To find medial straight lines commensurable in square only which contain a medial rectangle.

Set out the rational straight lines *A*, *B*, and *C* commensurable in square only. Take a mean proportional *D* between *A* and *B*. Let it be contrived that *B* is to *C* as *D* is to *E*.

Since *A* and *B* are rational straight lines commensurable in square only, therefore the rectangle *A* by *B*, that is, the square on *D*, is medial. Therefore *D* is medial.

And since *B* and *C* are commensurable in square only, and *B* is to *C* as *D* is to *E*, therefore *D* and *E* are also commensurable in square only.

But *D* is medial, therefore *E* is also medial.

Therefore *D* and *E* are medial straight lines commensurable in square only.

I say next that they also contain a medial rectangle.

Since *B* is to *C* as *D* is to *E*, therefore, alternately, *B* is to *D* as *C* is to *E*.

But *B* is to *D* as *D* is to *A*, therefore *D* is to *A* as *C* is to *E*. Therefore the rectangle *A* by *C* equals the rectangle *D* by *E*.

But the rectangle *A* by *C* is medial, therefore the rectangle *D* by *E* is also medial.

Therefore medial straight lines commensurable in square only have been found which contain a medial rectangle.

Q.E.D.

There has been no construction given in Book X for the required lines *A*, *B*, and *C*. One example would be where their lengths are 1, √2, and √ 3.

Set *A* to have length 1, *B* length √*b*, an irrational number where *b* is rational, and *C* length √*c*, an irrational number where *c* is rational, with *b*/*c* and irrational number. Then *D* has length *b*^{1/4} while *E* has length (√*c*)/*b*^{1/4}. Then *D* and *E* are medial and commensurable in square only, and the *b*^{1/4} by (√*c*)/*b*^{1/4} rectangle they contain has area √*c*, an irrational number.

This proposition is used in X.75.