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

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

Then, since *A* and *B* are rational and commensurable in square only, therefore the rectangle *A* by *B*, that is, the square on *C*, is medial. Therefore *C* is medial.

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

And *C* is medial, therefore *D* is also medial.

Therefore *C* and *D* are medial and commensurable in square only.

I say that they also contain a rational rectangle.

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

But *A* is to *C* as *C* is to *B*, therefore *C* is to *B* as *B* is to *D*. Therefore the rectangle *C* by *D* equals the square on *B*. But the square on *B* is rational, therefore the rectangle *C* by *D* is also rational.

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

Q.E.D.

If we set *A* to have length 1, and *B* length √*b*, an irrational number where *b* is rational, then *C* has length *b*^{1/4} and *D* has length *b*^{3/4}. Then *C* and *D* are medial and commensurable in square only, and the *b*^{1/4} by *b*^{3/4} rectangle they contain has area *b*, a rational number.

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