To apply a parallelogram equal to a given rectilinear figure to a given straight line but falling short by a parallelogram similar to a given one; thus the given rectilinear figure must not be greater than the parallelogram described on the half of the straight line and similar to the given parallelogram.

Let *C* be the given rectilinear figure, *AB* the given straight line, and *D* the given parallelogram, and let *C* not be greater than the parallelogram described on the half of *AB* similar to the given parallelogram *D.*

It is required to apply a parallelogram equal to the given rectilinear figure *C* to the given straight line *AB* but falling short by a parallelogram similar to *D.*

Bisect *AB* at the point *E.* Describe *EBFG* similar and similarly situated to *D* on *EB,* and complete the parallelogram *AG.*

If then *AG* equals *C,* that which was proposed is done, for the parallelogram *AG* equal to the given rectilinear figure *C* has been applied to the given straight line *AB* but falling short by a parallelogram *GB* similar to *D.*

But, if not, let *HE* be greater than *C.*

Now *HE* equals *GB,* therefore *GB* is also greater than *C.*

Construct *KLMN* equal to *GB* minus *C* and similar and similarly situated to *D.*

But *D* is similar to *GB,* therefore *KM* is also similar to *GB.*

Let, then, *KL* correspond to *GE,* and *LM* to *GF.*

Now, since *GB* equals *C* and *KM,* therefore *GB* is greater than *KM,* therefore also *GE* is greater than *KL,* and *GF* than *LM.*

Make *GO* equal to *KL,* and *GP* equal to *LM,* and let the parallelogram *OGPQ* be completed, therefore it is equal and similar to *KM.*

Therefore *GQ* is also similar to *GB,* therefore *GQ* is about the same diameter with *GB.*

Let *GQB* be their diameter, and describe the figure.

Then, since *BG* equals *C* and *KM,* and in them *GQ* equals *KM,* therefore the remainder, the gnomon *UWV,* equals the remainder *C.*

And, since *PR* equals *OS,* add *QB* to each, therefore the whole *PB* equals the whole *OB.*

But *OB* equals *TE,* since the side *AE* also equals the side *EB,* therefore *TE* also equals *PB.*

Add *OS* to each. Therefore the whole *TS* equals the whole, the gnomon *VWU.*

But the gnomon *VWU* was proved equal to *C,* therefore *TS* also equals *C.*

Therefore there the parallelogram *ST* equal to the given rectilinear figure *C* has been applied to the given straight line *AB* but falling short by a parallelogram *QB* similar to *D.*

Q.E.F.

The construction is as follows. Bisect AB at E, construct a square GFBE.
The next stage is to construct a square Complete the figure. We can understand the meaning of this construction more easily if we interpret it algebraically. Let |

In terms of the single variable *x,* the construction solves the quadratic equation *ax* – *x*^{2} = *C.*

The next proposition solves a similar quadratic equation: *ax* + *x*^{2} = *C.*