第10卷命题 31 · 求作两中项线满足特定条件

Let there be set out two rational straight lines A, B commensurable in square only and such that the square on A, being the greater, is greater than the square on B the less by the square on a straight line commensurable in length with A. [X. 29] And let the square on C be equal to the rectangle A, B. Now the rectangle A, B is medial; [X. 21] therefore the square on C is also medial; therefore C is also medial.

设两有理线段A、B仅平方可通约，且A较大，A上的正方形比B上的正方形大出一个与A长度可通约的线段上的正方形。

[X. 21] Let the rectangle C, D be equal to the square on B. Now the square on B is rational; therefore the rectangle C, D is also rational. And since, as A is to B, so is the rectangle A, B to the square on B, while the square on C is equal to the rectangle A, B, and the rectangle C, D is equal to the square on B, therefore, as A is to B, so is the square on C to the rectangle C, D.

作C上的正方形等于A、B所成矩形；因该矩形为中项线，故C也为中项线。

But, as the square on C is to the rectangle C, D, so is C to D; therefore also, as A is to B, so is C to D. But A is commensurable with B in square only; therefore C is also commensurable with D in square only. [X. 11] And C is medial; therefore D is also medial.

作C、D所成矩形等于B上的正方形；因B上的正方形为有理，故该矩形也为有理。

[X. 23, addition] And since, as A is to B, so is C to D, and the square on A is greater than the square on B by the square on a straight line commensurable with A, therefore also the square on C is greater than the square on D by the square on a straight line commensurable with C. [X. 14] Therefore two medial straight lines C, D, commensurable in square only and containing a rational rectangle, have been found, and the square on C is greater than the square on D by the square on a straight line commensurable in length with C.

由A比B等于C比D，且A与B仅平方可通约，得C与D也仅平方可通约；又A上的正方形比B上的正方形大出与A可通约的线段上的正方形，故C上的正方形比D上的正方形大出与C可通约的线段上的正方形。