Similar polygons (inscribed) in circles are to one another as the squares on the diameters (of the circles).
Let ABC and FGH be circles, and let ABCDE and FGHKL be similar polygons (inscribed) in them (respectively), and let BM and GN be the diameters of the circles (respectively).
I say that as the square on BM is to the square on GN, so polygon ABCDE (is) to polygon FGHKL.
For let BE, AM, GL, and FN have been joined.
And since polygon ABCDE (is) similar to polygon FGHKL, angle BAE is also equal to (angle) GFL, and as BA is to AE, so GF (is) to FL [Def. 6.1].
So, BAE and GFL are two triangles having one angle equal to one angle, (namely), BAE (equal) to GFL, and the sides around the equal angles proportional.
Triangle ABE is thus equiangular with triangle FGL [Prop. 6.6].
Thus, angle AEB is equal to (angle) FLG.
But, AEB is equal to AMB, and FLG to FNG, for they stand on the same circumference [Prop. 3.27].
Thus, AMB is also equal to FNG.
And the right-angle BAM is also equal to the right-angle GFN [Prop. 3.31].
Thus, the remaining (angle) is also equal to the remaining (angle) [Prop. 1.32].
Thus, triangle ABM is equiangular with triangle FGN.
Thus, proportionally, as BM is to GN, so BA (is) to GF [Prop. 6.4].
But, the (ratio) of the square on BM to the square on GN is the square of the ratio of BM to GN, and the (ratio) of polygon ABCDE to polygon FGHKL is the square of the (ratio) of BA to GF [Prop. 6.20].
And, thus, as the square on BM (is) to the square on GN, so polygon ABCDE (is) to polygon FGHKL.
Thus, similar polygons (inscribed) in circles are to one another as the squares on the diameters (of the circles).
(Which is) the very thing it was required to show.
Proposition 2
Circles are to one another as the squares on (their) diameters.
Let ABCD and EFGH be circles, and [let] BD and FH [be] their diameters.
I say that as circle ABCD is to circle EFGH, so the square on BD (is) to the square on FH.
For if the circle ABCD is not to the (circle) EFGH, as the square on BD (is) to the (square) on FH, then as the (square) on BD (is) to the (square) on FH, so circle ABCD will be to some area either less than, or greater than, circle EFGH.
Let it, first of all, be (in that ratio) to (some) lesser (area), S.
And let the square EFGH have been inscribed in circle EFGH [Prop. 4.6].
So the inscribed square is greater than half of circle EFGH, inasmuch as if we draw tangents to the circle through the points E, F, G, and H, then square EFGH is half of the square circumscribed about the circle [Prop. 1.47], and the circle is less than the circumscribed square.
Hence, the inscribed square EFGH is greater than half of circle EFGH.
Let the circumferences EF, FG, GH, and HE have been cut in half at points K, L, M, and N (respectively), and let EK, KF, FL, LG, GM, MH, HN, and NE have been joined.
And, thus, each of the triangles EKF, FLG, GMH, and HNE is greater than half of the segment of the circle about it, inasmuch as if we draw tangents to the circle through points K, L, M, and N, and complete the parallelograms on the straight-lines EF, FG, GH, and HE, then each of the triangles EKF, FLG, GMH, and HNE will be half of the parallelogram about it, but the segment about it is less than the parallelogram.
Hence, each of the triangles EKF, FLG, GMH, and HNE is greater than half of the segment of the circle about it.
So, by cutting the circumferences remaining behind in half, and joining straight-lines, and doing this continually, we will (eventually) leave behind some segments of the circle whose (sum) will be less than the excess by which circle EFGH exceeds the area S.
For we showed in the first theorem of the tenth book that if two unequal magnitudes are laid out, and if (a part) greater than a half is subtracted from the greater, and (if from) the remainder (a part) greater than a half (is subtracted), and this happens continually, then some magnitude will (eventually) be left which will be less than the lesser laid out magnitude [Prop. 10.1].
Therefore, let the (segments) have been left, and let the (sum of the) segments of the circle EFGH on EK, KF, FL, LG, GM, MH, HN, and NE be less than the excess by which circle EFGH exceeds area S.
Thus, the remaining polygon EKFLGMHN is greater than area S.
And let the polygon AOBPCQDR, similar to the polygon EKFLGMHN, have been inscribed in circle ABCD.
Thus, as the square on BD is to the square on FH, so polygon AOBPCQDR (is) to polygon EKFLGMHN [Prop. 12.1].
But, also, as the square on BD (is) to the square on FH, so circle ABCD (is) to area S.
And, thus, as circle ABCD (is) to area S, so polygon AOBPCQDR (is) to polygon EKFLGMHN [Prop. 5.11].
Thus, alternately, as circle ABCD (is) to the polygon (inscribed) within it, so area S (is) to polygon EKFLGMHN [Prop. 5.16].
And circle ABCD (is) greater than the polygon (inscribed) within it.
Thus, area S is also greater than polygon EKFLGMHN.
But, (it is) also less.
The very thing is impossible.
Thus, the square on BD is not to the (square) on FH, as circle ABCD (is) to some area less than circle EFGH.
So,similarly,wecan show that the (square) on FH (is) not to the (square) on BD as circle EFGH (is) to some area less than circle ABCD either.
So, I say that neither (is) the (square) on BD to the (square) on FH, as circle ABCD (is) to some area greater than circle EFGH.
For, if possible, let it be (in that ratio) to (some) greater (area), S.
Thus, inversely, as the square on FH [is] to the (square) on DB, so area S (is) to circle ABCD [Prop. 5.7 corr.].
But, as area S (is) to circle ABCD, so circle EFGH (is) to some area less than circle ABCD (see lemma).
And, thus, as the (square) on FH (is) to the (square) on BD, so circle EFGH (is) to some area less than circle ABCD [Prop. 5.11].
The very thing was shown (to be) impossible.
Thus, as the square on BD is to the (square) on FH, so circle ABCD (is) not to some area greater than circle EFGH.
And it was shown that neither (is it in that ratio) to (some) lesser (area).
Thus, as the square on BD is to the (square) on FH, so circle ABCD (is) to circle EFGH.
Thus, circles are to one another as the squares on (their) diameters.
(Which is) the very thing it was required to show.
Lemma
So, I say that, area S being greater than circle EFGH, as area S is to circle ABCD, so circle EFGH (is) to some area less than circle ABCD.
For let it have been contrived that as area S (is) to circle ABCD, so circle EFGH (is) to area T.
I say that area T is less than circle ABCD.
For since as area S is to circle ABCD, so circle EFGH (is) to area T, alternately, as area S is to circle EFGH, so circle ABCD (is) to area T [Prop. 5.16].
And area S (is) greater than circle EFGH.
Thus, circle ABCD (is) also greater than area T [Prop. 5.14].
Hence, as area S is to circle ABCD, so circle EFGH (is) to some area less than circle ABCD.
(Which is) the very thing it was required to show.
Proposition 3
Any pyramid having a triangular base is divided into two pyramids having triangular bases (which are) equal, similar to one another, and [similar] to the whole, and into two equal prisms.
And the (sum of the) two prisms is greater than half of the whole pyramid.
Let there be a pyramid whose base is triangle ABC, and (whose) apex (is) point D.
I say that pyramid ABCD is divided into two pyramids having triangular bases (which are) equal to one another, and similar to the whole, and into two equal prisms.
And the (sum of the) two prisms is greater than half of the whole pyramid.
For let AB, BC, CA, AD, DB, and DC have been cut in half at points E, F, G, H, K, and L (respectively).
And let HE, EG, GH, HK, KL, LH, KF, and FG have been joined.
Since AE is equal to EB, and AH to DH, EH is thus parallel to DB [Prop. 6.2].
So, for the same (reasons), HK is also parallel to AB.
Thus, HEBK is a parallelogram.
Thus, HK is equal to EB [Prop. 1.34].
But, EB is equal to EA.
Thus, AE is also equal to HK.
And AH is also equal to HD.
So the two (straight-lines) EA and AH are equal to the two (straight-lines) KH and HD, respectively.
And angle EAH (is) equal to angle KHD [Prop. 1.29].
Thus, base EH is equal to base KD [Prop. 1.4].
Thus, triangle AEH is equal and similar to triangle HKD [Prop. 1.4].
So, for the same (reasons), triangle AHG is also equal and similar to triangle HLD.
And since EH and HG are two straight-lines joining one another (which are respectively) parallel to two straight-lines joining one another, KD and DL, not being in the same plane, they will contain equal angles [Prop. 11.10].
Thus, angle EHG is equal to angle KDL.
And since the two straight-lines EH and HG are equal to the two straight-lines KD and DL, respectively, and angle EHG is equal to angle KDL, base EG [is] thus equal to base KL [Prop. 1.4].
Thus, triangle EHG is equal and similar to triangle KDL.
So, for the same (reasons), triangle AEG is also equal and similar to triangle HKL.
