To find two medial straight lines commensurable in square only, containing a medial rectangle, and 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.

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. [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.