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.