Thus, the pyramid whose base is triangle AEG, and apex the point H, is equal and similar to the pyramid whose base is triangle HKL, and apex the point D [Def. 11.10].
And since HK has been drawn parallel to one of the sides, AB, of triangle ADB, triangle ADB is equiangular to triangle DHK [Prop. 1.29], and they have proportional sides.
Thus, triangle ADB is similar to triangle DHK [Def. 6.1].
So, for the same (reasons), triangle DBC is also similar to triangle DKL, and ADC to DLH.
And since two straight-lines joining one another, BA and AC, are parallel to two straight-lines joining one another, KH and HL, not in the same plane, they will contain equal angles [Prop. 11.10].
Thus, angle BAC is equal to (angle) KHL.
And as BA is to AC, so KH (is) to HL.
Thus, triangle ABC is similar to triangle HKL [Prop. 6.6].
And, thus, the pyramid whose base is triangle ABC, and apex the point D, is similar to the pyramid whose base is triangle HKL, and apex the point D [Def. 11.9].
But, the pyramid whose base [is] triangle HKL, and apex the point D, was shown (to be) similar to the pyramid whose base is triangle AEG, and apex the point H.
Thus, each of the pyramids AEGH and HKLD is similar to the whole pyramid ABCD.
And since BF is equal to FC, parallelogram EBFG is double triangle GFC [Prop. 1.41].
And since, if two prisms (have) equal heights, and the former has a parallelogram as a base, and the latter a triangle, and the parallelogram (is) double the triangle, then the prisms are equal [Prop. 11.39], the prism contained by the two triangles BKF and EHG, and the three parallelograms EBFG, EBKH, and HKFG, is thus equal to the prism contained by the two triangles GFC and HKL, and the three parallelograms KFCL, LCGH, and HKFG.
And (it is) clear that each of the prisms whose base (is) parallelogram EBFG, and opposite (side) straight-line HK, and whose base (is) triangle GFC, and opposite (plane) triangle HKL, is greater than each of the pyramids whose bases are triangles AEG and HKL, and apexes the points H and D (respectively), inasmuch as, if we [also] join the straight-lines EF and EK then the prism whose base (is) parallelogram EBFG, and opposite (side) straight-line HK, is greater than the pyramid whose base (is) triangle EBF, and apex the point K.
But the pyramid whose base (is) triangle EBF, and apex the point K, is equal to the pyramid whose base is triangle AEG, and apex point H.
For they are contained by equal and similar planes.
And, hence, the prism whose base (is) parallelogram EBFG, and opposite (side) straight-line HK, is greater than the pyramid whose base (is) triangle AEG, and apex the point H.
And the prism whose base is parallelogram EBFG, and opposite (side) straight-line HK, (is) equal to the prism whose base (is) triangle GFC, and opposite (plane) triangle HKL.
And the pyramid whose base (is) triangle AEG, and apex the point H, is equal to the pyramid whose base (is) triangle HKL, and apex the point D.
Thus, the (sum of the) aforementioned two prisms is greater than the (sum of the) aforementioned two pyramids, whose bases (are) triangles AEG and HKL, and apexes the points H and D (respectively).
Thus, the whole pyramid, whose base (is) triangle ABC, and apex the point D, has been divided into two pyramids (which are) equal to one another [and similar to the whole], and into two equal prisms.
And the (sum of the) two prisms is greater than half of the whole pyramid.
(Which is) the very thing it was required to show.
Proposition 4
If there are two pyramids with the same height, having trianglular bases, and each of them is divided into two pyramids equal to one another, and similar to the whole, and into two equal prisms then as the base of one pyramid (is) to the base of the other pyramid, so (the sum of) all the prisms in one pyramid will be to (the sum of all) the equal number of prisms in the other pyramid.
Let there be two pyramids with the same height, having the triangular bases ABC and DEF, (with) apexes the points G and H (respectively).
And let each of them have been divided into two pyramids equal to one another, and similar to the whole, and into two equal prisms [Prop. 12.3].
I say that as base ABC is to base DEF, so (the sum of) all the prisms in pyramid ABCG (is) to (the sum of) all the equal number of prisms in pyramid DEFH.
For since BO is equal to OC, and AL to LC, LO is thus parallel to AB, and triangle ABC similar to triangle LOC [Prop. 12.3].
So, for the same (reasons), triangle DEF is also similar to triangle RVF.
And since BC is double CO, and EF (double) FV, thus as BC (is) to CO, so EF (is) to FV.
And the similar, and similarly laid out, rectilinear (figures) ABC and LOC have been described on BC and CO (respectively), and the similar, and similarly laid out, [rectilinear] (figures) DEF and RVF on EF and FV (respectively).
Thus, as triangle ABC is to triangle LOC, so triangle DEF (is) to triangle RVF [Prop. 6.22].
Thus, alternately, as triangle ABC is to [triangle] DEF, so [triangle] LOC (is) to triangle RVF [Prop. 5.16].
But, as triangle LOC (is) to triangle RVF, so the prism whose base [is] triangle LOC, and opposite (plane) PMN, (is) to the prism whose base (is) triangle RVF, and opposite (plane) STU (see lemma).
And, thus, as triangle ABC (is) to triangle DEF, so the prism whose base (is) triangle LOC, and opposite (plane) PMN, (is) to the prism whose base (is) triangle RVF, and opposite (plane) STU.
And as the aforementioned prisms (are) to one another, so the prism whose base (is) parallelogram KBOL, and opposite (side) straight-line PM, (is) to the prism whose base (is) parallelogram QEVR, and opposite (side) straight-line ST [Prop. 11.39] [Prop. 12.3].
Thus, also, (is) the (sum of the) two prisms—that whose base (is) parallelogram KBOL, and opposite (side) PM, and that whose base (is) LOC, and opposite (plane) PMN —to (the sum of) the (two) prisms—that whose base (is) QEVR, and opposite (side) straight-line ST, and that whose base (is) triangle RVF, and opposite (plane) STU [Prop. 5.12].
And, thus, as base ABC (is) to base DEF, so the (sum of the first) aforementioned two prisms (is) to the (sum of the second) aforementioned two prisms.
And, similarly, if pyramids PMNG and STUH are divided into two prisms, and two pyramids, as base PMN (is) to base STU, so (the sum of) the two prisms in pyramid PMNG will be to (the sum of) the two prisms in pyramid STUH.
But, as base PMN (is) to base STU, so base ABC (is) to base DEF.
For the triangles PMN and STU (are) equal to LOC and RVF, respectively.
And, thus, as base ABC (is) to base DEF, so (the sum of) the four prisms (is) to (the sum of) the four prisms [Prop. 5.12].
So, similarly, even if we divide the pyramids left behind into two pyramids and into two prisms, as base ABC (is) to base DEF, so (the sum of) all the prisms in pyramid ABCG will be to (the sum of) all the equal number of prisms in pyramid DEFH.
(Which is) the very thing it was required to show.
Lemma
And one may show, as follows, that as triangle LOC is to triangle RVF, so the prism whose base (is) triangle LOC, and opposite (plane) PMN, (is) to the prism whose base (is) [triangle] RVF, and opposite (plane) STU.
For, in the same figure, let perpendiculars have been conceived (drawn) from (points) G and H to the planes ABC and DEF (respectively).
These clearly turn out to be equal, on account of the pyramids being assumed (to be) of equal height.
And since two straight-lines, GC and the perpendicular from G, are cut by the parallel planes ABC and PMN they will be cut in the same ratios [Prop. 11.17].
And GC was cut in half by the plane PMN at N.
Thus, the perpendicular from G to the plane ABC will also be cut in half by the plane PMN.
So, for the same (reasons), the perpendicular from H to the plane DEF will also be cut in half by the plane STU.
And the perpendiculars from G and H to the planes ABC and DEF (respectively) are equal.
Thus, the perpendiculars from the triangles PMN and STU to ABC and DEF (respectively, are) also equal.
Thus, the prisms whose bases are triangles LOC and RVF, and opposite (sides) PMN and STU (respectively), [are] of equal height.
And, hence, the parallelepiped solids described on the aforementioned prisms [are] of equal height and (are) to one another as their bases [Prop. 11.32].
Likewise, the halves (of the solids) [Prop. 11.28].
Thus, as base LOC is to base RVF, so the aforementioned prisms (are) to one another.
(Which is) the very thing it was required to show.
Proposition 5
Pyramids which are of the same height, and have triangular bases, are to one another as their bases.
Let there be pyramids of the same height whose bases (are) the triangles ABC and DEF, and apexes the points G and H (respectively).
I say that as base ABC is to base DEF, so pyramid ABCG (is) to pyramid DEFH.
For if base ABC is not to base DEF, as pyramid ABCG (is) to pyramid DEFH, then base ABC will be to base DEF, as pyramid ABCG (is) to some solid either less than, or greater than, pyramid DEFH.
