The rectangle contained by medial straight lines commensurable in square only is either rational or medial.

Let the rectangle *AC* be contained by the medial straight lines *AB* and *BC* which are commensurable in square only.

I say that *AC* is either rational or medial.

Describe the squares *AD* and *BE* on *AB* and *BC*. Then each of the squares *AD* and *BE* is medial.

Set out a rational straight line *FG*. Apply the rectangular parallelogram *GH* to *FG* equal to *AD*, producing *FH* as breadth, apply the rectangular parallelogram *MK* to *HM* equal to *AC*, producing *HK* as breadth, and further apply similarly *NL* to *KN* equal to *BE*, producing *KL* as breadth. Then *FH*, *HK*, and *KL* are in a straight line.

Since each of the squares *AD* and *BE* is medial, and *AD* equals *GH* while *BE* equals *NL*, therefore each of the rectangles *GH* and *NL* is also medial.

And they are applied to the rational straight line *FG*, therefore each of the straight lines *FH* and *KL* is rational and incommensurable in length with *FG*.

And, since *AD* is commensurable with *BE*, therefore *GH* is commensurable with *NL*. And *GH* is to *NL* as *FH* is to *KL*, therefore *FH* is commensurable in length with *KL*.

Therefore *FH* and *KL* are rational straight lines commensurable in length, therefore the rectangle *FH* by *KL* is rational.

And, since *DB* equals *BA* while *OB* equals *BC*, therefore *DB* is to *BC* as *AB* is to *BO*.

But *DB* is to *BC* as *DA* is to *AC*, and *AB* is to *BO* as *AC* is to *CO*, therefore *DA* is to *AC* as *AC* is to *CO*.

But *AD* equals *GH*, *AC* equals *MK*, and *CO* equals *NL*, therefore *GH* is to *MK* as *MK* is to *NL*. Therefore *FH* is to *HK* as *HK* is to *KL*. Therefore the rectangle *FH* by *KL* equals the square on *HK*.

But the rectangle *FH* by *KL* is rational, therefore the square on *HK* is also rational.

Therefore *HK* is rational. And, if it is commensurable in length with *FG*, then *HN* is rational, but, if it is incommensurable in length with *FG*, then *KH* and *HM* are rational straight lines commensurable in square only, and therefore *HN* is medial.

Therefore *HN* is either rational or medial. But *HN* equals *AC*, therefore *AC* is either rational or medial.

Therefore, *the rectangle contained by medial straight lines commensurable in square only is either rational or medial.*

Q.E.D.

To say a rectangle with sides *a* and *b* is rational says that *ab* is a rational number. To say that it’s medial says that (*ab*)^{2} is a rational number but *ab* is an irrational number. Therefore, to say it is either rational or medial says only that (*ab*)^{2} is a rational number.

In this proposition we are given that *a* and *b* are medial, so both *a*^{4} and *b*^{4} are rational numbers, and we are given that *a*^{2}/*b*^{2} is a rational number. Therefore, (*ab*)^{2}, which equals *b*^{4}(*a*^{2}/*b*^{2}) is also a rational number.

This proposition is not used in the rest of the *Elements*.