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four Right Angles at the Center M: the Triangle * 6. 6.

BKT, shall be * similar to the Triangle FMO; and because it has been proved that B K is to KL, as FM is to MN, and BK is equal to KT, and FM to MO, it shall be as TK is to KL, so is O M to MN: and the proportional Sides are about equal Angles T KL, OMN, for they are Right Angles. Therefore the Triangle L K T shall be similar to the

Triangle MNO. And since, by the Similarity of the Triangles B KL, FMN, it is as L B is to BK, so is N F to FM; and, by the Similarity of the Triangle BKT, FMO, it is as KB-is to BT, so is M F to FO; it shall be (by Equality of Proportion) as L B is to BT, so is NÉ to FO. Again, fince by the Similarity of the Triangles LTĂ, NOM, it is as LT is to TK, so is NO to O M; and, by the Similarity of the Triangles KBT, OMF, it is as KT is to TB; fo is MO to OF. It shall 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 so the

Triangles LTB, NOF, are equiangular and similar to each other. And confequently the Pyramid, whose Base is the Triangle BKT, and Vertex the Point L, is similar to the Pyramid whose Base is the Triangle FMO, and Vertex the Point N; for they are con

tained under similar Planes equal in Multitude: But + 8 of this. fimilar Pyramids that have triangular Bases, are + to

one another in the triplicate Proportion of their homologous Sides. Therefore the Pyramid BKTL to the Pyramid F MON has a triplicate Proportion of that which BK has to FM. În like Manner, drawing Right Lines from the Points A, Q, D, V, C, Y to K, as also others, from the Points E, S, H, R, G, P, to M, and erecting Pyramids on the Triangles having the same Vertices as the Cones, we demonstrate 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 FH. But as one of the Antece$ 12. 5.

dents is to one of the Consequents, so are & all the Aritecedents to all the Consequents. Therefore as

the

the Pyramid BKTL is to the Pyramid FMON, so is the whole Pyramid whose Base is the Polygon A TBYCVDQ, and Vertex the Point L, to the whole Pyramid, whose Base is the Polygon EOFPGRHS, and Vertex the Point N. Wherefore the Pyramid, whose Base is the Polygon ATBYCVDQ, and Vertex the Point L, to the Pyramid whose 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, whose 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 also is greater than the Pyramid, whose Base is the Polygon EOFPGRHS, and Vertex the Point N; but it is also less, which is absurd. Therefore the Cone, whose Base is the Circle ABCD, and Vertex the Point L, to fome Solid less than the Cone, whose Base is the Circle EFGH, and Vertex the Point N, has not a triplicate Proportion of that which В D has to FH. In like Manner, we demonstrate that the Cone E F GHN, to fome Solid less than the Cone ABCDL, has not a triplicate Proportion of that which FH has to B D. Laftly, I say the Cone ABCDL, to a Solid greater than the Cone EFGHN, has not a triplicate Proportion of that which В D has to FH: For, if this be poffible, let it be fo to fome Solid Z greater than the Cone EFGHN. Then (by Inversion) 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 Conę EFGHN, the Solid Z shall be to the Cone ABCDL, as the Cone EFGHN is to some Solid less than the Cone ABCDL; and therefore the Cone EFGHN, to fome Solid less than the Conę ABCDL, hath a triplicate Proportion of that which FH has to BD, which has been proved to be imposible. Therefore the Cone ABCDL, to

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fome Solid greater than the Cone E FGHN, has not a triplicate Proportion of that which BD has to FH. It has been also demonstrated, that the Cone ABCDL, to fome Solid less than the Cone E F GHN, hath not a triplicate Proportion of that which BD has to FH. Wherefore the Cone ABCDL, to the Cone E F GHN, has a triplicate Proportion of that

which BD has FH. But as Cone is to Cone, fo is 15. 5.

Cylinder to Cylinder. For a Cylinder having the + 10 of this fame Base as a Cone, and the fame Altitude is + tri

ple of the Cone, since it is demonstrated, that every Cone is one third Part of a Cylinder, having the same Base and equal Altitude. Therefore also 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 Bases; which was to be demonstrated.

PROPOSITION XIII.