Let it, first of all, be (in this ratio) to (some) lesser (solid), W.
And let pyramid DEFH have been divided into two pyramids equal to one another, and similar to the whole, and into two equal prisms.
So, the (sum of the) two prisms is greater than half of the whole pyramid [Prop. 12.3].
And, again, let the pyramids generated by the division have been similarly divided, and let this be done continually until some pyramids are left from pyramid DEFH which (when added together) are less than the excess by which pyramid DEFH exceeds the solid W [Prop. 10.1].
Let them have been left, and, for the sake of argument, let them be DQRS and STUH.
Thus, the (sum of the) remaining prisms within pyramid DEFH is greater than solid W.
Let pyramid ABCG also have been divided similarly, and a similar number of times, as pyramid DEFH.
Thus, as base ABC is to base DEF, so the (sum of the) prisms within pyramid ABCG (is) to the (sum of the) prisms within pyramid DEFH [Prop. 12.4].
But, also, as base ABC (is) to base DEF, so pyramid ABCG (is) to solid W.
And, thus, as pyramid ABCG (is) to solid W, so the (sum of the) prisms within pyramid ABCG (is) to the (sum of the) prisms within pyramid DEFH [Prop. 5.11].
Thus, alternately, as pyramid ABCG (is) to the (sum of the) prisms within it, so solid W (is) to the (sum of the) prisms within pyramid DEFH [Prop. 5.16].
And pyramid ABCG (is) greater than the (sum of the) prisms within it.
Thus, solid W (is) also greater than the (sum of the) prisms within pyramid DEFH [Prop. 5.14].
But, (it is) also less.
This very thing is impossible.
Thus, as base ABC is to base DEF, so pyramid ABCG (is) not to some solid less than pyramid DEFH.
So, similarly, we can show that base DEF is not to base ABC, as pyramid DEFH (is) to some solid less than pyramid ABCG either.
So, I say that neither is base ABC to base DEF, as pyramid ABCG (is) to some solid greater than pyramid DEFH.
For, if possible, let it be (in this ratio) to some greater (solid), W.
Thus, inversely, as base DEF (is) to base ABC, so solid W (is) to pyramid ABCG.
And as solid W (is) to pyramid ABCG, so pyramid DEFH (is) to some (solid) less than pyramid ABCG, as shown before [Prop. 12.2 lem.].
And, thus, as base DEF (is) to base ABC, so pyramid DEFH (is) to some (solid) less than pyramid ABCG [Prop. 5.11].
The very thing was shown (to be) absurd.
Thus, base ABC is not to base DEF, as pyramid ABCG (is) to some solid greater than pyramid DEFH.
And, it was shown that neither (is it in this ratio) to a lesser (solid).
Thus, as base ABC is to base DEF, so pyramid ABCG (is) to pyramid DEFH.
(Which is) the very thing it was required to show.
Proposition 6
Pyramids which are of the same height, and have polygonal bases, are to one another as their bases.
Let there be pyramids of the same height whose bases (are) the polygons ABCDE and FGHKL, and apexes the points M and N (respectively).
I say that as base ABCDE is to base FGHKL, so pyramid ABCDEM (is) to pyramid FGHKLN.
For let AC, AD, FH, and FK have been joined.
Therefore, since ABCM and ACDM are two pyramids having triangular bases and equal height, they are to one another as their bases [Prop. 12.5].
Thus, as base ABC is to base ACD, so pyramid ABCM (is) to pyramid ACDM.
And, via composition, as base ABCD (is) to base ACD, so pyramid ABCDM (is) to pyramid ACDM [Prop. 5.18].
But, as base ACD (is) to base ADE, so pyramid ACDM (is) also to pyramid ADEM [Prop. 12.5].
Thus, via equality, as base ABCD (is) to base ADE, so pyramid ABCDM (is) to pyramid ADEM [Prop. 5.22].
And, again, via composition, as base ABCDE (is) to base ADE, so pyramid ABCDEM (is) to pyramid ADEM [Prop. 5.18].
So, similarly, it can also be shown that as base FGHKL (is) to base FGH, so pyramid FGHKLN (is) also to pyramid FGHN.
And since ADEM and FGHN are two pyramids having triangular bases and equal height, thus as base ADE (is) to base FGH, so pyramid ADEM (is) to pyramid FGHN [Prop. 12.5].
But, as base ADE (is) to base ABCDE, so pyramid ADEM (was) to pyramid ABCDEM.
Thus, via equality, as base ABCDE (is) to base FGH, so pyramid ABCDEM (is) also to pyramid FGHN [Prop. 5.22].
But, furthermore, as base FGH (is) to base FGHKL, so pyramid FGHN was also to pyramid FGHKLN.
Thus, via equality, as base ABCDE (is) to base FGHKL, so pyramid ABCDEM (is) also to pyramid FGHKLN [Prop. 5.22].
(Which is) the very thing it was required to show.
Proposition 7
Any prism having a triangular base is divided into three pyramids having triangular bases (which are) equal to one another.
Let there be a prism whose base (is) triangle ABC, and opposite (plane) DEF.
I say that prism ABCDEF is divided into three pyramids having triangular bases (which are) equal to one another.
For let BD, EC, and CD have been joined.
Since ABED is a parallelogram, and BD is its diagonal, triangle ABD is thus equal to triangle EBD [Prop. 1.34].
And, thus, the pyramid whose base (is) triangle ABD, and apex the point C, is equal to the pyramid whose base is triangle DEB, and apex the point C [Prop. 12.5].
But, the pyramid whose base is triangle DEB, and apex the point C, is the same as the pyramid whose base is triangle EBC, and apex the point D.
For they are contained by the same planes.
And, thus, the pyramid whose base is ABD, and apex the point C, is equal to the pyramid whose base is EBC and apex the point D.
Again, since FCBE is a parallelogram, and CE is its diagonal, triangle CEF is equal to triangle CBE [Prop. 1.34].
And, thus, the pyramid whose base is triangle BCE, and apex the point D, is equal to the pyramid whose base is triangle ECF, and apex the point D [Prop. 12.5].
And the pyramid whose base is triangle BCE, and apex the point D, was shown (to be) equal to the pyramid whose base is triangle ABD, and apex the point C.
Thus, the pyramid whose base is triangle CEF, and apex the point D, is also equal to the pyramid whose base [is] triangle ABD, and apex the point C.
Thus, the prism ABCDEF has been divided into three pyramids having triangular bases (which are) equal to one another.
And since the pyramid whose base is triangle ABD, and apex the point C, is the same as the pyramid whose base is triangle CAB, and apex the point D.
For they are contained by the same planes.
And the pyramid whose base (is) triangle ABD, and apex the point C, was shown (to be) a third of the prism whose base is triangle ABC, and opposite (plane) DEF, thus the pyramid whose base is triangle ABC, and apex the point D, is also a third of the pyramid having the same base, triangle ABC, and opposite (plane) DEF.
Corollary
And, from this, (it is) clear that any pyramid is the third part of the prism having the same base as it, and an equal height.
(Which is) the very thing it was required to show.
Proposition 8
Similar pyramids which also have triangular bases are in the cubed ratio of their corresponding sides.
Let there be similar, and similarly laid out, pyramids whose bases are triangles ABC and DEF, and apexes the points G and H (respectively).
I say that pyramid ABCG has to pyramid DEFH the cubed ratio of that BC (has) to EF.
For let the parallelepiped solids BGML and EHQP have been completed.
And since pyramid ABCG is similar to pyramid DEFH, angle ABC is thus equal to angle DEF, and GBC to HEF, and ABG to DEH.
And as AB is to DE, so BC (is) to EF, and BG to EH [Def. 11.9].
And since as AB is to DE, so BC (is) to EF, and (so) the sides around equal angles are proportional, parallelogram BM is thus similar to paralleleogram EQ.
So, for the same (reasons), BN is also similar to ER, and BK to EO.
Thus, the three (parallelograms) MB, BK, and BN are similar to the three (parallelograms) EQ, EO, ER (respectively).
But, the three (parallelograms) MB, BK, and BN are (both) equal and similar to the three opposite (parallelograms), and the three (parallelograms) EQ, EO, and ER are (both) equal and similar to the three opposite (parallelograms) [Prop. 11.24].
Thus, the solids BGML and EHQP are contained by equal numbers of similar (and similarly laid out) planes.
Thus, solid BGML is similar to solid EHQP [Def. 11.9].
And similar parallelepiped solids are in the cubed ratio of corresponding sides [Prop. 11.33].
Thus, solid BGML has to solid EHQP the cubed ratio that the corresponding side BC (has) to the corresponding side EF.
And as solid BGML (is) to solid EHQP, so pyramid ABCG (is) to pyramid DEFH, inasmuch as the pyramid is the sixth part of the solid, on account of the prism, being half of the parallelepiped solid [Prop. 11.28], also being three times the pyramid [Prop. 12.7].
