If an area be contained by a rational straight line and the sixth binomial, the “side” of the area is the irrational straight line called the side of the sum of two medial areas.
若一面积由一条有理线段和第六条二项线围成，则该面积的“边”是被称为两中项面积之和的边的无理线段。
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For let the area ABCD be contained by the rational straight line AB and the sixth binomial AD, divided into its terms at E, so that AE is the greater term; I say that the “side” of AC is the side of the sum of two medial areas. Let the same construction be made as before shown. It is then manifest that MO is the “side” of AC, and that MN is incommensurable in square with NO. Now, since EA is incommensurable in length with AB, therefore EA, AB are rational straight lines commensurable in square only; therefore AK, that is, the sum of the squares on MN, NO, is medial.
设面积ABCD由有理线段AB和第六条二项线AD围成，AD被分为两段，其中AE是较大的一段。
[X. 21] Again, since ED is incommensurable in length with AB, therefore FE is also incommensurable with EK; [X. 13] therefore FE, EK are rational straight lines commensurable in square only; therefore EL, that is, MR, that is, the rectangle MN, NO, is medial. [X. 21] And, since AE is incommensurable with EF, AK is also incommensurable with EL. [VI. 1, X. 11] But AK is the sum of the squares on MN, NO, and EL is the rectangle MN, NO; therefore the sum of the squares on MN, NO is incommensurable with the rectangle MN, NO.
作与之前相同的构造，则MO是AC的边，且MN与NO在平方上不可公度。
And each of them is medial, and MN, NO are incommensurable in square. Therefore MO is the side of the sum of two medial areas [X. 41], and is the “side” of AC. Q.
由于EA与AB长度不可公度，故EA、AB是仅平方可公度的有理线段，因此AK（即MN、NO上的正方形之和）是中项面。
E. D. [LEMMA. If a straight line be cut into unequal parts, the squares on the unequal parts are greater than twice the rectangle contained by the unequal parts. Let AB be a straight line, and let it be cut into unequal parts at C, and let AC be the greater; I say that the squares on AC, CB are greater than twice the rectangle AC, CB. For let AB be bisected at D. Since then a straight line has been cut into equal parts at D, and into unequal parts at C, therefore the rectangle AC, CB together with the square on CD is equal to the square on AD, [II. 5] so that the rectangle AC, CB is less than double of the square on AD.
又因ED与AB长度不可公度，故FE与EK仅平方可公度，因此EL（即MR，亦即MN、NO所成矩形）是中项面；且AK与EL不可公度，故MN、NO上的正方形之和与它们所成矩形不可公度，且两者均为中项面，MN与NO在平方上不可公度，因此MO是两中项面积之和的边，即AC的边。