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their Bafes are equal. Therefore the Cylinders EB, CM, will be alfo equal. And because the Cylinder FM is cut by a Plane CD, parallel to the oppofite Planes, it shall be as the Cylinder CM is to the Cylinder FD, fo 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 E B is to $15.5. the Cylinder FD, fo is the Cone ABG to the Cone * 10 of this. CDK; for the Cylinders are triple of the Cones. Therefore, as the Axis GH is to the Axis KL, fo is the Cone ABG to the Cone CDK, and so the Cylinder EB to the Cylinder F D. Wherefore, Cones and Cylinders being upon equal Bases, are to one another as their Altitudes; which was to be demonftrated.

*

PROPOSITION XV.

THEOREM.

The Bafes and Altitudes of equal Cones and Cylinders are reciprocally proportional; and Cones and Cylinders, whofe Bafes and Altitudes are reciprocally proportional, are equal to one another.

ET the Bafes of the equal Cones and Cylinders

be the Circles ABCD, EFGH, and their Diameters AC, EG; and Axis KL, MN; which are alfo the Altitudes of the Cones and Cylinders: And let the Cylinders AX, EO, be compleated. I fay, the Bafes and Altitudes of the Cylinders AX, EO, are reciprocally proportional, that is, the Bafe ABCD is to the Bafe EFGH, 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 A X, is equal to the Cylinder EO. But Cylinders and Cones that have the fame Altitude, * 11 of this are to one another as their Bases. Therefore the Bafe ABCD is equal to the Bafe EFGH. And confequently, as the Bafe ABCD is to the Bafe EFGH, fo is the Altitude MN to the Altitude KL.

But

*

But if the Altitude KL be not equal to the Altitude MN, let MN be the greater. And take PM equal to LK from MN; and let the Cylinder EO be cut thro' P by the Plane TYS, parallel to the oppofite Planes of the Circles EFGH, RO, and conceive ES to be a Cylinder, whofe Bafe is the Circle EFGH, and Altitude PM. Then, because the Cylinder AX is equal to the Cylinder EO, and ES is fome other Cylinder, the Cylinder AX to the Cylinder ES, fhall be as the Cylinder EO, is to the Cylinder ES. But as the Cylinder AX is to the Cylinder ES, fo is the Bafe ABCD to the Bafe II of this. EFGH; for the Cylinders AX, ES have the fame Altitude. And as the Cylinder EO is to the Cylinder ES, fo is the Altitude MN to the Altitude MP; † 13 of this. for the Cylinder EO is cut by the plane TYS parallel to the oppofite Planes. Therefore, as the Base ABCD is to the Bafe EFGN, fo is the Altitude MN to the Altitude MP. But the Altitude MP is equal to the Altitude KL. Wherefore as the Base ABCD is to the Base EF GH, fo is the Altitude MN to the Altitude KL. And therefore the Bases and Altitudes of the equal Cylinders AX, EO, are reciprocally proportional.

And if the Bafes and Altitudes of the Cylinders AX, EO, are reciprocally proportional, that is, if the Base ABCD be to the Bafe EFGH, as the Altitude MN is to the Altitude KL. I fay, the Cylinder AX is equal to the Cylinder EO. For the fame Construction remaining; because the Base ABCD is to the Bafe EFGH, as the Altitude MN is to the Altitude KL; and the Altitude KL is equal to the Altitude MP. It fhall be as the Base ABCD is to the Base EFGH, fo is the Altitude MN to the Altitude MP. But as the Base ABCD is to the Bafe EFGH, 'fo is the Cylinder AX to the Cylinder ES; for they have the fame Altitude. And as the Altitude MN is to the Altitude MP, fo is the Cylinder ‡ 11 of this, EO to the Cylinder ES. Therefore, as the Cylinder AX is to the Cylinder ES, fo is the Cylinder EO to the Cylinder ES. Wherefore the Cylinder AX is equal to the Cylinder EO. In like Manner we prove this in Cones; which was to be demonftrated.

PRO

16. 3.

+ 29.3.

PROPOSITION XVI.

PROBLEM.

Two Circles being about the fame Center, to infcribe in the greater a Polygon of equal Sides even in Number, that shall not touch the leffer Circle.

LET ABCD, EFGH, be two given Circles

about the Center K. It is required to inscribe a Polygon of equal Sides even in Number in the Circle ABCD, not touching the leffer Circle EFGH.

Draw the Right Line BD through the Center K, as alfo AG, from the Point G at Right Angles to BD, which produce to C; this Line will touch the Circle EFGH. Then bifecting the Circumference BAD, and again bifecting the half thereof, and doing this continually, we fhall have a Circumference left at laft lefs than AD. Let this Circumference be LD, and draw LM from the Point L perpendicular to BD, which produce to N; and join LD, DN. And then LD is † equal to DN. And fince LN is parallel to AC, and AC touches the Circle EFGH, LN will not touch the Circle EFGH. And much lefs do the Right Lines LD, DN, touch the Circle. And if Right Lines, each equal to LD, be applied round the Circle ABCD, we shall have a Polygon infcribed therein of equal Sides, even in Number, that does not touch the leffer Circle EFG. which was to be demonftrated.

