---
book: 12
number: 12
id: "XII.12"
kind: "theorem"
uses: ["[[book-4/proposition-6]]", "[[book-10/proposition-1]]", "[[book-5/proposition-16]]", "[[book-6/proposition-6]]", "[[book-5/proposition-22]]", "[[book-6/proposition-5]]", "[[book-12/proposition-8]]", "[[book-5/proposition-12]]", "[[book-12/proposition-10]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:12.prop.12"
license: "CC-BY-SA-4.0"
---

# XII.12

*Similar cones and cylinders are to one another in the triplicate ratio of the diameters in their bases.*

## Proof

Let there be similar cones and cylinders, let the circles *ABCD*, *EFGH* be their bases, *BD*, *FH* the diameters of the bases, and *KL*, *MN* the axes of the cones and cylinders; I say that the cone of which the circle *ABCD* is the base and the point *L* the vertex has to the cone of which the circle *EFGH* is the base and the point *N* the vertex the ratio triplicate of that which *BD* has to *FH*.

For, if the cone *ABCDL* has not to the cone *EFGHN* the ratio triplicate of that which *BD* has to *FH*, the cone *ABCDL* will have that triplicate ratio either to some solid less than the cone *EFGHN* or to a greater.

First, let it have that triplicate ratio to a less solid *O*.

Let the square *EFGH* be inscribed in the circle *EFGH*; [[book-4/proposition-6|IV. 6]] therefore the square *EFGH* is greater than the half of the circle *EFGH*.

Now let there be set up on the square *EFGH* a pyramid having the same vertex with the cone; therefore the pyramid so set up is greater than the half part of the cone.

Let the circumferences *EF*, *FG*, *GH*, *HE* be bisected at the points *P*, *Q*, *R*, *S*, and let *EP*, *PF*, *FQ*, *QG*, *GR*, *RH*, *HS*, *SE* be joined.

Therefore each of the triangles *EPF*, *FQG*, *GRH*, *HSE* is also greater than the half part of that segment of the circle *EFGH* which is about it.

Now on each of the triangles *EPF*, *FQG*, *GRH*, *HSE* let a pyramid be set up having the same vertex with the cone; therefore each of the pyramids so set up is also greater than the half part of that segment of the cone which is about it.

Thus, bisecting the circumferences so left, joining straight lines, setting up on each of the triangles pyramids having the same vertex with the cone, and doing this continually, we shall leave some segments of the cone which will be less than the excess by which the cone *EFGHN* exceeds the solid *O*. [[book-10/proposition-1|X. 1]]

Let such be left, and let them be the segments on *EP*, *PF*, *FQ*, *QG*, *GR*, *RH*, *HS*, *SE*; therefore the remainder, the pyramid of which the polygon *EPFQGRHS* is the base and the point *N* the vertex, is greater than the solid *O*.

Let there be also inscribed in the circle *ABCD* the polygon *ATBUCVDW* similar and similarly situated to the polygon *EPFQGRHS*, and let there be set up on the polygon *ATBUCVDW* a pyramid having the same vertex with the cone; of the triangles containing the pyramid of which the polygon *ATBUCVDW* is the base and the point *L* the vertex let *LBT* be one, and of the triangles containing the pyramid of which the polygon *EPFQGRHS* is the base and the point *N* the vertex let *NFP* be one; and let *KT*, *MP* be joined.

Now, since the cone *ABCDL* is similar to the cone *EFGHN*, therefore, as *BD* is to *FH*, so is the axis *KL* to the axis *MN*. [[book-11/definitions#Definition 24|XI. Def. 24]]

But, as *BD* is to *FH*, so is *BK* to *FM*; therefore also, as *BK* is to *FM*, so is *KL* to *MN*.

And, alternately, as *BK* is to *KL*, so is *FM* to *MN*. [[book-5/proposition-16|V. 16]]

And the sides are proportional about equal angles, namely the angles *BKL*, *FMN*; therefore the triangle *BKL* is similar to the triangle *FMN*. [[book-6/proposition-6|VI. 6]]

Again, since, as *BK* is to *KT*, so is *FM* to *MP*, and they are about equal angles, namely the angles *BKT*, *FMP*, inasmuch as, whatever part the angle *BKT* is of the four right angles at the centre *K*, the same part also is the angle *FMP* of the four right angles at the centre *M*; since then the sides are proportional about equal angles, therefore the triangle *BKT* is similar to the triangle *FMP*. [[book-6/proposition-6|VI. 6]]

Again, since it was proved that, as *BK* is to *KL*, so is *FM* to *MN*, while *BK* is equal to *KT*, and *FM* to *PM*, therefore, as *TK* is to *KL*, so is *PM* to *MN*; and the sides are proportional about equal angles, namely the angles *TKL*, *PMN*, for they are right; therefore the triangle *LKT* is similar to the triangle *NMP*. [[book-6/proposition-6|VI. 6]]

And since, owing to the similarity of the triangles *LKB*, *NMF*, as *LB* is to *BK*, so is *NF* to *FM*, and, owing to the similarity of the triangles *BKT*, *FMP*, as *KB* is to *BT*, so is *MF* to *FP*, therefore, *ex aequali*, as *LB* is to *BT*, so is *NF* to *FP*. [[book-5/proposition-22|V. 22]]

Again since, owing to the similarity of the triangles *LTK*, *NPM*, as *LT* is to *TK*, so is *NP* to *PM*, and, owing to the similarity of the triangles *TKB*, *PMF*, as *KT* is to *TB*, so is *MP* to *PF*; therefore, *ex aequali*, as *LT* is to *TB*, so is *NP* to *PF*. [[book-5/proposition-22|V. 22]]

But it was also proved that, as *TB* is to *BL*, so is *PF* to *FN*.

