---
book: 6
number: 20
id: "VI.20"
kind: "theorem"
uses: ["[[book-6/proposition-6]]", "[[book-6/proposition-4]]", "[[book-5/proposition-22]]", "[[book-1/proposition-32]]", "[[book-6/proposition-1]]", "[[book-5/proposition-12]]", "[[book-6/proposition-19]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:6.prop.20"
license: "CC-BY-SA-4.0"
---

# VI.20

*Similar polygons are divided into similar triangles*, *and into triangles equal in multitude and in the same ratio as the wholes*, *and the polygon has to the polygon a ratio duplicate of that which the corresponding side has to the corresponding* *side*.

## Proof

Let *ABCDE*, *FGHKL* be similar polygons, and let *AB* correspond to *FG*; I say that the polygons *ABCDE*, *FGHKL* are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes, and the polygon *ABCDE* has to the polygon *FGHKL* a ratio duplicate of that which *AB* has to *FG*.

Let *BE*, *EC*, *GL*, *LH* be joined.

Now, since the polygon *ABCDE* is similar to the polygon *FGHKL*, the angle *BAE* is equal to the angle *GFL*; and, as *BA* is to *AE*, so is *GF* to *FL*. [[book-6/definitions#Definition 1|VI. Def. 1]]

Since then *ABE*, *FGL* are two triangles having one angle equal to one angle and the sides about the equal angles proportional, therefore the triangle *ABE* is equiangular with the triangle *FGL*; [[book-6/proposition-6|VI. 6]] so that it is also similar; [[book-6/proposition-4|VI. 4 and Def. 1]] therefore the angle *ABE* is equal to the angle *FGL*.

But the whole angle *ABC* is also equal to the whole angle *FGH* because of the similarity of the polygons; therefore the remaining angle *EBC* is equal to the angle *LGH*.

And, since, because of the similarity of the triangles *ABE*, *FGL*, as *EB* is to *BA*, so is *LG* to *GF*, and moreover also, because of the similarity of the polygons, as *AB* is to *BC*, so is *FG* to *GH*, therefore, ex aequali, as *EB* is to *BC*, so is *LG* to *GH*; [[book-5/proposition-22|V. 22]] that is, the sides about the equal angles *EBC*, *LGH* are proportional; therefore the triangle *EBC* is equiangular with the triangle *LGH*, [[book-6/proposition-6|VI. 6]] so that the triangle *EBC* is also similar to the triangle *LGH*. [[book-6/proposition-4|VI. 4 and Def. 1]]

For the same reason the triangle *ECD* is also similar to the triangle *LHK*.

Therefore the similar polygons *ABCDE*, *FGHKL* have been divided into similar triangles, and into triangles equal in multitude.

I say that they are also in the same ratio as the wholes, that is, in such manner that the triangles are proportional, and *ABE*, *EBC*, *ECD* are antecedents, while *FGL*, *LGH*, *LHK* are their consequents, and that the polygon *ABCDE* has to the polygon *FGHKL* a ratio duplicate of that which the corresponding side has to the corresponding side, that is *AB* to *FG*.

For let *AC*, *FH* be joined.

Then since, because of the similarity of the polygons, the angle *ABC* is equal to the angle *FGH*, and, as *AB* is to *BC*, so is *FG* to *GH*, the triangle *ABC* is equiangular with the triangle *FGH*; [[book-6/proposition-6|VI. 6]] therefore the angle *BAC* is equal to the angle *GFH*, and the angle *BCA* to the angle *GHF*.

And, since the angle *BAM* is equal to the angle *GFN*, and the angle *ABM* is also equal to the angle *FGN*, therefore the remaining angle *AMB* is also equal to the remaining angle *FNG*; [[book-1/proposition-32|I. 32]] therefore the triangle *ABM* is equiangular with the triangle *FGN*.

Similarly we can prove that the triangle *BMC* is also equiangular with the triangle *GNH*.

Therefore, proportionally, as *AM* is to *MB*, so is *FN* to *NG*, and, as *BM* is to *MC*, so is *GN* to *NH*; so that, in addition, ex aequali, as *AM* is to *MC*, so is *FN* to *NH*.

But, as *AM* is to *MC*, so is the triangle *ABM* to *MBC*, and *AME* to *EMC*; for they are to one another as their bases. [[book-6/proposition-1|VI. 1]]

Therefore also, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [[book-5/proposition-12|V. 12]] therefore, as the triangle *AMB* is to *BMC*, so is *ABE* to *CBE*.

But, as *AMB* is to *BMC*, so is *AM* to *MC*; therefore also, as *AM* is to *MC*, so is the triangle *ABE* to the triangle *EBC*.

For the same reason also, as *FN* is to *NH*, so is the triangle *FGL* to the triangle *GLH*.

And, as *AM* is to *MC*, so is *FN* to *NH*; therefore also, as the triangle *ABE* is to the triangle *BEC*, so is the triangle *FGL* to the triangle *GLH*; and, alternately, as the triangle *ABE* is to the triangle *FGL*, so is the triangle *BEC* to the triangle *GLH*.

Similarly we can prove, if *BD*, *GK* be joined, that, as the triangle *BEC* is to the triangle *LGH*, so also is the triangle *ECD* to the triangle *LHK*.

And since, as the triangle *ABE* is to the triangle *FGL*, so is *EBC* to *LGH*, and further *ECD* to *LHK*, therefore also, as one of the antecedents is to one of the consequents so are all the antecedents to all the consequents; [[book-5/proposition-12|V. 12]] therefore, as the triangle *ABE* is to the triangle *FGL*, so is the polygon *ABCDE* to the polygon *FGHKL*.

But the triangle *ABE* has to the triangle *FGL* a ratio duplicate of that which the corresponding side *AB* has to the corresponding side *FG*; for similar triangles are in the duplicate ratio of the corresponding sides. [[book-6/proposition-19|VI. 19]]

Therefore the polygon *ABCDE* also has to the polygon *FGHKL* a ratio duplicate of that which the corresponding side *AB* has to the corresponding side *FG*.

Therefore etc.

Porism. Similarly also it can be proved in the case of quadrilaterals that they are in the duplicate ratio of the corresponding sides. And it was also proved in the case of triangles; therefore also, generally, similar rectilineal figures are to one another in the duplicate ratio of the corresponding sides. Q. E. D.