To apply a parallelogram equal to a given rectilinear figure to a given straight line but exceeding it by a parallelogram similar to a given one.

Let *C* be the given rectilinear figure, *AB* be the given straight line, and *D* the parallelogram to which the excess is required to be similar.

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

Bisect *AB* at *E.* Describe the parallelogram *BF* on *EB* similar and similarly situated to *D,* and construct *GH* equal to the sum of *BF* and *C* and similar and similarly situated to *D.*

Let *KH* correspond to *FL* and *KG* to *FE.*

Now, since *GH* is greater than *FB,* therefore *KH* is also greater than *FL,* and *KG* greater than*FE.*

Produce *FL* and *FE.* Make *FLM* equal to *KH,* and *FEN* equal to *KG.* Complete *MN.* Then *MN* is both equal and similar to *GH.*

But *GH* is similar to *EL,* therefore *MN* is also similar to *EL,* therefore *EL* is about the same diameter with *MN.*

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

Since *GH* equals the sum of *EL* and *C,* while *GH* equals *MN,* therefore *MN* also equals the sum of *EL* and *C.*

Subtract *EL* from each. Therefore the remainder, the gnomon *XWV,* equals *C.*

Now, since *AE* equals *EB,* therefore *AN* equals *NB,* that is, *LP.*

Add *EO* to each. Therefore the whole *AO* equals the gnomon *VWX.*

But the gnomon *VWX* equals *C,* therefore *AO* also equals *C.*

Therefore the parallelogram *AO* equal to the given rectilinear figure *C* has been applied to the given straight line *AB* but exceeding it by a parallelogram *QP* similar to *D,* since *PQ* is also similar to *EL.*

Q.E.F.

In that case of this proposition a rectangle AO equal to a given rectilinear figure C is applied to a given straight line AB but exceeds it by a square (BQOP in the figure). So the rectangle being laid alongside the line extends past the end of the line AB, but the part that extends beyond the end is a square.
For the construction, bisect |

As in the last proposition we can understand the meaning of this construction more easily if we interpret it algebraically. Let *a* stand for the known quantity *AB* and *x* stand for the unknown quantity *BP.* Then this
construction finds *x* so that (*a* + *x*) *x* = *C.* In other words it solves the quadratic equation *ax* + *x*^{2} = *C.*

The construction of this proposition is used in the next proposition to cut a straight line in extreme and mean ratio.