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KF and cylinder GD, any equimultiples whatever, viz. the axis Book XII. KM and cylinder GQ, and it has been demonstrated if the axis KL be greater than the axis KM, the cylinder PG is greater than the cylinder GQ; and if equal, equal, and if lefs, less. therefore a d. 5. Def. 5. the axis EK is to the axis KF, as the cylinder BG to the cylinder GD. Wherefore if a cylinder, &c. Q. E. D.

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CON

ONES and cylinders upon equal bases are to one
another as their altitudes.

Let the cylinders EB, FD be upon the equal bases AB, CD. as the cylinder EB to the cylinder FD, fo.is the axis GH to the axis KL.

Produce the axis KL to the point N, and make LN equal tò the axis GH, and let CM be a cylinder of which the base is CD, and axis LN. and because the cylinders EB, CM have the same altitude, they are to one another as their bases a. but their bases a. 11. 12. are equal, therefore also the cylinders EB, CM are equal. and be

F

K cause the cylinder FM is cut by the plane CD parallel to its opposite planes, as the cylinder CM to E

L

D the cylinder FD, fo is the axis

b. 13. 121 LN to the axis KL. but the cylinder CM is equal to the cylinder EB, and the axis LN to the axis GH. therefore as the cylin- A

H

B der EB to the cylinder FD, so is

M

N the axis GH to the axis KL. and as the cylinder EB to the cy- : linder FD, fo is the cone ABG to the cone CDK, because the C. 15. 50 cylinders are triple d of the cones. therefore also the axis GH is d. 10. 12. to the axis KL, as 'the cone ABG to the cone CDK, and the cylinder EB to the cylinder FD. Wherefore cones, &c. Q. E. D.

S

Book XII.

PROP. XV.

THEOR.

See N.

THE

"HE bases and altitudes of equal cones and cylinders

are reciprocally proportional; and if the bases and altitudes be reciprocally proportional, the cones and cylinders are equal to one another.

b, A. 5.

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Let the circles ABCD, EFGH, the diameters of which are AC, EG, be the bases, and KL, MN the axes, as also the altitudes, of equal cones and cylinders; and let ALC, ENG be the cones, and AX, EO the cylinders. the bases and altitudes of the cylinders AX, EO are reciprocally proportional; that is, as the base ABCD to the base EFGH, so is the altitude MN to the altitude KL.

Either the altitude MN is equal to the altitude KL, or these altitudes are not equal. First, let them be equal; and the cylinders

AX, EO being also equal, and cones and cylinders of the fame alti2. 11. 12. tude being to one another as their bases a. therefore the base ABCD

is equal to the base EFGH; and as the bafe ABCD is to the base
EFGH, so is the al-
titude MN to the al-

N
titude KL but let the

R
altitudes KL, MN be
unequal, and MN the

L
greater of the two,

X х
and from MN take
MP equal to KL, and,
thro' the point P, cut

E G the cylinder EO by K

M M the plane TYS paral

B

E lel to the opposite planes of the circles EFGH, RO; therefore the common fection of the plane TYS and the cylinder EO is a circle, and consequently ES is a cylinder, the base of which is the circle EFGH,

and altitude MP. and because the cylinder AX is equal to the C. 7. 5. cylinder EO, as AX is to the cylinder ES, fo is the cylinder EO to

the fame ES. but as the cylinder AX to the cylinder ES, so a is the base ABCD to the base EFGH; for the cylinders AX, ES are

of the fame altitude ; and as the cylinder EO to the cylinder ES, d. 13. 12. fod is the altitude MN to the altitude MP, because the cylinder

EO is cut by the plane TYS parallel to its opposite planes. there. Book XII. fore as the base ABCD to the bafe EFGH, so is the altitude MN Lampe to the altitude MP. but MP is equal to the altitude KL; wherefore as the bafe ABCD to the base EFGH, fo is the altitude MN to the altitude KL, that is, the bases and altitudes of the equal cylinders AX, EO are reciprocally proportional.

But let the bases and altitudes of the cylinders AX, EO, be reciprocally proportional, viz. the base ABCD to the base EFGH, as the altitude MN to the altitude KL. the cylinder AX is equal to the cylinder EO.

