If a rational and a medial area be added together, four irrational straight lines arise, namely a binomial or a first bimedial or a major or a side of a rational plus a medial area.
若一个有理面积与一个中项面积相加，则所得面积之边为以下四种无理线之一：二项线、第一双中项线、主线或有理中项面积边。
本页以“有理与中项面积和的无理线”整体图解辅助阅读；点、线、角、圆索引已按命题文字和证明步骤校订，可与证明和问答联动。
正文图形由校订坐标生成；点、线、角、圆可与证明和问答联动。
Let. AB be rational, and CD medial; I say that the “side” of the area AD is a binomial or a first bimedial or a major or a side of a rational plus a medial area. For AB is either greater or less than CD. First, let it be greater; let a rational straight line EF be set out, let there be applied to EF the rectangle EG equal to AB, producing EH as breadth, and let HI, equal to DC, be applied to EF, producing HK as breadth. Then, since AB is rational and is equal to EG, therefore EG is also rational. And it has been applied to EF, producing EH as breadth; therefore EH is rational and commensurable in length with EF. [X. 20] Again, since CD is medial and is equal to HI, therefore HI is also medial. And it is applied to the rational straight line EF, producing HK as breadth; therefore HK is rational and incommensurable in length with EF [X. 22] And, since CD is medial, while AB is rational, therefore AB is incommensurable with CD, so that EG is also incommensurable with HI.
设AB为有理面积，CD为中项面积，AD为两者之和。作有理线段EF，在其上作矩形EG等于AB，宽为EH；再作矩形HI等于CD，宽为HK。
But, as EG is to HI, so is EH to HK; [VI. 1] therefore EH is also incommensurable in length with HK. [X. 11] And both are rational; therefore EH, HK are rational straight lines commensurable in square only; therefore EK is a binomial straight line, divided at H. [X. 36] And, since AB is greater than CD, while AB is equal to EG and CD to HI, therefore EG is also greater than HI; therefore EH is also greater than HK. The square, then, on EH is greater than the square on HK either by the square on a straight line commensurable in length with EH or by the square on a straight line incommensurable with it. First, let the square on it be greater by the square on a straight line commensurable with itself. Now the greater straight line HE is commensurable in length with the rational straight line EF set out; therefore EK is a first binomial. [X. Deff. II. 1] But EF is rational; and, if an area be contained by a rational straight line and the first binomial, the side of the square equal to the area is binomial.
因AB有理，EG亦有理，故EH与EF可公度；又CD中项，HI亦中项，故HK与EF不可公度。且AB与CD不可公度，故EG与HI不可公度，从而EH与HK仅平方可公度，故EK为二项线，分于H。
[X. 54] Therefore the “side” of EI is binomial; so that the “side” of AD is also binomial. Next, let the square on EH be greater than the square on HK by the square on a straight line incommensurable with EH. Now the greater straight line EH is commensurable in length with the rational straight line EF set out; therefore EK is a fourth binomial. [X. Deff. II. 4] But EF is rational; and, if an area be contained by a rational straight line and the fourth binomial, the “side” of the area is the irrational straight line called major. [X. 57] Therefore the “side” of the area EI is major; so that the “side” of the area AD is also major. Next, let AB be less than CD; therefore EG is also less than HI, so that EH is also less than HK. Now the square on HK is greater than the square on EH either by the square on a straight line commensurable with HK or by the square on a straight line incommensurable with it.
若AB大于CD，则EH大于HK。若EH上的正方形大于HK上的正方形一可公度线段的正方形，则EK为第一二项线，故EI之边为二项线；若差为不可公度，则EK为第四二项线，故EI之边为主线。
First, let the square on it be greater by the square on a straight line commensurable in length with itself. Now the lesser straight line EH is commensurable in length with the rational straight line EF set out; therefore EK is a second binomial. [X. Deff. II. 2] But EF is rational, and, if an area be contained by a rational straight line and the second binomial, the side of the square equal to it is a first bimedial; [X. 55] therefore the “side” of the area EI is a first bimedial, so that the “side” of AD is also a first bimedial. Next, let the square on HK be greater than the square on HE by the square on a straight line incommensurable with HK. Now the lesser straight line EH is commensurable with the rational straight line EF set out; therefore EK is a fifth binomial. [X. Deff. II. 5] But EF is rational; and, if an area be contained by a rational straight line and the fifth binomial, the side of the square equal to the area is a side of a rational plus a medial area. [X. 58] Therefore the “side” of the area EI is a side of a rational plus a medial area, so that the “side” of the area AD is also a side of a rational plus a medial area.
若AB小于CD，则EH小于HK。若HK上的正方形大于EH上的正方形一可公度线段的正方形，则EK为第二二项线，故EI之边为第一双中项线；若差为不可公度，则EK为第五二项线，故EI之边为有理中项面积边。