Thus, pyramid ABCG also has to pyramid DEFH the cubed ratio that BC (has) to EF.
(Which is) the very thing it was required to show.
Corollary
So, from this, (it is) also clear that similar pyramids having polygonal bases (are) to one another as the cubed ratio of their corresponding sides.
For, dividing them into the pyramids (contained) within them which have triangular bases, with the similar polygons of the bases also being divided into similar triangles (which are) both equal in number, and corresponding, to the wholes [Prop. 6.20].
As one pyramid having a triangular base in the former (pyramid having a polygonal base is) to one pyramid having a triangular base in the latter (pyramid having a polygonal base), so (the sum of) all the pyramids having triangular bases in the former pyramid will also be to (the sum of) all the pyramids having triangular bases in the latter pyramid [Prop. 5.12] ---that is to say, the (former) pyramid itself having a polygonal base to the (latter) pyramid having a polygonal base.
And a pyramid having a triangular base is to a (pyramid) having a triangular base in the cubed ratio of corresponding sides [Prop. 12.8].
Thus, a (pyramid) having a polygonal base also has to to a (pyramid) having a similar base the cubed ratio of a (corresponding) side to a (corresponding) side.
Proposition 9
The bases of equal pyramids which also have triangular bases are reciprocally proportional to their heights.
And those pyramids which have triangular bases whose bases are reciprocally proportional to their heights are equal.
For let there be (two) equal pyramids having the triangular bases ABC and DEF, and apexes the points G and H (respectively).
I say that the bases of the pyramids ABCG and DEFH are reciprocally proportional to their heights, and (so) that as base ABC is to base DEF, so the height of pyramid DEFH (is) to the height of pyramid ABCG.
For let the parallelepiped solids BGML and EHQP have been completed.
And since pyramid ABCG is equal to pyramid DEFH, and solid BGML is six times pyramid ABCG (see previous proposition), and solid EHQP (is) six times pyramid DEFH, solid BGML is thus equal to solid EHQP.
And the bases of equal parallelepiped solids are reciprocally proportional to their heights [Prop. 11.34].
Thus, as base BM is to base EQ, so the height of solid EHQP (is) to the height of solid BGML.
But, as base BM (is) to base EQ, so triangle ABC (is) to triangle DEF [Prop. 1.34].
And, thus, as triangle ABC (is) to triangle DEF, so the height of solid EHQP (is) to the height of solid BGML [Prop. 5.11].
But, the height of solid EHQP is the same as the height of pyramid DEFH, and the height of solid BGML is the same as the height of pyramid ABCG.
Thus, as base ABC is to base DEF, so the height of pyramid DEFH (is) to the height of pyramid ABCG.
Thus, the bases of pyramids ABCG and DEFH are reciprocally proportional to their heights.
And so, let the bases of pyramids ABCG and DEFH be reciprocally proportional to their heights, and (thus) let base ABC be to base DEF, as the height of pyramid DEFH (is) to the height of pyramid ABCG.
I say that pyramid ABCG is equal to pyramid DEFH.
For, with the same construction, since as base ABC is to base DEF, so the height of pyramid DEFH (is) to the height of pyramid ABCG, but as base ABC (is) to base DEF, so parallelogram BM (is) to parallelogram EQ [Prop. 1.34], thus as parallelogram BM (is) to parallelogram EQ, so the height of pyramid DEFH (is) also to the height of pyramid ABCG [Prop. 5.11].
But, the height of pyramid DEFH is the same as the height of parallelepiped EHQP, and the height of pyramid ABCG is the same as the height of parallelepiped BGML.
Thus, as base BM is to base EQ, so the height of parallelepiped EHQP (is) to the height of parallelepiped BGML.
And those parallelepiped solids whose bases are reciprocally proportional to their heights are equal [Prop. 11.34].
Thus, the parallelepiped solid BGML is equal to the parallelepiped solid EHQP.
And pyramid ABCG is a sixth part of BGML, and pyramid DEFH a sixth part of parallelepiped EHQP.
Thus, pyramid ABCG is equal to pyramid DEFH.
Thus, the bases of equal pyramids which also have triangular bases are reciprocally proportional to their heights.
And those pyramids having triangular bases whose bases are reciprocally proportional to their heights are equal.
(Which is) the very thing it was required to show.
Proposition 10
Every cone is the third part of the cylinder which has the same base as it, and an equal height.
For let there be a cone (with) the same base as a cylinder, (namely) the circle ABCD, and an equal height.
I say that the cone is the third part of the cylinder—that is to say, that the cylinder is three times the cone.
For if the cylinder is not three times the cone then the cylinder will be either more than three times, or less than three times, (the cone).
Let it, first of all, be more than three times (the cone).
And let the square ABCD have been inscribed in circle ABCD [Prop. 4.6].
So, square ABCD is more than half of circle ABCD [Prop. 12.2].
And let a prism of equal height to the cylinder have been set up on square ABCD.
So, the prism set up is more than half of the cylinder, inasmuch as if we also circumscribe a square around circle ABCD [Prop. 4.7] then the square inscribed in circle ABCD is half of the circumscribed (square).
And the solids set up on them are parallelepiped prisms of equal height.
And parallelepiped solids having the same height are to one another as their bases [Prop. 11.32].
And, thus, the prism set up on square ABCD is half of the prism set up on the square circumscribed about circle ABCD.
And the cylinder is less than the prism set up on the square circumscribed about circle ABCD.
Thus, the prism set up on square ABCD of the same height as the cylinder is more than half of the cylinder.
Let the circumferences AB, BC, CD, and DA have been cut in half at points E, F, G, and H.
And let AE, EB, BF, FC, CG, GD, DH, and HA have been joined.
And thus each of the triangles AEB, BFC, CGD, and DHA is more than half of the segment of circle ABCD about it, as was shown previously [Prop. 12.2].
Let prisms of equal height to the cylinder have been set up on each of the triangles AEB, BFC, CGD, and DHA.
And each of the prisms set up is greater than the half part of the segment of the cylinder about it—inasmuch as if we draw (straight-lines) parallel to AB, BC, CD, and DA through points E, F, G, and H (respectively), and complete the parallelograms on AB, BC, CD, and DA, and set up parallelepiped solids of equal height to the cylinder on them, then the prisms on triangles AEB, BFC, CGD, and DHA are each half of the set up (parallelepipeds).
And the segments of the cylinder are less than the set up parallelepiped solids.
Hence, the prisms on triangles AEB, BFC, CGD, and DHA are also greater than half of the segments of the cylinder about them.
So (if) the remaining circumferences are cut in half, and straight-lines are joined, and prisms of equal height to the cylinder are set up on each of the triangles, and this is done continually, then we will (eventually) leave some segments of the cylinder whose (sum) is less than the excess by which the cylinder exceeds three times the cone [Prop. 10.1].
Let them have been left, and let them be AE, EB, BF, FC, CG, GD, DH, and HA.
Thus, the remaining prism whose base (is) polygon AEBFCGDH, and height the same as the cylinder, is greater than three times the cone.
But, the prism whose base is polygon AEBFCGDH, and height the same as the cylinder, is three times the pyramid whose base is polygon AEBFCGDH, and apex the same as the cone [Prop. 12.7 corr.].
And thus the pyramid whose base [is] polygon AEBFCGDH, and apex the same as the cone, is greater than the cone having (as) base circle ABCD.
But (it is) also less.
For it is encompassed by it.
The very thing (is) impossible.
Thus, the cylinder is not more than three times the cone.
So, I say that neither (is) the cylinder less than three times the cone.
For, if possible, let the cylinder be less than three times the cone.
Thus, inversely, the cone is greater than the third part of the cylinder.
So, let the square ABCD have been inscribed in circle ABCD [Prop. 4.6].
Thus, square ABCD is greater than half of circle ABCD.
And let a pyramid having the same apex as the cone have been set up on square ABCD.
Thus, the pyramid set up is greater than the half part of the cone, inasmuch as we showed previously that if we circumscribe a square about the circle [Prop. 4.7] then the square ABCD will be half of the square circumscribed about the circle [Prop. 12.2].
And if we set up on the squares parallelepiped solids—which are also called prisms—of the same height as the cone, then the (prism) set up on square ABCD will be half of the (prism) set up on the square circumscribed about the circle.
For they are to one another as their bases [Prop. 11.32].
Hence, (the same) also (goes for) the thirds.
Thus, the pyramid whose base is square ABCD is half of the pyramid set up on the square circumscribed about the circle [Prop. 12.7 corr.].
And the pyramid set up on the square circumscribed about the circle is greater than the cone.
For it encompasses it.
Thus, the pyramid whose base is square ABCD, and apex the same as the cone, is greater than half of the cone.
Let the circumferences AB, BC, CD, and DA have been cut in half at points E, F, G, and H (respectively).
And let AE, EB, BF, FC, CG, GD, DH, and HA have been joined.
