the Square EFGH of the fame Altitude with the Cone, then that Pyramid is greater than one half of the Cone. And fo let the Circumferences EF, FG, GH, HE, be bifected in the Points O, P, R, S, and join EO, OF, FP, PG, GR, RH, HS, SE; then each of the Triangles EOF, FPG, GRH, HSE, is greater than one half of the Segment of the Circle EFGH, in which it is; and erect a Pyramid upon each of the Triangles EOF, EPG, GRH, HSE, having the fame Altitude as the Cone: Then each of the Pyramids thus erected, is greater than half its correfponding Segment of the Cone. Wherefore bifecting the remaining Circumferences joining Right Lines, and erecting Pyramids upon each of the Triangles, having the fame Vertex as the Cone; and doing this continually, we shall leave at last certain Segments of the Cone that shall be less than the Excefs by which the Cone EFGHN exceeds the Solid X. Let these be the Segments that ftand on EO, OF, FP, PG, GR, RH, HS, SE; then the remaining Pyramid whose Base is the Polygon EOFPGRHS, and Vertex the Point N, is greater than the Solid X. Alfo let the Polygon ATBYCVDQ be defcribed in the Circle ABCD, fimilar and alike fituate to the Polygon EOFPGRHS; upon which erect a Pyramid having the fame Altitude as the Cone; and let LBT be one of the Triangles containing the Pyramid, whose Base is the Polygon ATBYCVDQ, and Vertex the Point L; as likewife NFO one of the Triangles containing the Pyramid EOFPGRHS, and Vertex the Point N, and let KT, MO, be joined. Then because the Cone ABCDL is fimilar to the Cone EFGHN, it fhall be as BD is to FH, fo is the Axis KL to the Axis MN; but as BD is to FH, fo is BK to FM; and as BK is to FM, confe- * 15. 5. quently fo is KL to MN; and (by Alternation) as BK is to KL, fo is F M to MN. And fince each is perpendicular, and the Sides about the equal Angles BKL, FMN, are proportional, the Triangle BK L shall be + fimilar to the Triangle FMN. Again, be- † 6. 6. cause BK is to KT, as F M is to MO, the Sides are proportional about equal Angles BKT, FMO, for the Angle BKT is the fame Part of the four Right Angles at the Center K, as the Angle F MO is of the

* 6.6.

four Right Angles at the Center M: the Triangle BKT, fhall be* fimilar to the Triangle FMO; and because it has been proved that BK is to KL, as FM is to MN, and BK is equal to KT, and F M to MO, it fhall be as TK is to KL, fo is OM to MN: and the proportional Sides are about equal Angles TK L, OMN, for they are Right Angles. Therefore the Triangle LKT fhall be fimilar to the Triangle MNO. And fince, by the Similarity of the Triangles BKL, FMN, it is as LB is to BK, fo is NF to FM; and, by the Similarity of the Triangle BKT, FMO, it is as KB is to BT, fo is MF to FO; it fhall be (by Equality of Proportion) as L B is to BT, fo is NF to FO. Again, fince by the Similarity of the Triangles LTK, NOM, it is as LT is to TK, fo is NO to OM; and, by the Similarity of the Triangles KBT, OMF, it is as KT is to TB; fo is MO to OF. It fhall be (by Equality of Proportion) as LT is to TB, fo is NO to OF: But it has been proved that TB is to BL, as OF is to FN. Wherefore, again (by Equality of Proportion) as TL is to LB, fo is ON to NF; and therefore the Sides of the Triangles LTB, NOF, are proportional; and fo the Triangles LTB, NO F, are equiangular and fimilar to each other. And confequently the Pyramid, whofe Bafe is the Triangle B KT, and Vertex the Point L, is fimilar to the Pyramid whose Base is the Triangle FMO, and Vertex the Point N; for they are contained under fimilar Planes equal in Multitude: But +8 of this. fimilar Pyramids that have triangular Bafes, are + to one another in the triplicate Proportion of their homologous Sides. Therefore the Pyramid BK TL to the Pyramid F MON has a triplicate Proportion of that which BK has to FM. In like Manner, drawing Right Lines from the Points A, Q, D, V, C, Y to K, as alfo others, from the Points E, S, H, R, G, P, to M, and erecting Pyramids on the Triangles having the fame Vertices as the Cones, we demonftrate that every Pyramid of one Cone, to every one of the other Cone, has a triplicate Proportion of that which the Side BK has to the homologous Side MF, that is, which BD has to F H. But as one of the Antecedents is to one of the Confequents, so are all the Antecedents to all the Confequents. Therefore as the

12.5.

