For from the medial straight line AB let there be subtracted the medial straight line CB which is commensurable with the whole AB in square only and such that the rectangle AB, BC, which it contains with the whole AB, is medial; [X. 28] I say that the remainder AC is irrational; and let it be called a second apotome of a medial straight line. For let a rational straight line DI be set out, let DE equal to the squares on AB, BC be applied to DI, producing DG as breadth, and let DH equal to twice the rectangle AB, BC be applied to DI, producing DF as breadth; therefore the remainder FE is equal to the square on AC. [II. 7] Now, since the squares on AB, BC are medial and commensurable, therefore DE is also medial. [X. 15 and 23, Por.] And it is applied to the rational straight line DI, producing DG as breadth; therefore DG is rational and incommensurable in length with DI. [X. 22] Again, since the rectangle AB, BC is medial, therefore twice the rectangle AB, BC is also medial. [X. 23, Por.] And it is equal to DH; therefore DH is also medial. And it has been applied to the rational straight line DI, producing DF as breadth; therefore DF is rational and incommensurable in length with DI. [X. 22] And, since AB, BC are commensurable in square only, therefore AB is incommensurable in length with BC; therefore the square on AB is also incommensurable with the rectangle AB, BC. [X. 11] But the squares on AB, BC are commensurable with the square on AB, [X. 15] and twice the rectangle AB, BC is commensurable with the rectangle AB, BC; [X. 6] therefore twice the rectangle AB, BC is incommensurable with the squares on AB, BC. [X. 13] But DE is equal to the squares on AB, BC, and DH to twice the rectangle AB, BC; therefore DE is incommensurable with DH. But, as DE is to DH, so is GD to DF; [VI. 1] therefore GD is incommensurable with DF. [X. 11] And both are rational; therefore GD, DF are rational straight lines commensurable in square only; therefore FG is an apotome. [X. 73] But DI is rational, and the rectangle contained by a rational and an irrational straight line is irrational, [deduction from X. 20] and its ‘side’ is irrational. And AC is the ‘side’ of FE; therefore AC is irrational.
从一条中项线段AB减去另一条中项线段CB，CB与AB仅平方可通约，且AB与CB所成矩形为中项面，则余线段AC为无理线，称为第二中项线段的余线。
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设DI为有理线段，在DI上作矩形DE等于AB、BC上的正方形和，得宽DG；再作矩形DH等于二倍AB、BC所成矩形，得宽DF；则余量FE等于AC上的正方形。
由于AB、BC上的正方形均为中项面且可通约，故DE为中项面，且贴于有理线段DI上，得宽DG为有理线且与DI长度不可通约。
同理，二倍AB、BC所成矩形为中项面，故DH为中项面，得宽DF为有理线且与DI长度不可通约。又因AB与BC仅平方可通约，故AB上的正方形与AB、BC所成矩形不可通约，从而DE与DH不可通约，进而GD与DF不可通约。
GD与DF均为有理线且仅平方可通约，故FG为余线。由于DI为有理线，而有理线与无理线所成矩形为无理，其边为无理，故AC为无理线。