KF and cylinder GD, any equimultiples whatever, viz. the axis Book XII. KM and cylinder GQ; and it has been demonftrated 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 less, less. therefore d 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.

PROP. XIV. THEOR.

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

C

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

KL.

Produce the axis KL to the point N, and make LN equal to 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 fame altitude, they are to one another as their bases a. but their bases a. 11. 12. are equal, therefore also the cylin

the axis GH to the axis KL. and as the cylinder EB to the cy

linder FD, so is the cone ABG to the cone CDK, because the c. 15. 5

cylinders are triple d of the cones. therefore alfo 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.

See N.

a. 11. 12.

b. A. 5.

PROP. XV. THEOR.

HE bases and altitudes of

THE

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.

1

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 altitude being to one another as their bases. therefore the base ABCD is equal to the base EFGH; and as the base ABCD is to the base EFGH, so is the al

C. 7.5.

planes of the circles EFGH, RO; therefore the common section 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 cylinder EO, as AX is to the cylinder ES, so is the cylinder EO to the fame ES. but as the cylinder AX to the cylinder ES, so a is the bafe 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 base EFGH, so is the altitude MN to the altitude MP. but MP is equal to the altitude KL; wherefore as the bafe ABCD to the base EFGH, so 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 because 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 same 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 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. therefore 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 fame reasoning holds in cones. Q. E. D.

T

PROP. XVI. PRO В.

O describe in the greater of two circles that have
the fame center, a polygon of an even number of

equal fides, that shall not meet the lesser circle.

Let ABCD, EFGH be two given circles having the fame center K. it is required to infcribe in the greater circle ABCD a polygon of an even number of equal fides, that shall not meet the leffer circle.

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

:

a. II. 12.

Book XII. GA at right angles to BD, and produce it to C; therefore AC touches the circle EFGH. then if the circumference BAD be

a. 16. 3. bisected, 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

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 fides not meeting the leffer circle. Which was to be done.

I

F two

LEMMA II.

trapeziums ABCD, EFGH be inscribed in the circles the centers of which are the points K, L; and if the fides AB, DC be parallel, as also EF, HG; and the other-four fides AD, BC, EH, FG be all equal to one another; but the side 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. therefore because in two equal circles, AD, BC in the one are equal to

8. 28, 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

equal, therefore the circumference AD is greater than MP; for the same 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 fame 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 fides of which are equal to AD, BC, but its base less than AB the greater of the two fides AB, DC; the straight line KA may, in the fame manner, be demonftrated to be greater than the straight line drawn from the center to the circumference of the circle described about the triangle.

S3