If two medial straight lines commensurable in square only and containing a medial rectangle are added together, then the whole is irrational; let it be called the *second bimedial* straight line.

Let two medial straight lines *AB* and *BC* commensurable in square only and containing a medial rectangle be added together.

I say that *AC* is irrational.

Set out a rational straight line *DE*, and apply parallelogram *DF* to *DE* equal to the square on *AC*, producing *DG* as breadth.

Since the square on *AC* equals the sum of the squares on *AB* and *BC* and twice the rectangle *AB* by *BC*, apply *EH*, equal to the sum of the squares on *AB* and *BC*, to *DE*. Then the remainder *HF* equals twice the rectangle *AB* by *BC*.

Since each of the straight lines *AB* and *BC* is medial, therefore the squares on *AB* and *BC* are also medial. But, by hypothesis, twice the rectangle *AB* by *BC* is also medial. And *EH* equals the sum of the squares on *AB* and *BC*, while *FH* equals twice the rectangle *AB* by *BC*, therefore each of the rectangles *EH* and *HF* is medial.

And they are applied to the rational straight line *DE*, therefore each of the straight lines *DH* and *HG* is rational and incommensurable in length with *DE*.

Since *AB* is incommensurable in length with *BC*, and *AB* is to *BC* as the square on *AB* is to the rectangle *AB* by *BC*, therefore the square on *AB* is incommensurable with the rectangle *AB* by *BC*.

But the sum of the squares on *AB* and *BC* is commensurable with the square on *AB*, and twice the rectangle *AB* by *BC* is commensurable with the rectangle *AB* by *BC*.

Therefore the sum of the squares on *AB* and *BC* is incommensurable with twice the rectangle *AB* by *BC*.

But *EH* equals the sum of the squares on *AB* and *BC*, and *HF* equals twice the rectangle *AB* by *BC*.

Therefore *EH* is incommensurable with *HF*, so that *DH* is also incommensurable in length with *HG*.

Therefore *DH* and *HG* are rational straight lines commensurable in square only, so that *DG* is irrational.

But *DE* is rational, and the rectangle contained by an irrational and a rational straight line is irrational, therefore the area *DF* is irrational, and the side of the square equal to it is irrational.

But *AC* is the side of the square equal to *DF*, therefore *AC* is irrational. Let it be called a *second bimedial* straight line.

Q.E.D.