Therefore, ex aequali, as *TL* is to *LB*, so is *PN* to *NF*. [[book-5/proposition-22|V. 22]]

Therefore in the triangles *LTB*, *NPF* the sides are proportional; therefore the triangles *LTB*, *NPF* are equiangular; [[book-6/proposition-5|VI. 5]] hence they are also similar. [[book-6/definitions#Definition 1|VI. Def. I]]

Therefore the pyramid of which the triangle *BKT* is the base and the point *L* the vertex is also similar to the pyramid of which the triangle *FMP* is the base and the point *N* the vertex, for they are contained by similar planes equal in multitude. [[book-11/definitions#Definition 9|XI. Def. 9]]

But similar pyramids which have triangular bases are to one another in the triplicate ratio of their corresponding sides. [[book-12/proposition-8|XII. 8]]

Therefore the pyramid *BKTL* has to the pyramid *FMPN* the ratio triplicate of that which *BK* has to *FM*.

Similarly, by joining straight lines from *A*, *W*, *D*, *V*, *C*, *U* to *K*, and from *E*, *S*, *H*, *R*, *G*, *Q* to *M*, and setting up on each of the triangles pyramids which have the same vertex with the cones, we can prove that each of the similarly arranged pyramids will also have to each similarly arranged pyramid the ratio triplicate of that which the corresponding side *BK* has to the corresponding side *FM*, that is, which *BD* has to *FH*.

And, 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 also, as the pyramid *BKTL* is to the pyramid *FMPN*, so is the whole pyramid of which the polygon *ATBUCVDW* is the base and the point *L* the vertex to the whole pyramid of which the polygon *EPFQGRHS* is the base and the point *N* the vertex; hence also the pyramid of which *ATBUCVDW* is the base and the point *L* the vertex has to the pyramid of which the polygon *EPFQGRHS* is the base and the point *N* the vertex the ratio triplicate of that which *BD* has to *FH*.

But, by hypothesis, the cone of which the circle *ABCD* is the base and the point *L* the vertex has also to the solid *O* the ratio triplicate of that which *BD* has to *FH*; therefore, as the cone of which the circle *ABCD* is the base and the point *L* the vertex is to the solid *O*, so is the pyramid of which the polygon *ATBUCVDW* is the base and *L* the vertex to the pyramid of which the polygon *EPFQGRHS* is the base and the point *N* the vertex; therefore, alternately, as the cone of which the circle *ABCD* is the base and *L* the vertex is to the pyramid contained in it of which the polygon *ATBUCVDW* is the base and *L* the vertex, so is the solid *O* to the pyramid of which the polygon *EPFQGRHS* is the base and *N* the vertex. [[book-5/proposition-16|V. 16]]

But the said cone is greater than the pyramid in it; for it encloses it.

Therefore the solid *O* is also greater than the pyramid of which the polygon *EPFQGRHS* is the base and *N* the vertex.

But it is also less: which is impossible.

Therefore the cone of which the circle *ABCD* is the base and *L* the vertex has not to any solid less than the cone of which the circle *EFGH* is the base and the point *N* the vertex the ratio triplicate of that which *BD* has to *FH*:

Similarly we can prove that neither has the cone *EFGHN* to any solid less than the cone *ABCDL* the ratio triplicate of that which *FH* has to *BD*.

I say next that neither has the cone *ABCDL* to any solid greater than the cone *EFGHN* the ratio triplicate of that which *BD* has to *FH*.

For, if possible, let it have that ratio to a greater solid *O*.

Therefore, inversely, the solid *O* has to the cone *ABCDL* the ratio triplicate of that which *FH* has to *BD*.

But, as the solid *O* is to the cone *ABCDL*, so is the cone *EFGHN* to some solid less than the cone *ABCDL*.

Therefore the cone *EFGHN* also has to some solid less than the cone *ABCDL* the ratio triplicate of that which *FH* has to *BD*: which was proved impossible.

Therefore the cone *ABCDL* has not to any solid greater than the cone *EFGHN* the ratio triplicate of that which *BD* has to *FH*.

But it was proved that neither has it this ratio to a less solid than the cone *EFGHN*.

Therefore the cone *ABCDL* has to the cone *EFGHN* the ratio triplicate of that which *BD* has to *FH*.

But, as the cone is to the cone, so is the cylinder to the cylinder, for the cylinder which is on the same base as the cone and of equal height with it is triple of the cone; [[book-12/proposition-10|XII. 10]] therefore the cylinder also has to the cylinder the ratio triplicate of that which *BD* has to *FH*.

Therefore etc. Q. E. D.