第10卷命题 78 · 产生中面中面余线的无理量

For from the straight line AB let there be subtracted the straight line BC incommensurable in square with AB and fulfilling the given conditions; [X. 35] I say that the remainder AC is the irrational straight line called that which produces with a medial area a medial whole. For let a rational straight line DI be set out, to DI let there be applied DE equal to the squares on AB, BC, producing DG as breadth, and let DH equal to twice the rectangle AB, BC be subtracted. Therefore the remainder FE is equal to the square on AC, [II. 7] so that AC is the “side” of FE. Now, since the sum of the squares on AB, BC is medial and is equal to DE, therefore DE is medial. 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 twice the rectangle AB, BC is medial and is equal to DH, therefore DH is medial. And it is applied to the rational straight line DI, producing DF as breadth; therefore DF is also rational and incommensurable in length with DI. [X. 22] And, since the squares on AB, BC are incommensurable with twice the rectangle AB, BC, therefore DE is also incommensurable with DH. But, as DE is to DH, so also is DG to DF; [VI. 1] therefore DG 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] And FH is rational; but the rectangle contained by a rational straight line and an apotome is irrational, [deduction from X. 20] and its “side” is irrational. And AC is the “side” of FE; therefore AC is irrational.

从线段AB减去与AB平方不可通约且满足给定条件的线段BC，则余线段AC是被称为“与中面一起产生中面余线”的无理线段。