第10卷命题 52 · 求第五二项线

Let two numbers AC, CB be set out such that AB has not to either of them the ratio which a square number has to a square number; let any rational straight line D be set out, and let EF be commensurable with D; therefore EF is rational. Let it be contrived that, as CA is to AB, so is the square on EF to the square on FG. [X. 6, Por.] But CA has not to AB the ratio which a square number has to a square number; therefore neither has the square on EF to the square on FG the ratio which a square number has to a square number.

设两数AC、CB，使得AB与其中任一个之比不是平方数比平方数；取定有理线段D，作EF与D可公度，则EF为有理。

Therefore EF, FG are rational straight lines commensurable in square only; [X. 9] therefore EG is binomial. [X. 36] I say next that it is also a fifth binomial straight line.

令CA比AB等于EF上的正方形比FG上的正方形。因CA与AB之比不是平方数比平方数，故EF上的正方形与FG上的正方形之比也不是平方数比平方数，因此EF、FG为仅平方可公度的有理线段，故EG为二项线。

For since, as CA is to AB, so is the square on EF to the square on FG, inversely, as BA is to AC, so is the square on FG to the square on FE; therefore the square on GF is greater than the square on FE. Let then the squares on EF, H be equal to the square on GF; therefore, convertendo, as the number AB is to BC, so is the square on GF to the square on H. [V. 19, Por.] But AB has not to BC the ratio which a square number has to a square number; therefore neither has the square on FG to the square on H the ratio which a square number has to a square number.

又因BA比AC等于FG上的正方形比FE上的正方形，故GF上的正方形大于FE上的正方形。设GF上的正方形等于EF上的正方形与H上的正方形之和，则由换比，AB比BC等于GF上的正方形比H上的正方形。

Therefore FG is incommensurable in length with H; [X. 9] so that the square on FG is greater than the square on FE by the square on a straight line incommensurable with FG. And GF, FE are rational straight lines commensurable in square only, and the lesser term EF is commensurable in length with the rational straight line D set out.

因AB与BC之比不是平方数比平方数，故FG上的正方形与H上的正方形之比也不是平方数比平方数，因此FG与H长度不可公度，即FG上的正方形比FE上的正方形大出一个与FG不可公度的线段上的正方形。且GF、FE为仅平方可公度的有理线段，较小项EF与所取有理线段D可公度。