PROPOSITION XVII.

PROBLEM.

To defcribe a folid Polyhedron, in the greater of two Spheres, having the fame Center, which shall not touch the Superficies of the leffer Sphere.

LET two Spheres be fuppofed about the fame

Center A. It is required to defcribe a folid Polyhedron in the greater Sphere, not touching the Superficies of the leffer Sphere,

Let

Let the Spheres be cut by fome Plane paffing thro' the Center. Then the Sections will be Circles; for

because a Sphere is * made by the turning of a Semi- * Def. 14circle about the Diameter which is at reft: In what- II. foever Pofition the Semicircle is conceived to be, the Plane in which it is fhall make a Circle in the Superficies of the Sphere. It is alfo manifest that this Circle is a great Circle, fince the Diameter of the Sphere, which is likewife the Diameter of the Semicircle, is +greater than all Right Lines that are drawn in the † 15. 3. Circle, or Sphere. Now, let BCDE be that Circle of the greater Sphere, and F G H of the leffer Sphere; and let BD, CE be two of their Diameters drawn at Right Angles to one another. Let BD meet the lef fer Circle in the Point G, from which to AG let GL be drawn at Right Angles, and AL joined. Then bifecting the Circumference EB, as alfo the half thereof, and doing thus continually, we shall have left at last a certain Circumference lefs than that Part of the Circumference of the Circle BCD, which is fubtended by a Right Line equal to GL. Let this be the Circumference BK. Then the Right Line BK is lefs than GL; and BK fhall be the Side of a Polygon of equal Sides, even in Number, not touching the leffer Circle. Now, let the Sides of the Polygon in the Quadrant of the Circle BE, be the Right Lines BK, KL, LM, ME; and produce the Line joining the Points K, A, to N: And raise ‡ AX ‡ 12. II. from the Point A, perpendicular to the Plane of the Circle BCDE, meeting the Superficies of the Sphere in the Point X, and let Planes be drawn thro' AX, and BD, and thro' AX, and KN, which from what has been faid will make great Circles in the Superficies of the Sphere. And let BXD, KXN, be Semicircles on the Diameters BD, KN. Then because XA is perpendicular to the Plane of the Circle BCDE, all Planes that pafs thro' XA fhall alfo * be perpendicular to that fame Plane. Therefore the Semicircles BXD, KXN are perpendicular to that fame Plane. And because the Semicircles BED, BXD, KXN, are equal; for they ftand upon equal Diameters BD, KN; their Quadrants BE, BX, KX, fhall be alfo equal. And therefore, as many Sides as the Polygon in the Quadrant BE has, fo many Sides may there

* 18. II.

be

[blocks in formation]

*

be in the Quadrants BX, KX, equal to the Sides
BK, KL, LM, ME. Let thofe Sides be BO,
OP, PR, RX, KS, ST, TY, YX: And join
SO, TP, YR; and let Perpendiculars be drawn
from O, S, to the Plane of the Circle BCDE.
Thefe will fall on BD, KN, the common Sections
of the Planes; becaufe the Planes of the Semicircles
BXD, KXN, are perpendicular to the Plane of the
Circle BCDE. Let the faid Perpendiculars be OV,
SQ, and join VQ. Then fince the equal Circumfe-
rences BŎ, SK, are taken in the equal Semicircles
BXD, KXN, and OV, SQ are Perpendiculars,
OV fhall be equal to SQ, and BV to KQ. But
the Whole BA is equal to the Whole KA. There-
fore the Part remaining VA, is equal to the Part re-
maining QA. Therefore as BV is to VA, fo is KQ
to QA: And fo. VQ is ‡ parallel to BK. And fince
OV and SQ are both perpendicular to the Plane of
the Circle BCDE, OV fhall be parallel to SQ.
But it has also been proved equal to it. Wherefore
QV, SO are + equal and parallel. And because QV is
parallel to SO, and alfo parallel to KB, SO fhall be
alfo parallel to KB: But BO, KS, join them.
Therefore KBOS is a quadrilateral Figure in one
Plane For if two Right Lines be parallel, and
Points be taken in both of them, a Right Line join-
ing the faid Points is in the fame Plane as the Paral-
lels are.
And for the fame Reafon, each of the qua
drilateral Figures SOPT, TPRY, are in one Plane.
And the Triangle YRX, is + in one Plane. There
fore, if Right Lines be fuppofed to be drawn from
the Points O, S, P, T, R, Y, to the Point A, there
will be conftituted a certain folid polyhedrous Figure
within the Circumferences BX, KX, compofed of
Pyramids, whofe Bafes are the quadrilateral Figures
KBOS, SOPT, TPRY, and the Triangle YRX;
and Vertices the Point A. And if there be made
the fame Conftruction on each of the Sides KL, LM,
ME, like as we have done on the Side KB, and
also in the other three Quadrants, and the other He-
misphere, there will be conftituted a polyhedrous Fi-
gure described in the Sphere, compofed of Pyramids
whofe Bafes are the aforefaid quadrilateral Figures,
and the Triangle YRX, being of the fame Order,

and

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