And, thus, each of the triangles AEB, BFC, CGD, and DHA is greater than the half part of the segment of circle ABCD about it [Prop. 12.2].
And let pyramids having the same apex as the cone have been set up on each of the triangles AEB, BFC, CGD, and DHA.
And, thus, in the same way, each of the pyramids set up is more than the half part of the segment of the cone about it.
So, (if) the remaining circumferences are cut in half, and straightlines are joined, and pyramids having the same apex as the cone are set up on each of the triangles, and this is done continually, then we will (eventually) leave some segments of the cone whose (sum) is less than the excess by which the cone exceeds the third part of the cylinder [Prop. 10.1].
Let them have been left, and let them be the (segments) on AE, EB, BF, FC, CG, GD, DH, and HA.
Thus, the remaining pyramid whose base is polygon AEBFCGDH, and apex the same as the cone, is greater than the third part of the cylinder.
But, the pyramid whose base is polygon AEBFCGDH, and apex the same as the cone, is the third part of the prism whose base is polygon AEBFCGDH, and height the same as the cylinder [Prop. 12.7 corr.].
Thus, the prism whose base is polygon AEBFCGDH, and height the same as the cylinder, is greater than the cylinder whose base is circle ABCD.
But, (it is) also less.
For it is encompassed by it.
The very thing is impossible.
Thus, the cylinder is not less than three times the cone.
And it was shown that neither (is it) greater than three times (the cone).
Thus, the cylinder (is) three times the cone.
Hence, the cone is the third part of the cylinder.
Thus, every cone is the third part of the cylinder which has the same base as it, and an equal height.
(Which is) the very thing it was required to show.
Proposition 11
Cones and cylinders having the same height are to one another as their bases.
Let there be cones and cylinders of the same height whose bases [are] the circles ABCD and EFGH, axes KL and MN, and diameters of the bases AC and EG (respectively).
I say that as circle ABCD is to circle EFGH, so cone AL (is) to cone EN.
For if not, then as circle ABCD (is) to circle EFGH, so cone AL will be to some solid either less than, or greater than, cone EN.
Let it, first of all, be (in this ratio) to (some) lesser (solid), O.
And let solid X be equal to that (magnitude) by which solid O is less than cone EN.
Thus, cone EN is equal to (the sum of) solids O and X.
Let the square EFGH have been inscribed in circle EFGH [Prop. 4.6].
Thus, the square is greater than half of the circle [Prop. 12.2].
Let a pyramid of the same height as the cone have been set up on square EFGH.
Thus, the pyramid set up is greater than half of the cone, inasmuch as, if we circumscribe a square about the circle [Prop. 4.7], and set up on it a pyramid of the same height as the cone, then the inscribed pyramid is half of the circumscribed pyramid.
For they are to one another as their bases [Prop. 12.6].
And the cone (is) less than the circumscribed pyramid.
Let the circumferences EF, FG, GH, and HE have been cut in half at points P, Q, R, and S.
And let HP, PE, EQ, QF, FR, RG, GS, and SH have been joined.
Thus, each of the triangles HPE, EQF, FRG, and GSH is greater than half of the segment of the circle about it [Prop. 12.2].
Let pyramids of the same height as the cone have been set up on each of the triangles HPE, EQF, FRG, and GSH.
And, thus, each of the pyramids set up is greater than half of the segment of the cone about it [Prop. 12.10].
So, (if) the remaining circumferences are cut in half, and straight-lines are joined, and pyramids of equal height to the cone are set up on each of the triangles, and this is done continually, then we will (eventually) leave some segments of the cone (the sum of) which is less than solid X [Prop. 10.1].
Let them have been left, and let them be the (segments) on HPE, EQF, FRG, and GSH.
Thus, the remaining pyramid whose base is polygon HPEQFRGS, and height the same as the cone, is greater than solid O [Prop. 6.18].
And let the polygon DTAUBVCW, similar, and similarly laid out, to polygon HPEQFRGS, have been inscribed in circle ABCD.
And on it let a pyramid of the same height as cone AL have been set up.
Therefore, since as the (square) on AC is to the (square) on EG, so polygon DTAUBVCW (is) to polygon HPEQFRGS [Prop. 12.1], and as the (square) on AC (is) to the (square) on EG, so circle ABCD (is) to circle EFGH [Prop. 12.2], thus as circle ABCD (is) to circle EFGH, so polygon DTAUBVCW also (is) to polygon HPEQFRGS.
And as circle ABCD (is) to circle EFGH, so cone AL (is) to solid O.
And as polygon DTAUBVCW (is) to polygon HPEQFRGS, so the pyramid whose base is polygon DTAUBVCW, and apex the point L, (is) to the pyramid whose base is polygon HPEQFRGS, and apex the point N [Prop. 12.6].
And, thus, as cone AL (is) to solid O, so the pyramid whose base is DTAUBVCW, and apex the point L, (is) to the pyramid whose base is polygon HPEQFRGS, and apex the point N [Prop. 5.11].
Thus, alternately, as cone AL is to the pyramid within it, so solid O (is) to the pyramid within cone EN [Prop. 5.16].
But, cone AL (is) greater than the pyramid within it.
Thus, solid O (is) also greater than the pyramid within cone EN [Prop. 5.14].
But, (it is) also less.
The very thing (is) absurd.
Thus, circle ABCD is not to circle EFGH, as cone AL (is) to some solid less than cone EN.
So, similarly, we can show that neither is circle EFGH to circle ABCD, as cone EN (is) to some solid less than cone AL.
So, I say that neither is circle ABCD to circle EFGH, as cone AL (is) to some solid greater than cone EN.
For, if possible, let it be (in this ratio) to (some) greater (solid), O.
Thus, inversely, as circle EFGH is to circle ABCD, so solid O (is) to cone AL [Prop. 5.7 corr.].
But, as solid O (is) to cone AL, so cone EN (is) to some solid less than cone AL [Prop. 12.2 lem.].
And, thus, as circle EFGH (is) to circle ABCD, so cone EN (is) to some solid less than cone AL.
The very thing was shown (to be) impossible.
Thus, circle ABCD is not to circle EFGH, as cone AL (is) to some solid greater than cone EN.
And, it was shown that neither (is it in this ratio) to (some) lesser (solid).
Thus, as circle ABCD is to circle EFGH, so cone AL (is) to cone EN.
But, as the cone (is) to the cone, (so) the cylinder (is) to the cylinder.
For each (is) three times each [Prop. 12.10].
Thus, circle ABCD (is) also to circle EFGH, as (the ratio of the cylinders) on them (having) the same height.
Thus, cones and cylinders having the same height are to one another as their bases.
(Which is) the very thing it was required to show.
Proposition 12
Similar cones and cylinders are to one another in the cubed ratio of the diameters of their bases.
Let there be similar cones and cylinders of which the bases (are) the circles ABCD and EFGH, the diameters of the bases (are) BD and FH, and the axes of the cones and cylinders (are) KL and MN (respectively).
I say that the cone whose base [is] circle ABCD, and apex the point L, has to the cone whose base [is] circle EFGH, and apex the point N, the cubed ratio that BD (has) to FH.
For if cone ABCDL does not have to cone EFGHN the cubed ratio that BD (has) to FH then cone ABCDL will have the cubed ratio to some solid either less than, or greater than, cone EFGHN.
Let it, first of all, have (such a ratio) to (some) lesser (solid), O.
And let the square EFGH have been inscribed in circle EFGH [Prop. 4.6].
Thus, square EFGH is greater than half of circle EFGH [Prop. 12.2].
And let a pyramid having the same apex as the cone have been set up on square EFGH.
Thus, the pyramid set up is greater than the half part of the cone [Prop. 12.10].
So, let the circumferences EF, FG, GH, and HE have been cut in half at points P, Q, R, and S (respectively).
And let EP, PF, FQ, QG, GR, RH, HS, and SE have been joined.
And, thus, each of the triangles EPF, FQG, GRH, and HSE is greater than the half part of the segment of circle EFGH about it [Prop. 12.2].
And let a pyramid having the same apex as the cone have been set up on each of the triangles EPF, FQG, GRH, and HSE.
And thus each of the pyramids set up is greater than the half part of the segment of the cone about it [Prop. 12.10].
So, (if) the the remaining circumferences are cut in half, and straight-lines are joined, and pyramids having the same apex as the cone are set up on each of the triangles, and this is done continually, then we will (eventually) leave some segments of the cone whose (sum) is less than the excess by which cone EFGHN exceeds solid O [Prop. 10.1].
Let them have been left, and let them be the (segments) on EP, PF, FQ, QG, GR, RH, HS, and SE.
Thus, the remaining pyramid whose base is polygon EPFQGRHS, and apex the point N, is greater than solid O.
And let the polygon ATBUCVDW, similar, and similarly laid out, to polygon EPFQGRHS, have been inscribed in circle ABCD [Prop. 6.18].
