If the square on a straight line be five times the square on a segment of it, then, when the double of the said segment is cut in extreme and mean ratio, the greater segment is the remaining part of the original straight line.
如果一条线段上的正方形是其一段上的正方形的五倍，那么当该段的两倍被分成中外比时，较大的一段是原线段剩余的部分。
底线 A→C→B→D：AC 满足 AB² = 5·AC²（B 介于 C 与 D 之间），CD = 2·AC。AB 上立正方形 AF，CD 上立正方形 CG；过 B 的竖线 BE 贯穿两正方形。H、K、L、M、N、O、Q 为拐尺形 MNO 与各分割线的交点。
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 the square on the straight line AB be five times the square on the segment AC of it, and let CD be double of AC; I say that, when CD is cut in extreme and mean ratio, the greater segment is CB. Let the squares AF, CG be described on AB, CD respectively, let the figure in AF be drawn, and let BE be drawn through. Now, since the square on BA is five times the square on AC, AF is five times AH. Therefore the gnomon MNO is quadruple of AH. And, since DC is double of CA, therefore the square on DC is quadruple of the square on CA, that is, CG is quadruple of AH. But the gnomon MNO was also proved quadruple of AH; therefore the gnomon MNO is equal to CG.
设线段AB上的正方形是AC上的正方形的五倍，且CD是AC的两倍。需证：当CD被分成中外比时，较大的一段是CB。
And, since DC is double of CA, while DC is equal to CK, and AC to CH, therefore KB is also double of BH. [VI. 1] But LH, HB are also double of HB; therefore KB is equal to LH, HB. But the whole gnomon MNO was also proved equal to the whole CG; therefore the remainder HF is equal to BG. And BG is the rectangle CD, DB, for CD is equal to DG; and HF is the square on CB; therefore the rectangle CD, DB is equal to the square on CB. Therefore, as DC is to CB, so is CB to BD. But DC is greater than CB; therefore CB is also greater than BD.
在AB和CD上分别作正方形AF和CG，并完成图形。由于BA上的正方形是AC上的正方形的五倍，故AF是AH的五倍，因此拐尺形MNO是AH的四倍。
Therefore, when the straight line CD is cut in extreme and mean ratio, CB is the greater segment. Therefore etc. Q. E. D. LEMMA.
因为DC是CA的两倍，所以DC上的正方形是CA上的正方形的四倍，即CG是AH的四倍。但已证拐尺形MNO也是AH的四倍，故拐尺形MNO等于CG。
That the double of AC is greater than BC is to be proved thus. If not, let BC be, if possible, double of CA. Therefore the square on BC is quadruple of the square on CA; therefore the squares on BC, CA are five times the square on CA. But, by hypothesis, the square on BA is also five times the square on CA; therefore the square on BA is equal to the squares on BC, CA: which is impossible. [II. 4] Therefore CB is not double of AC. Similarly we can prove that neither is a straight line less than CB double of CA; for the absurdity is much greater.
由于DC是CA的两倍，且DC等于CK，AC等于CH，故KB是BH的两倍。因此KB等于LH与HB之和，从而剩余部分HF等于BG。而BG是CD与DB所成矩形，HF是CB上的正方形，故CD:CB = CB:BD。因此CD被分成中外比，CB是较大的一段。