the Pyramid BKTL is to the Pyramid F MON, fo is the whole Pyramid whofe Bafe is the Polygon ATBY CVDQ, and Vertex the Point L, to the whole Pyramid, whofe Bafe is the Polygon EOFPGRHS, and Vertex the Point N. Wherefore the Pyramid, whofe Bafe is the Polygon AT BYCVDQ, and Vertex the Point L, to the Pyramid whofe Bafe is the Polygon EOFPGRHS, and Vertex the Point N, has a triplicate Proportion of that which BD hath to FH. But the Cone whose Base is the Circle ABCD, and Vertex the Point L, is fuppofed to have to the Solid X a triplicate Proportion of that which BD has to FH. Therefore as the Cone, whofe Base is the Circle ABCD, and Vertex the Point L, is to the Solid X, fo is the Pyramid whose Base is the Polygon ATBYCVDQ, and Vertex the Point L, to the Pyramid whose Base is the Polygon EOFPGRHS, and Vertex the Point N. But the faid Cone is greater than the Pyramid that is in it, for it comprehends it. Therefore the Solid X alfo is greater than the Pyramid, whofe Bafe is the Polygon EOFPGRHS, and Vertex the Point N; but it is alfo lefs, which is abfurd. Therefore the Cone, whofe Bafe is the Circle ABCD, and Vertex the Point L, to fome Solid lefs than the Cone, whose Base is the Circle EF GH, and Vertex the Point N, has not a triplicate Proportion of that which BD has to FH. In like Manner, we demonstrate that the Cone EFGHN, to fome Solid lefs than the Cone ABCDL, has not a triplicate Proportion of that which FH has to B D. Laftly, I fay the Cone A B CDL, to a Solid greater than the Cone EFGHN, has not a triplicate Proportion of that which BD has to FH: For, if this be poffible, let it be fo to fome Solid Z greater than the Cone EFGHN. Then (by Inverfion) the Solid Z, to the Cone ABCDL, has a triplicate Proportion of that which FH has to BD. But fince the Solid Z is greater than the Cone EFGHN, the Solid Z shall be to the Cone ABCDL, as the Cone EFGHN is to fome Solid lefs than the Cone ABCDL; and therefore the Cone EFGHN, to fome Solid lefs than the Cone ABCDL, hath a triplicate Proportion of that which FH has to BD, which has been proved to be impoffible. Therefore the Cone ABCDL, to

S 3

fome

* 15. 5. t10 of this.

fome Solid greater than the Cone EFGHN, has not a triplicate Proportion of that which BD has to FH. It has been alfo demonftrated, that the Cone ABCDL, to fome Solid lefs than the Cone EFGHN, hath not a triplicate Proportion of that which BD has to F H. Wherefore the Cone ABCDL, to the Cone EFGHN, has a triplicate Proportion of that which BD has F H. But as Cone is to Cone, fo is Cylinder to Cylinder. For a Cylinder having the fame Base as a Cone, and the fame Altitude is + triple of the Cone, fince it is demonftrated, that every Cone is one third Part of a Cylinder, having the fame Bafe and equal Altitude. Therefore alfo a Cylinder to Cylinder has a triplicate Proportion of that which BD has to FH. Therefore, fimilar Cones and Cylinders are to one another in a triplicate Proportion of the Diameters of their Bafes; which was to be demonftrated.

*

PROPOSITION XIII.

THEOREM.

If a Cylinder be divided by a Plane parallel to the oppofite Planes, then as one Cylinder is to the other Cylinder, fo is the Axis to the Axis.

LET the Cylinder AD be divided by the Plane

GH, parallel to the oppofite Planes AB, CD, and meeting the Axis EF in the Point K. I fay, as the Cylinder BG is to the Cylinder GD, fo is the Axis E K to the Axis K F.

For let the Axis EF be both ways produced to L and M, and put any Number of EN, NL, &c. each equal to the Axis EK; and any Number of FX, XM, &c. each equal to F K. And thro' the Points L, N, X, M, let Planes parallel to AB, CD país. And in those Planes from L, N, X, M, as Centers, defcribe the Circles, OP, RS, TY, VQ, each equal to AB, CD, and conceive the Cylinders PR, RB, DT, TQ, to be compleated. Then because the Axis LN, NE, EK, are equal to each other, the *11 of this. Cylinders PR, RB, BG will be* to one another as their Bafes. And therefore the Cylinders PR, RB, BG, are equal. And fince the Axis LN, NE, EK,

are

1

are equal to each other, as alfo the Cylinders PR, RB, BG; and the Number of LN, NE, EK, is equal to the Number of PR, RB, BG: The Axis KL fhall be the fame Multiple of the Axis E K, as the Cylinder PG, is of the Cylinder G B. For the fame Reason, the Axis M K is the fame Multiple of the Axis KF, as the Cylinder GQ is of the Cylinder GD. Now, if the Axis KL be equal to the Axis KM, the Cylinder PG fhall be equal to the Cylinder GQ; if the Axis LK be greater than the Axis KM, the Cylinder PG fhall be likewife greater than the Cylinder GQ; and if lefs, lefs. Therefore, because there are four Magnitudes, viz. The Axis EK, K F, and the Cylinders BG, GD, and there are taken their Equimultiples, namely, the Axis KL and the Cylinder PG, the Equimultiples of the Axis EK, and the Cylinder B G; and the Axis K M, and the Cylinder GQ, the Equimultiples of the Axis KF, and the Cylinder GD: And it is demonstrated, that if the Axis L K exceeds the Axis K M, the Cylinder PG will exceed the Cylinder GQ; and if it be equal, equal, and lefs, lefs. Therefore, as the Axis EK is to the Axis KF, fo* is the Cylinder B G to the * Def. 5. 5. Cylinder, GD. Wherefore, if a Cylinder be divided by a Plane parallel to the oppofite Planes, then as one Cylinder is to the other Cylinder, fo is the Axis to the Axis; which was to be demonftrated.

PROPOSITION XIV.

THEOREM.

Cones and Cylinders being upon equal Bafes, are to one another as their Altitudes.

L

ET the Cylinders EB, FD, ftand upon equal Bases A B, CD. I fay, as the Cylinder E B is to the Cylinder FD, fo is the Axis GH to the Axis KL.

For produce the Axis KL to the Point N; and put LN, equal to the Axis GH; and let a Cylinder CM be conceived about the Axis LN. Then because the Cylinders E B, CM, have the fame Altitude, they are * to one another as their Bafes. But* 11 of this,

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their