Let AB be bimedial, and let CD be commensurable in length with AB.
I say that CD is bimedial and the same in order with AB.
Since AB is bimedial, divide it into its medials at E.
Then AE and EB are medial straight lines commensurable in square only.
Let it be contrived that AB is to CD as AE is to CF. Then the remainder EB is to the remainder FD as AB is to CD.
But AB is commensurable in length with CD, therefore AE and EB are commensurable with CF and FD respectively.
But AE and EB are medial, therefore CF and FD are also medial.
Since AE is to EB as CF is to FD, and AE and EB are commensurable in square only, therefore CF and FD are also commensurable in square only.
But they were also proved medial, therefore CD is bimedial.
I say next that it is also the same in order with AB.
Since AE is to EB as CF is to FD, therefore the square on AE is to the rectangle AE by EB as the square on CF is to the rectangle CF by FD. Therefore, alternately, the square on AE is to the square on CF as the rectangle AE by EB is to the rectangle CF by FD.
But the square on AE is commensurable with the square on CF, therefore the rectangle AE by EB is commensurable with the rectangle CF by FD.
Therefore if the rectangle AE by EB is rational, then the rectangle CF by FD is also rational, and for this reason CD is a first bimedial, but if medial, medial, and each of the straight lines AB and CD is a second bimedial. And for this reason CD is the same in order with AB.
Therefore, a straight line commensurable with a bimedial straight line is itself also bimedial and the same in order.