If to any straight line there is applied a parallelogram but falling short by a square, then the applied parallelogram equals the rectangle contained by the segments of the straight line resulting from the application.

Apply to the straight line *AB* the parallelogram *AD* but falling short by the square *DB*.

I say that *AD* equals the rectangle *AC* by *CB*.

This is indeed at once clear, for, since *DB* is a square, *DC* equals *CB*, and *AD* is the rectangle *AC* by *CD*, that is, the rectangle *AC* by *CB*.

Q.E.D.

If there are two unequal straight lines, and to the greater there is applied a parallelogram equal to the fourth part of the square on the less minus a square figure, and if it divides it into parts commensurable in length, then the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater. And if the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater, and if there is applied to the greater a parallelogram equal to the fourth part of the square on the less minus a square figure, then it divides it into parts commensurable in length.

Let *A* and *BC* be two unequal straight lines, of which *BC* is the greater, and let there be applied to *BC* a parallelogram equal to the fourth part of the square on the less, *A*, that is, equal to the square on the half of *A* but falling short by a square figure. Let this be the rectangle *BD* by *DC*, and let *BD* be commensurable in length with *DC*.

I say that the square on *BC* is greater than the square on *A* by the square on a straight line commensurable with *BC*.

Bisect *BC* at the point *E*, and make *EF* equal to *DE*.

Therefore the remainder *DC* equals *BF*. And, since the straight line *BC* was cut into equal parts at *E*, and into unequal parts at *D*, therefore the rectangle *BD* by *DC*, together with the square on *ED*, equals the square on *EC*.

And the same is true of their quadruples, therefore four times the rectangle *BD* by *DC*, together with four times the square on *DE*, equals four times the square on *EC*.

But the square on *A* equals four times the rectangle *BD* by *DC*, and the square on *DF* equals four times the square on *DE*, for *DF* is double *DE*. And the square on *BC* equals four times the square on *EC*, for again *BC* is double *CE*.

Therefore the sum of the squares on *A* and *DF* equals the square on *BC*, so that the square on *BC* is greater than the square on *A* by the square on *DF*.

It is to be proved that *BC* is also commensurable with *DF*.

Since *BD* is commensurable in length with *DC*, therefore *BC* is also commensurable in length with *CD*.

But *CD* is commensurable in length with *CD* and *BF*, for *CD* equals *BF*.

Therefore *BC* is also commensurable in length with *BF* and *CD*, so that *BC* is also commensurable in length with the remainder *FD*. Therefore the square on *BC* is greater than the square on *A* by the square on a straight line commensurable with *BC*.

Next, let the square on *BC* be greater than the square on *A* by the square on a straight line commensurable with *BC*. Apply to *BC* a parallelogram equal to the fourth part of the square on *A* but falling short by a square figure, and let it be the rectangle *BD* by *DC*.

It is to be proved that *BD* is commensurable in length with *DC*.

With the same construction, we can prove similarly that the square on *BC* is greater than the square on *A* by the square on *FD*.

But the square on *BC* is greater than the square on *A* by the square on a straight line commensurable with *BC*.

Therefore *BC* is commensurable in length with *FD*, so that *BC* is also commensurable in length with the remainder, the sum of *BF* and *DC*.

But the sum of *BF* and *DC* is commensurable with *DC*, so that *BC* is also commensurable in length with *CD*, and therefore, taken separately, *BD* is commensurable in length with *DC*.

Therefore, *if there are two unequal straight lines, and to the greater there is applied a parallelogram equal to the fourth part of the square on the less minus a square figure, and if it divides it into parts commensurable in length, then the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater. And if the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater, and if there is applied to the greater a parallelogram equal to the fourth part of the square on the less minus a square figure, then it divides it into parts commensurable in length.*

Q.E.D.

The construction solves the equation

where *b* denotes denotes the greater line *BC* and *a* denotes the lesser line *A*. The solution is the line *DC*, which is

Then the proposition asserts that the ratio *b* : *x* is a numeric ratio if and only if the ratio √(*b*^{2} – *a*^{2}) : *b* is a numeric ratio. Since *x*/*b* is equal to 1/2 – √(*b*^{2} – *a*^{2})/*b*, it’s clear why one would be a numeric ratio if and only if the other is.

The lemma is also used in the next proposition. The proposition is used in several times in Book X starting with X.54.