If there are two straight lines, then the first is to the second as the square on the first is to the rectangle contained by the two straight lines.

Let *FE* and *EG* be two straight lines.

I say that *FE* is to *EG* as the square on *FE* is to the rectangle *FE* by *EG*.

Describe the square *DF* on *FE*, and complete *GD*.

Since then *FE* is to *EG* as *FD* is to *DG*, and *FD* is the square on *FE*, and *DG* the rectangle *DE* by *EG*, that is, the rectangle *FE* by *EG*, therefore *FE* is to *EG* as the square on *FE* is to the rectangle *FE* by *EG*. Similarly the rectangle *GE* by *EF* is to the square on *EF*, that is *GD* is to *FD*, as *GE* is to *EF*.

Q.E.D.

The square on a medial straight line, if applied to a rational straight line, produces as breadth a straight line rational and incommensurable in length with that to which it is applied.

Let *A* be medial and *CB* rational, and let a rectangular area *BD* equal to the square on *A* be applied to *BC*, producing *CD* as breadth.

I say that *CD* is rational and incommensurable in length with *CB*.

Since *A* is medial, the square on it equals a rectangular area contained by rational straight lines commensurable in square only.

Let the square on it equal *GF*. But the square on it also equals *BD*, therefore *BD* equals *GF*.

But it is also equiangular with it, and in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional, therefore, *BC* is to *EG* as *EF* is to *CD*.

Therefore the square on *BC* is to the square on *EG* as the square on *EF* is to the square on *CD*.

But the square on *CB* is commensurable with the square on *EG*, for each of these straight lines is rational, therefore the square on *EF* is also commensurable with the square on *CD*.

But the square on *EF* is rational, therefore the square on *CD* is also rational. Therefore *CD* is rational.

And since *EF* is incommensurable in length with *EG*, for they are commensurable in square only, while *EF* is to *EG* as the square on *EF* is to the rectangle *FE* by *EG*, therefore the square on *EF* is incommensurable with the rectangle *FE* by *EG*.

But the square on *CD* is commensurable with the square on *EF*, for the straight lines are rational in square, and the rectangle *DC* by *CB* is commensurable with the rectangle *FE* by *EG*, for they equal the square on *A*, therefore the square on *CD* is incommensurable with the rectangle *DC* by *CB*.

But the square on *CD* is to the rectangle *DC* by *CB* as *DC* is to *CB*, therefore *DC* is incommensurable in length with *CB*.

Therefore *CD* is rational and incommensurable in length with *CB*.

Therefore, *the square on a medial straight line, if applied to a rational straight line, produces as breadth a straight line rational and incommensurable in length with that to which it is applied.*

Q.E.D.

Given medial line *A* and rational line *BC*, then *CD* = *A*^{2}/(*BC*) is a rational line incommensurable to *BC*.

To understand this, let *a* be the length of *A* (relative to the standard unit length), *b* the length of *BC*, and *c* the length of *CD*, so that *c* = *a*^{2}/*b*.

We’re given a medial number *a* (*a*^{4} is a rational number but *a*^{2} is an irrational number), and *b*^{2} is a rational number. Then it follows that *c*^{2} is a rational number (being *a*^{4}/*b*^{2}), but *c*/*b* is an irrational number (being *a*^{2}/*b*^{2}). That follows since the product or quotient of two rational numbers is a rational number, but the product or quotient of an irrational number and a rational number is an irrational number.

This proposition is used frequently in Book X starting with the next proposition.