And let a pyramid having the same apex as the cone have been set up on polygon ATBUCVDW.
And let LBT be one of the triangles containing the pyramid whose base is polygon ATBUCVDW, and apex the point L.
And let NFP be one of the triangles containing the pyramid whose base is triangle EPFQGRHS, and apex the point N.
And let KT and MP have been joined.
And since cone ABCDL is similar to cone EFGHN, thus as BD is to FH, so axis KL (is) to axis MN [Def. 11.24].
And as BD (is) to FH, so BK (is) to FM.
And, thus, as BK (is) to FM, so KL (is) to MN.
And, alternately,as BK (is) to KL, so FM (is) to MN [Prop. 5.16].
And the sides around the equal angles BKL and FMN are proportional.
Thus, triangle BKL is similar to triangle FMN [Prop. 6.6].
Again, since as BK (is) to KT, so FM (is) to MP, and (they are) about the equal angles BKT and FMP, inasmuch as whatever part angle BKT is of the four right-angles at the center K, angle FMP is also the same part of the four right-angles at the center M.
Therefore, since the sides about equal angles are proportional, triangle BKT is thus similar to traingle FMP [Prop. 6.6].
Again, since it was shown that as BK (is) to KL, so FM (is) to MN, and BK (is) equal to KT, and FM to PM, thus as TK (is) to KL, so PM (is) to MN.
And the sides about the equal angles TKL and PMN ---for (they are both) right-angles---are proportional.
Thus, triangle LKT (is) similar to triangle NMP [Prop. 6.6].
And since, on account of the similarity of triangles LKB and NMF, as LB (is) to BK, so NF (is) to FM, and, on account of the similarity of triangles BKT and FMP, as KB (is) to BT, so MF (is) to FP [Def. 6.1], thus, via equality, as LB (is) to BT, so NF (is) to FP [Prop. 5.22].
Again, since, on account of the similarity of triangles LTK and NPM, as LT (is) to TK, so NP (is) to PM, and, on account of the similarity of triangles TKB and PMF, as KT (is) to TB, so MP (is) to PF, thus, via equality, as LT (is) to TB, so NP (is) to PF [Prop. 5.22].
And it was shown that as TB (is) to BL, so PF (is) to FN.
Thus, via equality, as TL (is) to LB, so PN (is) to NF [Prop. 5.22].
Thus, the sides of triangles LTB and NPF are proportional.
Thus, triangles LTB and NPF are equiangular [Prop. 6.5].
And, hence, (they are) similar [Def. 6.1].
And, thus, the pyramid whose base is triangle BKT, and apex the point L, is similar to the pyramid whose base is triangle FMP, and apex the point N.
For they are contained by equal numbers of similar planes [Def. 11.9].
And similar pyramids which also have triangular bases are in the cubed ratio of corresponding sides [Prop. 12.8].
Thus, pyramid BKTL has to pyramid FMPN the cubed ratio that BK (has) to FM.
So, similarly, joining straight-lines from (points) A, W, D, V, C, and U to (center) K, and from (points) E, S, H, R, G, and Q to (center) M, and setting up pyramids having the same apexes as the cones on each of the triangles (so formed), we can also show that each of the pyramids (on base ABCD taken) in order will have to each of the pyramids (on base EFGH taken) in order the cubed ratio that the corresponding side BK (has) to the corresponding side FM —that is to say, that BD (has) to FH.
And (for two sets of proportional magnitudes) as one of the leading (magnitudes is) to one of the following, so (the sum of) all of the leading (magnitudes is) to (the sum of) all of the following (magnitudes) [Prop. 5.12].
And, thus, as pyramid BKTL (is) to pyramid FMPN, so the whole pyramid whose base is polygon ATBUCVDW, and apex the point L, (is) to the whole pyramid whose base is polygon EPFQGRHS, and apex the point N.
And, hence, the pyramid whose base is polygon ATBUCVDW, and apex the point L, has to the pyramid whose base is polygon EPFQGRHS, and apex the point N, the cubed ratio that BD (has) to FH.
And it was also assumed that the cone whose base is circle ABCD, and apex the point L, has to solid O the cubed ratio that BD (has) to FH.
Thus, as the cone whose base is circle ABCD, and apex the point L, is to solid O, so the pyramid whose base (is) [polygon] ATBUCVDW, and apex the point L, (is) to the pyramid whose base is polygon EPFQGRHS, and apex the point N.
Thus, alternately, as the cone whose base (is) circle ABCD, and apex the point L, (is) to the pyramid within it whose base (is) the polygon ATBUCVDW, and apex the point L, so the [solid] O (is) to the pyramid whose base is polygon EPFQGRHS, and apex the point N [Prop. 5.16].
And the aforementioned cone (is) greater than the pyramid within it.
For it encompasses it.
Thus, solid O (is) also greater than the pyramid whose base is polygon EPFQGRHS, and apex the point N.
But, (it is) also less.
The very thing is impossible.
Thus, the cone whose base (is) circle ABCD, and apex the [point] L, does not have to some solid less than the cone whose base (is) circle EFGH, and apex the point N, the cubed ratio that BD (has) to FH.
So, similarly, we can show that neither does cone EFGHN have to some solid less than cone ABCDL the cubed ratio that FH (has) to BD.
So, I say that neither does cone ABCDL have to some solid greater than cone EFGHN the cubed ratio that BD (has) to FH.
For, if possible, let it have (such a ratio) to a greater (solid), O.
Thus, inversely, solid O has to cone ABCDL the cubed ratio that FH (has) to BD [Prop. 5.7 corr.].
And as solid O (is) to cone ABCDL, so cone EFGHN (is) to some solid less than cone ABCDL [12.2 lem.].
Thus, cone EFGHN also has to some solid less than cone ABCDL the cubed ratio that FH (has) to BD.
The very thing was shown (to be) impossible.
Thus, cone ABCDL does not have to some solid greater than cone EFGHN the cubed ratio than BD (has) to FH.
And it was shown that neither (does it have such a ratio) to a lesser (solid).
Thus, cone ABCDL has to cone EFGHN the cubed ratio that BD (has) to FH.
And as the cone (is) to the cone, so the cylinder (is) to the cylinder.
For a cylinder is three times a cone on the same base as the cone, and of the same height as it [Prop. 12.10].
Thus, the cylinder also has to the cylinder the cubed ratio that BD (has) to FH.
Thus, similar cones and cylinders are in the cubed ratio of the diameters of their bases.
(Which is) the very thing it was required to show.
Proposition 13
If a cylinder is cut by a plane which is parallel to the opposite planes (of the cylinder) then as the cylinder (is) to the cylinder, so the axis will be to the axis.
For let the cylinder AD have been cut by the plane GH which is parallel to the opposite planes (of the cylinder), AB and CD.
And let the plane GH have met the axis at point K.
I say that as cylinder BG is to cylinder GD, so axis EK (is) to axis KF.
For let axis EF have been produced in each direction to points L and M.
And let any number whatsoever (of lengths), EN and NL, equal to axis EK, be set out (on the axis EL), and any number whatsoever (of lengths), FO and OM, equal to (axis) FK, (on the axis KM).
And let the cylinder PW, whose bases (are) the circles PQ and VW, have been conceived on axis LM.
And let planes parallel to AB, CD, and the bases of cylinder PW, have been produced through points N and O, and let them have made the circles RS and TU around the centers N and O (respectively).
And since axes LN, NE, and EK are equal to one another, the cylinders QR, RB, and BG are to one another as their bases [Prop. 12.11].
But the bases are equal.
Thus, the cylinders QR, RB, and BG (are) also equal to one another.
Therefore, since the axes LN, NE, and EK are equal to one another, and the cylinders QR, RB, and BG are also equal to one another, and the number (of the former) is equal to the number (of the latter), thus as many multiples as axis KL is of axis EK, so many multiples is cylinder QG also of cylinder GB.
And so, for the same (reasons), as many multiples as axis MK is of axis KF, so many multiples is cylinder WG also of cylinder GD.
And if axis KL is equal to axis KM then cylinder QG will also be equal to cylinder GW, and if the axis (is) greater than the axis then the cylinder (will also be) greater than the cylinder, and if (the axis is) less then (the cylinder will also be) less.
So, there are four magnitudes—the axes EK and KF, and the cylinders BG and GD —and equal multiples have been taken of axis EK and cylinder BG —(namely), axis LK and cylinder QG —and of axis KF and cylinder GD —(namely), axis KM and cylinder GW.
And it has been shown that if axis KL exceeds axis KM then cylinder QG also exceeds cylinder GW, and if (the axes are) equal then (the cylinders are) equal, and if ( KL is) less then ( QG is) less.
Thus, as axis EK is to axis KF, so cylinder BG (is) to cylinder GD [Def. 5.5].
(Which is) the very thing it was required to show.
Proposition 14
Cones and cylinders which are on equal bases are to one another as their heights.
