方原正方形与上方加倍图形 RS。F、G、H、K、L、M、N 是 D、C 处水平/竖直辅助线与各正方形边的交点；O、P、Q 标注拐尺形 OPQ 的角点。
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 any straight line AB be cut in extreme and mean ratio at the point C, let AC be the greater segment, and let AC be bisected at D; I say that the square on BD is five times the square on DC. For let the square AE be described on AB, and let the figure be drawn double. Since AC is double of DC, therefore the square on AC is quadruple of the square on DC, that is, RS is quadruple of FG. And, since the rectangle AB, BC is equal to the square on AC, and CE is the rectangle AB, BC, therefore CE is equal to RS.
设线段AB被点C分成中外比，AC为较大段，D为AC中点。需证BD上的正方形是DC上的正方形的五倍。
But RS is quadruple of FG; therefore CE is also quadruple of FG. Again, since AD is equal to DC, HK is also equal to KF. Hence the square GF is also equal to the square HL.
在AB上作正方形AE，并完成图形。因AC是DC的两倍，故AC上的正方形是DC上的正方形的四倍，即RS是FG的四倍。
Therefore GK is equal to KL, that is, MN to NE; hence MF is also equal to FE. But MF is equal to CG; therefore CG is also equal to FE. Let CN be added to each; therefore the gnomon OPQ is equal to CE. But CE was proved quadruple of GF; therefore the gnomon OPQ is also quadruple of the square FG.
由于AB与BC所成矩形等于AC上的正方形，且CE为AB与BC所成矩形，故CE等于RS。但RS是FG的四倍，因此CE也是FG的四倍。
Therefore the gnomon OPQ and the square FG are five times FG. But the gnomon OPQ and the square FG are the square DN. And DN is the square on DB, and GF the square on DC.
因AD等于DC，故HK等于KF，从而GF等于HL，GK等于KL，即MN等于NE，故MF等于FE。又MF等于CG，故CG等于FE。加CN于两者，得拐尺形OPQ等于CE。已证CE是GF的四倍，故拐尺形OPQ也是FG的四倍，因此拐尺形OPQ与FG之和为FG的五倍。但拐尺形OPQ与FG构成正方形DN，而DN是DB上的正方形，GF是DC上的正方形。