---
book: 6
number: 28
id: "VI.28"
kind: "construction"
uses: ["[[book-6/proposition-18]]", "[[book-6/proposition-25]]", "[[book-6/proposition-21]]", "[[book-6/proposition-26]]", "[[book-1/proposition-36]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:6.prop.28"
license: "CC-BY-SA-4.0"
---

# VI.28

*To a given straight line to apply a parallelogram equal to a given rectilineal figure and deficient by a parallelogrammic figure similar to a given one : thus the given rectilineal figure must not be greater than the parallelogram described on the half of the straight line and similar to the defect*.

## Proof

Let *AB* be the given straight line, *C* the given rectilineal figure to which the figure to be applied to *AB* is required to be equal, not being greater than the parallelogram described on the half of *AB* and similar to the defect, and *D* the parallelogram to which the defect is required to be similar; thus it is required to apply to the given straight line *AB* a parallelogram equal to the given rectilineal figure *C* and deficient by a parallelogrammic figure which is similar to *D*.

Let *AB* be bisected at the point *E*, and on *EB* let *EBFG* be described similar and similarly situated to *D*; [[book-6/proposition-18|VI. 18]] let the parallelogram *AG* be completed.

If then *AG* is equal to *C*, that which was enjoined will have been done; for there has been applied to the given straight line *AB* the parallelogram *AG* equal to the given rectilineal figure *C* and deficient by a parallelogrammic figure *GB* which is similar to *D*.

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

Now *HE* is equal to *GB*; therefore *GB* is also greater than *C*.

Let *KLMN* be constructed at once equal to the excess by which *GB* is greater than *C* and similar and similarly situated to *D*. [[book-6/proposition-25|VI. 25]]

But *D* is similar to *GB*; therefore *KM* is also similar to *GB*. [[book-6/proposition-21|VI. 21]]

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

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

Let *GO* be made 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*; [[book-6/proposition-21|VI. 21]] therefore *GQ* is about the same diameter with *GB*. [[book-6/proposition-26|VI. 26]]

Let *GQB* be their diameter, and let the figure be described.

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

And, since *PR* is equal to *OS*, let *QB* be added to each; therefore the whole *PB* is equal to the whole *OB*.

But *OB* is equal to *TE*, since the side *AE* is also equal to the side *EB*; [[book-1/proposition-36|I. 36]] therefore *TE* is also equal to *PB*.

Let *OS* be added to each; therefore the whole *TS* is equal to the whole, the gnomon *VWU*.

But the gnomon *VWU* was proved equal to *C*; therefore *TS* is also equal to *C*.

Therefore to the given straight line *AB* there has been applied the parallelogram *ST* equal to the given rectilineal figure *C* and deficient by a parallelogrammic figure *QB* which is similar to *D*. Q. E. F.