---
book: 6
number: 25
id: "VI.25"
kind: "construction"
uses: ["[[book-1/proposition-44]]", "[[book-1/proposition-45]]", "[[book-6/proposition-13]]", "[[book-6/proposition-18]]", "[[book-6/proposition-19]]", "[[book-6/proposition-1]]", "[[book-5/proposition-16]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:6.prop.25"
license: "CC-BY-SA-4.0"
---

# VI.25

*To construct one and the same figure similar to a given rectilineal figure and equal to another given rectilineal figure*.

## Proof

Let *ABC* be the given rectilineal figure to which the figure to be constructed must be similar, and *D* that to which it must be equal; thus it is required to construct one and the same figure similar to *ABC* and equal to *D*.

Let there be applied to *BC* the parallelogram *BE* equal to the triangle *ABC* [[book-1/proposition-44|I. 44]], and to *CE* the parallelogram *CM* equal to *D* in the angle *FCE* which is equal to the angle *CBL*. [[book-1/proposition-45|I. 45]]

Therefore *BC* is in a straight line with *CF*, and *LE* with *EM*.

Now let *GH* be taken a mean proportional to *BC*, *CF* [[book-6/proposition-13|VI. 13]], and on *GH* let *KGH* be described similar and similarly situated to *ABC*. [[book-6/proposition-18|VI. 18]]

Then, since, as *BC* is to *GH*, so is *GH* to *CF*, and, if three straight lines be proportional, as the first is to the third, so is the figure on the first to the similar and similarly situated figure described on the second, [[book-6/proposition-19|VI. 19, Por.]] therefore, as *BC* is to *CF*, so is the triangle *ABC* to the triangle *KGH*.

But, as *BC* is to *CF*, so also is the parallelogram *BE* to the parallelogram *EF*. [[book-6/proposition-1|VI. 1]]

Therefore also, as the triangle *ABC* is to the triangle *KGH*, so is the parallelogram *BE* to the parallelogram *EF*; therefore, alternately, as the triangle *ABC* is to the parallelogram *BE*, so is the triangle *KGH* to the parallelogram *EF*. [[book-5/proposition-16|V. 16]]

But the triangle *ABC* is equal to the parallelogram *BE*; therefore the triangle *KGH* is also equal to the parallelogram *EF*.

But the parallelogram *EF* is equal to *D*; therefore *KGH* is also equal to *D*.

And *KGH* is also similar to *ABC*.

Therefore one and the same figure *KGH* has been constructed similar to the given rectilineal figure *ABC* and equal to the other given figure *D*. Q. E. D.