If a rational area is applied to a rational straight line, then it produces as breadth a straight line rational and commensurable in length with the straight line to which it is applied.

Let the rational area *AC* be applied to *AB*, a straight line once more rational in any of the aforesaid ways, producing *BC* as breadth.

I say that *BC* is rational and commensurable in length with *BA*.

Describe the square *AD* on *AB*. Then *AD* is rational.

But *AC* is also rational, therefore *DA* is commensurable with *AC*. And *DA* is to *AC* as *DB* is to *BC*. Therefore *DB* is also commensurable with *BC*, and *DB* equals *BA*. Therefore *AB* is also commensurable with *BC*.

But *AB* is rational, *C* therefore *BC* is also rational and commensurable in length with *AB*.

Therefore, *if a rational area is applied to a rational straight line, then it produces as breadth a straight line rational and commensurable in length with the straight line to which it is applied.*

Q.E.D.

This proposition is a converse of the last, except that it’s preceded by applying an area to a straight line to get the rectangle. That would be more evident if it read “If one side of a rational rectangle is rational, then the other side is rational and commensurable with the first.”

It’s often easier to understand the statement and proof they’re translated into modern language. Let *a* be the length of *AB*, and *b* be the length of *BC*. Since the rectangle *AC* is rational, that means its area *ab* is a rational number. Since *AB* is a rational line, that means *a*^{2} is a rational number. Therefore, *a*/*b* is a rational number, that is, lines *AB* and *BC* are commensurable. Also, *b*^{2} is a rational number, that is, line *BC* is a rational line.

This proposition is used frequently in Book X starting with X.26.