For let EB and FD be cylinders on equal bases, (namely) the circles AB and CD (respectively).
I say that as cylinder EB is to cylinder FD, so axis GH (is) to axis KL.
For let the axis KL have been produced to point N.
And let LN be made equal to axis GH.
And let the cylinder CM have been conceived about axis LN.
Therefore, since cylinders EB and CM have the same height they are to one another as their bases [Prop. 12.11].
And the bases are equal to one another.
Thus, cylinders EB and CM are also equal to one another.
And since cylinder FM has been cut by the plane CD, which is parallel to its opposite planes, thus as cylinder CM is to cylinder FD, so axis LN (is) to axis KL [Prop. 12.13].
And cylinder CM is equal to cylinder EB, and axis LN to axis GH.
Thus, as cylinder EB is to cylinder FD, so axis GH (is) to axis KL.
And as cylinder EB (is) to cylinder FD, so cone ABG (is) to cone CDK [Prop. 12.10].
Thus, also, as axis GH (is) to axis KL, so cone ABG (is) to cone CDK, and cylinder EB to cylinder FD.
(Which is) the very thing it was required to show.
Proposition 15
The bases of equal cones and cylinders are reciprocally proportional to their heights.
And, those cones and cylinders whose bases (are) reciprocally proportional to their heights are equal.
Let there be equal cones and cylinders whose bases are the circles ABCD and EFGH, and the diameters of (the bases) AC and EG, and (whose) axes (are) KL and MN, which are also the heights of the cones and cylinders (respectively).
And let the cylinders AO and EP have been completed.
I say that the bases of cylinders AO and EP are reciprocally proportional to their heights, and (so) as base ABCD is to base EFGH, so height MN (is) to height KL.
For height LK is either equal to height MN, or not.
Let it, first of all, be equal.
And cylinder AO is also equal to cylinder EP.
And cones and cylinders having the same height are to one another as their bases [Prop. 12.11].
Thus, base ABCD (is) also equal to base EFGH.
And, hence, reciprocally, as base ABCD (is) to base EFGH, so height MN (is) to height KL.
And so, let height LK not be equal to MN, but let MN be greater.
And let QN, equal to KL, have been cut off from height MN.
And let the cylinder EP have been cut, through point Q, by the plane TUS (which is) parallel to the planes of the circles EFGH and RP.
And let cylinder ES have been conceived, with base the circle EFGH, and height NQ.
And since cylinder AO is equal to cylinder EP, thus, as cylinder AO (is) to cylinder ES, so cylinder EP (is) to cylinder ES [Prop. 5.7].
But, as cylinder AO (is) to cylinder ES, so base ABCD (is) to base EFGH.
For cylinders AO and ES (have) the same height [Prop. 12.11].
And as cylinder EP (is) to (cylinder) ES, so height MN (is) to height QN.
For cylinder EP has been cut by a plane which is parallel to its opposite planes [Prop. 12.13].
And, thus, as base ABCD is to base EFGH, so height MN (is) to height QN [Prop. 5.11].
And height QN (is) equal to height KL.
Thus, as base ABCD is to base EFGH, so height MN (is) to height KL.
Thus, the bases of cylinders AO and EP are reciprocally proportional to their heights.
And, so, let the bases of cylinders AO and EP be reciprocally proportional to their heights, and (thus) let base ABCD be to base EFGH, as height MN (is) to height KL.
I say that cylinder AO is equal to cylinder EP.
For, with the same construction, since as base ABCD is to base EFGH, so height MN (is) to height KL, and height KL (is) equal to height QN, thus, as base ABCD (is) to base EFGH, so height MN will be to height QN.
But, as base ABCD (is) to base EFGH, so cylinder AO (is) to cylinder ES.
For they are the same height [Prop. 12.11].
And as height MN (is) to [height] QN, so cylinder EP (is) to cylinder ES [Prop. 12.13].
Thus, as cylinder AO is to cylinder ES, so cylinder EP (is) to (cylinder) ES [Prop. 5.11].
Thus, cylinder AO (is) equal to cylinder EP [Prop. 5.9].
In the same manner, (the proposition can) also (be demonstrated) for the cones.
(Which is) the very thing it was required to show.
Proposition 16
There being two circles about the same center, to inscribe an equilateral and even-sided polygon in the greater circle, not touching the lesser circle.
Let ABCD and EFGH be the given two circles, about the same center, K.
So, it is necessary to inscribe an equilateral and even-sided polygon in the greater circle ABCD, not touching circle EFGH.
Let the straight-line BKD have been drawn through the center K.
And let GA have been drawn, at right-angles to the straight-line BD, through point G, and let it have been drawn through to C.
Thus, AC touches circle EFGH [Prop. 3.16 corr.].
So, (by) cutting circumference BAD in half, and the half of it in half, and doing this continually, we will (eventually) leave a circumference less than AD [Prop. 10.1].
Let it have been left, and let it be LD.
And let LM have been drawn, from L, perpendicular to BD, and let it have been drawn through to N.
And let LD and DN have been joined.
Thus, LD is equal to DN [Prop. 3.3] [Prop. 1.4].
And since LN is parallel to AC [Prop. 1.28], and AC touches circle EFGH, LN thus does not touch circle EFGH.
Thus, even more so, LD and DN do not touch circle EFGH.
And if we continuously insert (straight-lines) equal to straight-line LD into circle ABCD [Prop. 4.1] then an equilateral and even-sided polygon, not touching the lesser circle EFGH, will have been inscribed in circle ABCD.
(Which is) the very thing it was required to do.
Proposition 17
There being two spheres about the same center, to inscribe a polyhedral solid in the greater sphere, not touching the lesser sphere on its surface.
Let two spheres have been conceived about the same center, A.
So, it is necessary to inscribe a polyhedral solid in the greater sphere, not touching the lesser sphere on its surface.
Let the spheres have been cut by some plane through the center.
So, the sections will be circles, inasmuch as a sphere is generated by the diameter remaining behind, and a semi-circle being carried around [Def. 11.14].
And, hence, whatever position we conceive (of for) the semi-circle, the plane produced through it will make a circle on the surface of the sphere.
And (it is) clear that (it is) also a great (circle), inasmuch as the diameter of the sphere, which is also manifestly the diameter of the semi-circle and the circle, is greater than all of the (other) [straight-lines] drawn across in the circle or the sphere [Prop. 3.15].
Therefore, let BCDE be the circle in the greater sphere, and FGH the circle in the lesser sphere.
And let two diameters of them have been drawn at right-angles to one another, (namely), BD and CE.
And there being two circles about the same center—(namely), BCDE and FGH —let an equilateral and even-sided polygon have been inscribed in the greater circle, BCDE, not touching the lesser circle, FGH [Prop. 12.16], of which let the sides in the quadrant BE be BK, KL, LM, and ME.
And, KA being joined, let it have been drawn across to N.
And let AO have been set up at point A, at right-angles to the plane of circle BCDE.
And let it meet the surface of the (greater) sphere at O.
And let planes have been produced through AO and each of BD and KN.
So, according to the aforementioned (discussion), they will make great circles on the surface of the (greater) sphere.
Let them make (great circles), of which let BOD and KON be semi-circles on the diameters BD and KN (respectively).
And since OA is at right-angles to the plane of circle BCDE, all of the planes through OA are thus also at right-angles to the plane of circle BCDE [Prop. 11.18].
And, hence, the semi-circles BOD and KON are also at right-angles to the plane of circle BCDE.
And since semi-circles BED, BOD, and KON are equal—for (they are) on the equal diameters BD and KN [Def. 3.1] —the quadrants BE, BO, and KO are also equal to one another.
Thus, as many sides of the polygon as are in quadrant BE, so many are also in quadrants BO and KO equal to the straight-lines BK, KL, LM, and ME.
Let them have been inscribed, and let them be BP, PQ, QR, RO, KS, ST, TU, and UO.
And let SP, TQ, and UR have been joined.
And let perpendiculars have been drawn from P and S to the plane of circle BCDE [Prop. 11.11].
So, they will fall on the common sections of the planes BD and KN (with BCDE), inasmuch as the planes of BOD and KON are also at right-angles to the plane of circle BCDE [Def. 11.4].
Let them have fallen, and let them be PV and SW.
And let WV have been joined.
And since BP and KS are equal (circumferences) having been cut off in the equal semi-circles BOD and KON [Def. 3.28], and PV and SW are perpendiculars having been drawn (from them), PV is [thus] equal to SW, and BV to KW [Prop. 3.27] [Prop. 1.26].
And the whole of BA is also equal to the whole of KA.
And, thus, as BV is to VA, so KW (is) to WA.
WV is thus parallel to KB [Prop. 6.2].
And since PV and SW are each at right-angles to the plane of circle BCDE, PV is thus parallel to SW [Prop. 11.6].
And it was also shown (to be) equal to it.
And, thus, WV and SP are equal and parallel [Prop. 1.33].
