---
book: 12
number: 1
id: "XII.1"
kind: "theorem"
uses: ["[[book-6/proposition-6]]", "[[book-3/proposition-27]]", "[[book-3/proposition-31]]", "[[book-1/proposition-32]]", "[[book-6/proposition-4]]", "[[book-6/proposition-20]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:12.prop.1"
license: "CC-BY-SA-4.0"
---

# XII.1

*Similar polygons inscribed in circles are to one another as the squares on the diameters.*

## Proof

Let *ABC*, *FGH* be circles, let *ABCDE*, *FGHKL* be similar polygons inscribed in them, and let *BM*, *GN* be diameters of the circles; I say that, as the square on *BM* is to the square on *GN*, so is the polygon *ABCDE* to the polygon *FGHKL*.

For let *BE*, *AM*, *GL*, *FN* 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*. [VI. Def. I]

Thus *BAE*, *GFL* are two triangles which have one angle equal to one angle, namely the angle *BAE* to the angle *GFL*, and the sides about the equal angles proportional; therefore the triangle *ABE* is equiangular with the triangle *FGL*. [[book-6/proposition-6|VI. 6]]

Therefore the angle *AEB* is equal to the angle *FLG*.

But the angle *AEB* is equal to the angle *AMB*, for they stand on the same circumference; [[book-3/proposition-27|III. 27]] and the angle *FLG* to the angle *FNG*; therefore the angle *AMB* is also equal to the angle *FNG*.

But the right angle *BAM* is also equal to the right angle *GFN*; [[book-3/proposition-31|III. 31]] therefore the remaining angle is equal to the remaining angle. [[book-1/proposition-32|I. 32]]

Therefore the triangle *ABM* is equiangular with the triangle *FGN*.

Therefore, proportionally, as *BM* is to *GN*, so is *BA* to *GF*. [[book-6/proposition-4|VI. 4]]

But the ratio of the square on *BM* to the square on *GN* is duplicate of the ratio of *BM* to *GN*, and the ratio of the polygon *ABCDE* to the polygon *FGHKL* is duplicate of the ratio of *BA* to *GF*; [[book-6/proposition-20|VI. 20]] therefore also, as the square on *BM* is to the square on *GN*, so is the polygon *ABCDE* to the polygon *FGHKL*.

Therefore etc. Q. E. D.