The rectangle contained by medial straight lines commensurable in square only is either rational or medial.
由仅平方可通约的两条中项线段所成的矩形，要么是有理矩形，要么是中项矩形。
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For let the rectangle AC be contained by the medial straight lines AB, BC which are commensurable in square only; I say that AC is either rational or medial. For on AB, BC let the squares AD, BE be described; therefore each of the squares AD, BE is medial. Let a rational straight line FG be set out, to FG let there be applied the rectangular parallelogram GH equal to AD, producing FH as breadth, to HM let there be applied the rectangular parallelogram MK equal to AC, producing HK as breadth, and further to KN let there be similarly applied NL equal to BE, producing KL as breadth; therefore FH, HK, KL are in a straight line. Since then each of the squares AD, BE is medial, and AD is equal to GH, and BE to NL, therefore each of the rectangles GH, NL is also medial.
设矩形AC由中项线段AB、BC围成，且AB与BC仅平方可通约。在AB、BC上作正方形AD、BE，两者均为中项。
And they are applied to the rational straight line FG; therefore each of the straight lines FH, KL is rational and incommensurable in length with FG. [X. 22] And, since AD is commensurable with BE, therefore GH is also commensurable with NL. And, as GH is to NL, so is FH to KL; [VI. 1] therefore FH is commensurable in length with KL. [X. 11] Therefore FH, KL are rational straight lines commensurable in length; therefore the rectangle FH, KL is rational.
取有理线段FG，作矩形GH等于AD，得宽FH；作矩形MK等于AC，得宽HK；作矩形NL等于BE，得宽KL。则FH、HK、KL共线。
[X. 19] And, since DB is equal to BA, and OB to BC, therefore, as DB is to BC, so is AB to BO. But, as DB is to BC, so is DA to AC, [VI. 1] and, as AB is to BO, so is AC to CO; [id.] therefore, as DA is to AC, so is AC to CO. But AD is equal to GH, AC to MK and CO to NL; therefore, as GH is to MK, so is MK to NL; therefore also, as FH is to HK, so is HK to KL; [VI. 1, V. 11] therefore the rectangle FH, KL is equal to the square on HK. [VI. 17] But the rectangle FH, KL is rational; therefore the square on HK is also rational.
由于AD、BE均为中项且分别等于GH、NL，故GH、NL均为中项，且它们贴在有理线段FG上，因此FH、KL均为有理线段且与FG长度不可通约。又AD与BE可通约，故GH与NL可通约，从而FH与KL长度可通约，因此矩形FH、KL为有理。
Therefore HK is rational. And, if it is commensurable in length with FG, HN is rational; [X. 19] but, if it is incommensurable in length with FG, KH, HM are rational straight lines commensurable in square only, and therefore HN is medial. [X. 21] Therefore HN is either rational or medial. But HN is equal to AC; therefore AC is either rational or medial.
由比例关系可得GH:MK = MK:NL，从而FH:HK = HK:KL，故矩形FH、KL等于HK上的正方形。因矩形FH、KL为有理，故HK上的正方形为有理，HK为有理线段。若HK与FG长度可通约，则HN为有理；若不可通约，则HN为中项。因此HN即AC为有理或中项。