A straight line commensurable with a bimedial straight line is itself also bimedial and the same in order.

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.*

Q.E.D.