And since WV is parallel to SP, but WV is parallel to KB, SP is thus also parallel to KB [Prop. 11.1].
And BP and KS join them.
Thus, the quadrilateral KBPS is in one plane, inasmuch as if there are two parallel straight-lines, and a random point is taken on each of them, then the straight-line joining the points is in the same plane as the parallel (straight-lines) [Prop. 11.7].
So, for the same (reasons), each of the quadrilaterals SPQT and TQRU is also in one plane.
And triangle URO is also in one plane [Prop. 11.2].
So, if we conceive straight-lines joining points P, S, Q, T, R, and U to A then some solid polyhedral figure will have been constructed between the circumferences BO and KO, being composed of pyramids whose bases (are) the quadrilaterals KBPS, SPQT, TQRU, and the triangle URO, and apex the point A.
And if we also make the same construction on each of the sides KL, LM, and ME, just as on BK, and, further, (repeat the construction) in the remaining three quadrants, then some polyhedral figure which has been inscribed in the sphere will have been constructed, being contained by pyramids whose bases (are) the aforementioned quadrilaterals, and triangle URO, and the (quadrilaterals and triangles) similarly arranged to them, and apex the point A.
So, I say that the aforementioned polyhedron will not touch the lesser sphere on the surface on which the circle FGH is (situated).
Let the perpendicular (straight-line) AX have been drawn from point A to the plane KBPS, and let it meet the plane at point X [Prop. 11.11].
And let XB and XK have been joined.
And since AX is at right-angles to the plane of quadrilateral KBPS, it is thus also at right-angles to all of the straight-lines joined to it which are also in the plane of the quadrilateral [Def. 11.3].
Thus, AX is at right-angles to each of BX and XK.
And since AB is equal to AK, the (square) on AB is also equal to the (square) on AK.
And the (sum of the squares) on AX and XB is equal to the (square) on AB.
For the angle at X (is) a right-angle [Prop. 1.47].
And the (sum of the squares) on AX and XK is equal to the (square) on AK [Prop. 1.47].
Thus, the (sum of the squares) on AX and XB is equal to the (sum of the squares) on AX and XK.
Let the (square) on AX have been subtracted from both.
Thus, the remaining (square) on BX is equal to the remaining (square) on XK.
Thus, BX (is) equal to XK.
So, similarly, we can show that the straight-lines joined from X to P and S are equal to each of BX and XK.
Thus, a circle drawn (in the plane of the quadrilateral) with center X, and radius one of XB or XK, will also pass through P and S, and the quadrilateral KBPS will be inside the circle.
And since KB is greater than WV, and WV (is) equal to SP, KB (is) thus greater than SP.
And KB (is) equal to each of KS and BP.
Thus, KS and BP are each greater than SP.
And since quadrilateral KBPS is in a circle, and KB, BP, and KS are equal (to one another), and PS (is) less (than them), and BX is the radius of the circle, the (square) on KB is thus greater than double the (square) on BX.
Let the perpendicular KY have been drawn from K to BV.
And since BD is less than double DY, and as BD is to DY, so the (rectangle contained) by DB and BY (is) to the (rectangle contained) by DY and YB —a square being described on BY, and a (rectangular) parallelogram (with short side equal to BY) completed on YD —the (rectangle contained) by DB and BY is thus also less than double the (rectangle contained) by DY and YB.
And, KD being joined, the (rectangle contained) by DB and BY is equal to the (square) on BK, and the (rectangle contained) by DY and YB equal to the (square) on KY [Prop. 3.31] [Prop. 6.8 corr.].
Thus, the (square) on KB is less than double the (square) on KY.
But, the (square) on KB is greater than double the (square) on BX.
Thus, the (square) on KY (is) greater than the (square) on BX.
And since BA is equal to KA, the (square) on BA is equal to the (square) on AK.
And the (sum of the squares) on BX and XA is equal to the (square) on BA, and the (sum of the squares) on KY and YA (is) equal to the (square) on KA [Prop. 1.47].
Thus, the (sum of the squares) on BX and XA is equal to the (sum of the squares) on KY and YA, of which the (square) on KY (is) greater than the (square) on BX.
Thus, the remaining (square) on YA is less than the (square) on XA.
Thus, AX (is) greater than AY.
Thus, AX is much greater than AG.
And AX is (a perpendicular) on one of the bases of the polyhedron, and AG (is a perpendicular) on the surface of the lesser sphere.
Hence, the polyhedron will not touch the lesser sphere on its surface.
Thus, there being two spheres about the same center, a polyhedral solid has been inscribed in the greater sphere which does not touch the lesser sphere on its surface.
(Which is) the very thing it was required to do.
Corollary
And, also, if a similar polyhedral solid to that in sphere BCDE is inscribed in another sphere then the polyhedral solid in sphere BCDE has to the polyhedral solid in the other sphere the cubed ratio that the diameter of sphere BCDE has to the diameter of the other sphere.
For if the solids are divided into similarly numbered, and similarly situated, pyramids, then the pyramids will be similar.
And similar pyramids are in the cubed ratio of corresponding sides [Prop. 12.8 corr.].
Thus, the pyramid whose base is quadrilateral KBPS, and apex the point A, will have to the similarly situated pyramid in the other sphere the cubed ratio that a corresponding side (has) to a corresponding side.
That is to say, that of radius AB of the sphere about center A to the radius of the other sphere.
And, similarly, each pyramid in the sphere about center A will have to each similarly situated pyramid in the other sphere the cubed ratio that AB (has) to the radius of the other sphere.
And as one of the leading (magnitudes is) to one of the following (in two sets of proportional magnitudes), so (the sum of) all the leading (magnitudes is) to (the sum of) all of the following (magnitudes) [Prop. 5.12].
Hence, the whole polyhedral solid in the sphere about center A will have to the whole polyhedral solid in the other [sphere] the cubed ratio that (radius) AB (has) to the radius of the other sphere.
That is to say, that diameter BD (has) to the diameter of the other sphere.
(Which is) the very thing it was required to show.
Proposition 18
Spheres are to one another in the cubed ratio of their respective diameters.
Let the spheres ABC and DEF have been conceived, and (let) their diameters (be) BC and EF (respectively).
I say that sphere ABC has to sphere DEF the cubed ratio that BC (has) to EF.
For if sphere ABC does not have to sphere DEF the cubed ratio that BC (has) to EF then sphere ABC will have to some (sphere) either less than, or greater than, sphere DEF the cubed ratio that BC (has) to EF.
Let it, first of all, have (such a ratio) to a lesser (sphere), GHK.
And let DEF have been conceived about the same center as GHK.
And let a polyhedral solid have been inscribed in the greater sphere DEF, not touching the lesser sphere GHK on its surface [Prop. 12.17].
And let a polyhedral solid, similar to the polyhedral solid in sphere DEF, have also been inscribed in sphere ABC.
Thus, the polyhedral solid in sphere ABC has to the polyhedral solid in sphere DEF the cubed ratio that BC (has) to EF [Prop. 12.17 corr.].
And sphere ABC also has to sphere GHK the cubed ratio that BC (has) to EF.
Thus, as sphere ABC is to sphere GHK, so the polyhedral solid in sphere ABC (is) to the polyhedral solid in sphere DEF.
[Thus], alternately, as sphere ABC (is) to the polyhedral solid within it, so sphere GHK (is) to the polyhedral solid within sphere DEF [Prop. 5.16].
And sphere ABC (is) greater than the polyhedron within it.
Thus, sphere GHK (is) also greater than the polyhedron within sphere DEF [Prop. 5.14].
But, (it is) also less.
For it is encompassed by it.
Thus, sphere ABC does not have to (a sphere) less than sphere DEF the cubed ratio that diameter BC (has) to EF.
So, similarly, we can show that sphere DEF does not have to (a sphere) less than sphere ABC the cubed ratio that EF (has) to BC either.
So, I say that sphere ABC does not have to some (sphere) greater than sphere DEF the cubed ratio that BC (has) to EF either.
For, if possible, let it have (the cubed ratio) to a greater (sphere), LMN.
Thus, inversely, sphere LMN (has) to sphere ABC the cubed ratio that diameter EF (has) to diameter BC [Prop. 5.7 corr.].
And as sphere LMN (is) to sphere ABC, so sphere DEF (is) to some (sphere) less than sphere ABC, inasmuch as LMN is greater than DEF, as was shown before [Prop. 12.2 lem.].
And, thus, sphere DEF has to some (sphere) less than sphere ABC the cubed ratio that EF (has) to BC.
The very thing was shown (to be) impossible.
Thus, sphere ABC does not have to some (sphere) greater than sphere DEF the cubed ratio that BC (has) to EF.
And it was shown that neither (does it have such a ratio) to a lesser (sphere).
Thus, sphere ABC has to sphere DEF the cubed ratio that BC (has) to EF.
(Which is) the very thing it was required to show.