First, let the base ABCD be equal to the base EFGH, then becaufe as the base ABCD is to the base EFGH, so is the altitude MN to the altitude KL; MN is equal to KL, and therefore the b. A. 5. cylinder AX is equal a to the cylinder EO.

But let the bases ABCD, EFGH be unequal, and let ABCD be the greater; and because as ABCD is to the bafe EFGH, so is the altitude MN to the altitude KL, therefore MN is greater than KL; then, the fame construction being made as before, because as the base ABCD to the base EFGH, so is the altitude MN to the altitude KL; and because the altitude KL is equal to the altitude MP; therefore the base ABCD is a to the base EFGH, as the cylinder AX to the cylinder ES ; and as the altitude MN to the altitude MP or KL, so is the cylinder EO to the cylinder ES. there-, fore the cylinder AX is to the cylinder ES, as the cylinder EO is to the fame "S. whence the cylinder AX is equal to the cylinder EO. and the same reasoning holds in cones. Q. E. D.

a. II. I 2.

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To describe in the greater of two circles that have

the same center, a polygon of an even number of equal fides, that shall not meet the lefser circle.

T:

Let ABCD, EFGH be two given circles having the same center K. it is required to inscribe in the greater circle ABCD a polygon of an even number of equal sides, that shall not meet the lefser circle.

Thro' the center K draw the straight line BD, and from the point G, where it meets the circumference of the lefser circle, draw

Book XII. GA at right angles to BD, and produce it to C; therefore AC

touches a the circle EFGH. then if the circumference BAD be 2. 16. 3. bifected, and the half of it be again bisected, and so on, there b. Lemma. must at length remain a circumference less than AD. let this be

LD; and from the point L draw
LM perpendicular to BD, and pro-

A
duce it to N; and join LD, DN.

H
therefore LD is equal to DN, and
because LN is parallel to AC, and B EA Ķ GM
that AC touches the circle EFGH;
therefore LN does not meet the
circle EFGH. and much less shall

E
the straight lines LD, DN meet the
circle EFGH. so that if straight lines equal to LD be applied in
the circle ABCD from the point L around to N, there shall be
described in the circle a polygon of an even number of equal sides
not meeting the lesser circle. Which was to be done.

LEMMA II.

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(F two trapeziums ABCD, EFGH be inscribed in the

circles the centers of which are the points K, L; and if the sides AB, DC be parallel, as also EF, HG; and the other-four sides AD, BC, EH, FG be all equal to one another; but the fide AB greater than EF, and DC greater than HG. the straight line KA from the center of the circle in which the greater fides are, is greater than the straight line LE drawn from the center to the circumference of the other circle.

If it be possible, let KA be not greater than LE; then KA must be either equal to it, or less. First, let KA be equal to LE. there

fore because in two equal circles, AD, BC in the one are equal to 2. 23. 3. EH, FG in the other, the circumferences AD, BC are equal a to

the circumferences EH, FG; but because the straight lines AB,
DC are respectively greater than EF, GH, the circumferences AB,
DC are greater than EF, HG. therefore the whole circumference
ABCD is greater than the whole EFGH; but it is also equal to it,

which is impossible. therefore the straight line KA is not equal Book XII. to LE.

But let KA be less than LE, and make LM equal to KA, and from the center L, and distance LM defcribe the circle MNOP, meeting the straight lines LE, LF, LG, LH, in M, N, O, P; and join MN, NO, OP, PM which are respectively parallel b to, and b. 2. 6. less than EF, FG, GH, HE. then, because EH is greater than MP, AD is greater than MP, and the circles ABCD, MNOP are

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equal, therefore the circumference AD is greater than MP; for the fame reason, the circumference BC is greater than NO; and because the straight line AB is greater than EF which is greater than MN, much more is AB greater than MN. therefore the circumference AB is greater than MN; and for the same reason, the circumference DC is greater than PO. therefore the whole circumference ABCD is greater than the whole MNOP; but it is likewise equal to it, which is impossible. therefore KA is not less than LE; nor is it equal to it; the straight line KA must therefore be greater than LE. Q. E. D.

Cor. And if there be an Isosceles triangle the sides of which are equal to AD, BC, but its base less than AB the greater of the two sides AB, DC; the straight line KA may, in the same manner, be demonstrated to be greater than the straight line drawn from the center to the circumference of the circle described about the triangle.

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