第10卷命题 32 · 求作两中项线平方可通

Let there be set out three rational straight lines A, B, C commensurable in square only, and such that the square on A is greater than the square on C by the square on a straight line commensurable with A, [X. 29] and let the square on D be equal to the rectangle A, B. Therefore the square on D is medial; therefore D is also medial. [X. 21] Let the rectangle D, E be equal to the rectangle B, C. Then since, as the rectangle A, B is to the rectangle B, C, so is A to C; while the square on D is equal to the rectangle A, B, and the rectangle D, E is equal to the rectangle B, C, therefore, as A is to C, so is the square on D to the rectangle D, E. But, as the square on D is to the rectangle D, E, so is D to E; therefore also, as A is to C, so is D to E. But A is commensurable with C in square only; therefore D is also commensurable with E in square only.

取三条仅平方可通的有理线段A、B、C，使A上的正方形比C上的正方形大出一个与A可通长的线段上的正方形。

[X. 11] But D is medial; therefore E is also medial. [X. 23, addition] And, since, as A is to C, so is D to E, while the square on A is greater than the square on C by the square on a straight line commensurable with A, therefore also the square on D will be greater than the square on E by the square on a straight line commensurable with D. [X. 14] I say next that the rectangle D, E is also medial. For, since the rectangle B, C is equal to the rectangle D, E, while the rectangle B, C is medial, [X. 21] therefore the rectangle D, E is also medial. Therefore two medial straight lines D, E, commensurable in square only, and containing a medial rectangle, have been found such that the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater.

作D上的正方形等于A、B所成矩形，则D上的正方形为中项，故D为中项线。作矩形D、E等于B、C所成矩形。

Similarly again it can be proved that the square on D is greater than the square on E by the square on a straight line incommensurable with D, when the square on A is greater than the square on C by the square on a straight line incommensurable with A. [X. 30] LEMMA. Let ABC be a right-angled triangle having the angle A right, and let the perpendicular AD be drawn; I say that the rectangle CB, BD is equal to the square on BA, the rectangle BC, CD equal to the square on CA, the rectangle BD, DC equal to the square on AD, and, further, the rectangle BC, AD equal to the rectangle BA, AC. And first that the rectangle CB, BD is equal to the square on BA. For, since in a right-angled triangle AD has been drawn from the right angle perpendicular to the base, therefore the triangles ABD, ADC are similar both to the whole ABC and to one another.

由A比C等于D上的正方形比矩形D、E，且等于D比E，故D与E仅平方可通。又D为中项，故E也为中项。

[VI. 8] And since the triangle ABC is similar to the triangle ABD, therefore, as CB is to BA, so is BA to BD; [VI. 4] therefore the rectangle CB, BD is equal to the square on AB. [VI. 17] For the same reason the rectangle BC, CD is also equal to the square on AC. And since, if in a right-angled triangle a perpendicular be drawn from the right angle to the base, the perpendicular so drawn is a mean proportional between the segments of the base, [VI. 8, Por.] therefore, as BD is to DA, so is AD to DC; therefore the rectangle BD, DC is equal to the square on AD. [VI. 17] I say that the rectangle BC, AD is also equal to the rectangle BA, AC. For since, as we said, ABC is similar to ABD, therefore, as BC is to CA, so is BA to AD.

由A上的正方形比C上的正方形大出与A可通长的线段上的正方形，得D上的正方形比E上的正方形大出与D可通长的线段上的正方形。且矩形D、E等于中项矩形B、C，故也为中项。