Of all the parallelograms applied to the same straight line falling short by parallelogrammic figures similar and similarly situated to that described on the half of the straight line, that parallelogram is greatest which is applied to the half of the straight line and is similar to the difference.

Let *AB* be a straight line and let it be bisected at *C.* Let there be applied to the straight line *AB* the parallelogram *AD* falling short by the parallelogrammic figure *DB* described on the half of *AB,* that is, *CB.*

I say that, of all the parallelograms applied to *AB* falling short by parallelogrammic figures similar and similarly situated to *DB, AD* is greatest.

Let there be applied to the straight line *AB* the parallelogram *AF* falling short by the parallelogrammic figure *FB* similar and similarly situated to *DB.*

I say that *AD* is greater than *AF.*

Since the parallelogram *DB* is similar to the parallelogram *FB,* therefore they are about the same diameter.

Draw their diameter *DB,* and describe the figure.

Then, since *CF* equals *FE,* and *FB* is common, therefore the whole *CH* equals the whole *KE.*

But *CH* equals *CG,* since *AC* also equals *CB.*

Therefore *CG* also equals *KE.*

Add *CF* to each. Therefore the whole *AF* equals the gnomon *LMN,* so that the parallelogram *DB,* that is, *AD,* is greater than the parallelogram *AF.*

Therefore, *of all the parallelograms applied to the same straight line falling short by parallelogrammic figures similar and similarly situated to that described on the half of the straight line, that parallelogram is greatest which is applied to the half of the straight line and is similar to the difference.*

Q.E.D.

When that proposition is applied, the part which falls short is usually a square, not just any parallelogram, and this and the next proposition are much more easily understood in that case. In that case, the next proposition applies a rectangle equal to a given area to a line but falling short by a square. And this proposition implies that can only be done if the given area is at least the square on half the line, since that square is the greatest rectangle that can be so applied.