THEOREM.
If a Cylinder be divided by a Plane parallel to the

opposite Planes, then as one Cylinder is to the
other Cylinder, so is the Axis to the Axis.
ET the Cylinder AD be divided by the Plane

GH, parallel to the opposite Planes AB, CD, and meeting the Axis E F in the Point K. I say, as the Cylinder BG is to the Cylinder GD, fo is the Axis É K to the Axis KF.

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 FK. And thro' the Points L, N, X, M, let Planes parallel to AB, CD pass. And in those Planes from L, N, X, M, as Centers, describe 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 * 37 of this. Cylinders PR, RB, B G will be * to one another as

their Bases. And therefore the Cylinders PR, RB, BG, are equal. And fince the Axis LN, NE, EK,

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are

are equal to each other, as also the Cylinders PR, RB, BG; and the Number of LN, NE, EK, is equal to the Number of PR, RB, BG: The Axis KL shall be the same Multiple of the Axis EK, as the Cylinder PG, is of the Cylinder GB. For the fame Reason, the Axis M K is the same 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 P G shall be likewise greater than the Cylinder GQ; and if less, less. Therefore, because there are four Magnitudes, viz. The Axis EK, KF, 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 BG; and the Axis KM, and the Cylinder Ġ Q, the Equimultiples of the Axis KF, and the Cylinder GD: And it is demonstrated, that if the Axis L K exceeds the Axis KM, the Cylinder PG will exceed the Cylinder GQ; and if it be equal, equal, and less, less. Therefore, as the Axış E K is to the Axis KF, so * is the Cylinder. BG to the * Def. 5. 5, Cylinder, GD. Wherefore, if a Cylinder be divided by a Plane parallel to the opposite Planes, then as one Cylinder is to the other Cylinder, fo is the Axis to the Axis; which was to be demonstrated.

PROPOSITION XIV.

THEOREM.

Cones and Cylinders being upon equal Bases, are

to one another as their Altitudes.

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ET the Cylinders EB, FD, stand upon equal

Bases A B, CD. I fay, as the Cylinder E B is to the Cylinder FD, so is the Axis Ģ H 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 same Altitude, they are to one another as their Bases. But * 11 of this, $ 4

their

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their Bases are equal. Therefore the Cylinders EB, CM, will be also equal. And because the Cylinder FM is cut by'a Plane CD, parallel to the opposite Planes, it shall be as the Cylinder CM is to the Cylinder FD, so is the Axis LN, to the Axis KL. But the Cylinder CM is equal to the Cylinder EB; and the Axis LN to the Axis GH. Therefore the Cylinder E B is to the Cylinder FD, as the Axis GH

is to the Axis KL. And as the Cylinder EB is to | 15. 5.

the Cylinder FD, so is I the Cone ABG to the Cone . 10 of this. CDK; for the Cylinders are * triple of the Cones.

Therefore, as the Axis G H is to the Axis KL, lo is the Cone ABG to the Cone CDK, and so the Cylinder E B to the Cylinder F D. Wherefore, Cones and Cylinders being upon equal Bases, are to one another as their Altitudes; which was to be demonstrated.

PROPOSITION XV.

THEOREM.
The Bases and Altitudes of equal Cones and Cylin-

ders are reciprocally proportional; and Cones
and Cylinders, whose Bases and Altitudes are
reciprocally proportional, are equal to one an-
otber.

be the Circles ABCD, EFGH, and their Diameters AC, EG; and Axis KL, MN; which are also the Altitudes of the Cones and Cylinders: And let the Cylinders AX, EO, be compleated. I say, the Bases and Altitudes of the Cylinders AX, EO, are reciprocally proportional, that is, the Base ABCD is to the Base EF GH, as the Altitude MN is to the Altitude KL.

For, the Altitude KL is either equal to the Altitude MN, or not equal. First, let it be equal ; and the Cylinder AX, is equal to the Cylinder EO.

But Cylinders and Cones that have the fame Altitude, * 11 tbs. are * to one another as their Bases. Therefore the

Base ABCD is equal to the Base E F G H. And consequently, as the Base ABCD is to the Base EFGH, so is the Altitude MN to the Altitude KL.

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