To find the first binomial straight line.
求作第一二项线段。
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Let two numbers AC, CB be set out such that the sum of them AB has to BC the ratio which a square number has to a square number, but has not to CA the ratio which a square number has to a square number; [Lemma I after X. 28] let any rational straight line D be set out, and let EF be commensurable in length with D. Therefore EF is also rational. Let it be contrived that, as the number BA is to AC, so is the square on EF to the square on FG. [X. 6, Por.] But AB has to AC the ratio which a number has to a number; therefore the square on EF also has to the square on FG the ratio which a number has to a number, so that the square on EF is commensurable with the square on FG.
设两数AC、CB，使其和AB与BC之比为平方数比平方数，但AB与CA之比不为平方数比平方数。
[X. 6] And EF is rational; therefore FG is also rational. And, since BA has not to AC the ratio which a square number has to a square number. neither, therefore, has the square on EF to the square on FG the ratio which a square number has to a square number; therefore EF is incommensurable in length with FG. [X. 9] Therefore EF, FG are rational straight lines commensurable in square only; therefore EG is binomial.
取任意有理线段D，作EF与D长度可公度，故EF为有理。令BA比AC等于EF上的正方形比FG上的正方形。
[X. 36] I say that it is also a first binomial straight line. For since, as the number BA is to AC, so is the square on EF to the square on FG, while BA is greater than AC, therefore the square on EF is also greater than the square on FG. Let then the squares on FG, H be equal to the square on EF. Now since, as BA is to AC, so is the square on EF to the square on FG, therefore, convertendo, as AB is to BC, so is the square on EF to the square on H.
因BA与AC之比不为平方数比平方数，故EF与FG长度不可公度，仅平方可公度，故EG为二项线。
[V. 19, Por.] But AB has to BC the ratio which a square number has to a square number; therefore the square on EF also has to the square on H the ratio which a square number has to a square number. Therefore EF is commensurable in length with H; [X. 9] therefore the square on EF is greater than the square on FG by the square on a straight line commensurable with EF. And EF, FG are rational, and EF is commensurable in length with D.
因BA大于AC，故EF上的正方形大于FG上的正方形。作H使FG、H上的正方形和等于EF上的正方形，由反比得AB比BC等于EF上的正方形比H上的正方形，且该比为平方数比，故EF与H长度可公度，即EF上的正方形比FG上的正方形大一个与EF可公度的线段上的正方形，且EF与D可公度，故EG为第一二项线。