---
book: 12
number: 2
id: "XII.2"
kind: "theorem"
uses: ["[[book-12/proposition-1]]", "[[book-5/proposition-11]]", "[[book-5/proposition-16]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:12.prop.2"
license: "CC-BY-SA-4.0"
---

# XII.2

*Circles are to one another as the squares on the diameters.*

## Proof

Let *ABCD*, *EFGH* be circles, and *BD*, *FH* their diameters; I say that, as the circle *ABCD* is to the circle *EFGH*, so is the square on *BD* to the square on *FH*.

For, if the square on *BD* is not to the square on *FH* as the circle *ABCD* is to the circle *EFGH*, then, as the square on *BD* is to the square on *FH*, so will the circle *ABCD* be either to some less area than the circle *EFGH*, or to a greater.

First, let it be in that ratio to a less area *S*.

Let the square *EFGH* be inscribed in the circle *EFGH*; then the inscribed square is greater than the half of the circle *EFGH*, inasmuch as, if through the points *E*, *F*, *G*, *H* we draw tangents to the circle, the square *EFGH* is half the square circumscribed about the circle, and the circle is less than the circumscribed square; hence the inscribed square *EFGH* is greater than the half of the circle *EFGH*.

Let the circumferences *EF*, *FG*, *GH*, *HE* be bisected at the points *K*, *L*, *M*, *N*, and let *EK*, *KF*, *FL*, *LG*, *GM*, *MH*, *HN*, *NE* be joined; therefore each of the triangles *EKF*, *FLG*, *GMH*, *HNE* is also greater than the half of the segment of the circle about it, inasmuch as, if through the points *K*, *L*, *M*, *N* we draw tangents to the circle and complete the parallelograms on the straight lines *EF*, *FG*, *GH*, *HE*, each of the triangles *EKF*, *FLG*, *GMH*, *HNE* will be half of the parallelogram about it, while the segment about it is less than the parallelogram; hence each of the triangles *EKF*, *FLG*, *GMH*, *HNE* is greater than the half of the segment of the circle about it.

Thus, by bisecting the remaining circumferences and joining straight lines, and by doing this continually, we shall leave some segments of the circle which will be less than the excess by which the circle *EFGH* exceeds the area *S*.

For it was proved in the first theorem of the tenth book that, if two unequal magnitudes be set out, and if from the greater there be subtracted a magnitude greater than the half, and from that which is left a greater than the half, and if this be done continually, there will be left some magnitude which will be less than the lesser magnitude set out.

Let segments be left such as described, and let the segments of the circle *EFGH* on *EK*, *KF*, *FL*, *LG*, *GM*, *MH*, *HN*, *NE* be less than the excess by which the circle *EFGH* exceeds the area *S*.

Therefore the remainder, the polygon *EKFLGMHN*, is greater than the area *S*.

Let there be inscribed, also, in the circle *ABCD* the polygon *AOBPCQDR* similar to the polygon *EKFLGMHN*; therefore, as the square on *BD* is to the square on *FH*, so is the polygon *AOBPCQDR* to the polygon *EKFLGMHN*. [[book-12/proposition-1|XII. 1]]

But, as the square on *BD* is to the square on *FH*, so also is the circle *ABCD* to the area *S*; therefore also, as the circle *ABCD* is to the area *S*, so is the polygon *AOBPCQDR* to the polygon *EKFLGMHN*; [[book-5/proposition-11|V. 11]] therefore, alternately, as the circle *ABCD* is to the polygon inscribed in it, so is the area *S* to the polygon *EKFLGMHN*. [[book-5/proposition-16|V. 16]]

But the circle *ABCD* is greater than the polygon inscribed in it; therefore the area *S* is also greater than the polygon *EKFLGMHN*.

But it is also less: which is impossible.

Therefore, as the square on *BD* is to the square on *FH*, so is not the circle *ABCD* to any area less than the circle *EFGH*.

Similarly we can prove that neither is the circle *EFGH* to any area less than the circle *ABCD* as the square on *FH* is to the square on *BD*.

I say next that neither is the circle *ABCD* to any area greater than the circle *EFGH* as the square on *BD* is to the square on *FH*.

For, if possible, let it be in that ratio to a greater area *S*.

Therefore, inversely, as the square on *FH* is to the square on *DB*, so is the area *S* to the circle *ABCD*.

But, as the area *S* is to the circle *ABCD*, so is the circle *EFGH* to some area less than the circle *ABCD*; therefore also, as the square on *FH* is to the square on *BD*, so is the circle *EFGH* to some area less than the circle *ABCD*: [[book-5/proposition-11|V. 11]] which was proved impossible.

Therefore, as the square on *BD* is to the square on *FH*, so is not the circle *ABCD* to any area greater than the circle *EFGH*.

And it was proved that neither is it in that ratio to any area less than the circle *EFGH*; therefore, as the square on *BD* is to the square on *FH*, so is the circle *ABCD* to the circle *EFGH*.

Therefore etc. Q. E. D.

Lemma. I say that, the area *S* being greater than the circle *EFGH*, as the area *S* is to the circle *ABCD*, so is the circle *EFGH* to some area less than the circle *ABCD*.

For let it be contrived that, as the area *S* is to the circle *ABCD*, so is the circle *EFGH* to the area *T*.

I say that the area *T* is less than the circle *ABCD*.

For since, as the area *S* is to the circle *ABCD*, so is the circle *EFGH* to the area *T*, therefore, alternately, as the area *S* is to the circle *EFGH*, so is the circle *ABCD* to the area *T*. [[book-5/proposition-16|V. 16]]

But the area *S* is greater than the circle *EFGH*; therefore the circle *ABCD* is also greater than the area *T*.

Hence, as the area *S* is to the circle *ABCD*, so is the circle *EFGH* to some area less than the circle *ABCD*. Q. E. D.