The Elements of Euclid

ABCDL, so is the cone EFGHN to some solid, which must be less (14.5.) than the cone ABCDL, because the solid Z is greater than the cone EFGHN: therefore the cone EFGHN has to a solid which is less than the cone ABCDL, the triplicate ratio of that which EG has to AC, which was demonstrated to be impossible: therefore the cone ABCDL has not to any solid greater than the cone EFGHN, the triplicate ratio of that which AC has to EG; and it was demonstrated that it could not have that ratio to any solid less than the cone EFGHN: therefore the cone ABCDL has to the cone EFGHN the triplicate ratio of that which AC has to EG: but as the cone is to the cone, so (15. 5.) the cylinder to the cylinder; for every cone is the third part of the cylinder upon the same base, and of the same altitude: therefore also the cylinder has to the cylinder the triplicate ratio of that which AC has to EG. Wherefore similar cones, &c. Q. E. D.

PROP. XIII. THEOR.

Ir a cylinder be cut by a plane, parallel to its opposite planes, or bases, it divides the cylinder into two cylinders, one of which is to the other as the axis of the first to the axis of the other.*

א:

Let the cylinder AD be cut by the plane GH, parallel to the opposite planes AB, CD, O meeting the axis EF in the point K, and let the line GH be the common section of the plane GH and the surface of the cylinder AD: let AEFC be the parallelogram, in any position of R it, by the revolution of which about the straight line EF the cylinder AD is described; and let GK be the common section of the plane GH, and the plane AEFC: and because the paral- A lel planes AB, GH are cut by the plane AEKG, AE, KG, their common sections with it, are parallel (16. 11.): wherefore AK is a parallelogram, and GK equal to EA the straight G line from the centre of the circle AB: for the same reason each of the straight lines drawn C from the point K to the line GH may be proved to be equal to those which are drawn from the centre of the circle AB to its circumference, and are therefore all equal to one another. Therefore the line GH is the circumference of a circle, (15. def. 1.) of which the centre is the point K: therefore the plane GH divides the cylinder AD into the cylinders AH, GD; for they are the same which would be described by the revolution of the parallelograms AK, GF, about the straight lines EK, KF: and

* See Note.

it is to be shown, that the cylinder AH is to the cylinder HC, as the axis EK to the axis KF.

Produce the axis EF both ways; and take any number of straight lines EN, NL, each equal to EK; and any number FX, XM each equal to FK; and let planes parallel to AB, CD pass through the points L, N, X, M; there- O fore the common sections of these planes with the cylinder produced are circles the centres of which are the points L, N, X, M, as was R proved of the plane GH: and these planes cut off the cylinders PR, RB, DT, TQ: and because the axes LN, NE, EK are all equal, therefore the cylinders PR, RB, BG are (11. 12.) to A one another as their bases: but their bases are equal, and therefore the cylinders PR, RB, BG are equal: and because the axes LN, NE, EK are equal to one another, as also the cylinders PR, RP, BG, and that there are as many axes as cylinders; therefore, whatever multiple the C axis KL is of the axis KE, the same multiple is the cylinder PG of the cylinder GB: for the same reason, whatever multiple the axis MK T is of the axis KF, the same multiple is the cylinder QG of the cylinder GD: and if the axis V KL be equal to the axis KM, the cylinder PG is equal to the cylinder GQ; and if the axis KL be greater than the axis KM, the cylinder PG is greater than the cylinder QG; and if less, less since therefore there are four magnitudes, viz. the axes EK, KF, and the cylinders BG, GD, and that of the axis EK and cylinder BG, there has been taken any equimultiples whatever, viz. the axis KL and cylinder PG; and of the axis KF and cylinder GD, any equimultiples whatever, viz. the axis 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 less, less therefore (5. def. 5.) the axis EK is to the axis FK, as the cylinder BG to the cylinder GD. Wherefore, if a cylinder, &c. Q. E. D.

PROP. XIV. THEOR.

CONES 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, 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 same altitude, they are to one another as their bases: (11. 12.) but their bases are equal:

therefore also the cylinders EB, CM are equal. And because the

cylinder FM is cut by the plane CD

parallel to its opposite planes, as the

cylinder CM to the cylinder FD, so is (13. 12.) the axis LN to the axis KL. But the cylinder CM is equal to the cylinder EB, and the axis LN E to the axis GH: therefore as the cylinder EB to the cylinder FD, so is the axis GH to the axis KL: and as the cylinder EB to the cylinder FD, so is (15. 5.) the cone ABG to the cone CDK, because the cylinders are triple (10. 12.) of the cones: A therefore also the axis GH is 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.

H B

PROP. XV. THEOR.

THE 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.*

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.

R N

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 same altitude being to one another as their bases, (11. 12.) therefore the base ABCD is equal (A. 5.) to the base EFGH; and as the base ABCD is to the base EFGH, so is the altitude MN to the altitude KL. But let the altitudes KL, MN be unequal, and MN the greater of the two, and from MN take MP, equal to KL, and, through the point P, cut the cylinder EO A by the plane TYS, parallel to the

opposite planes of the circles

EFGH, RO; therefore the com

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mon section of the plane TYS and the cylinder EO is a circle, and

* See Note.

consequently ES is a cylinder, the base of which is the circle EFHG, and altitude MP: and because the cylinder AX is equal to the cylinder EO, as AX is to the cylinder ES, so (7. 5.) is the cylinder EO to the same ES. But as the cylinder AX to the cylinder ES, so (11. 12.) is the base ABCD to the base EFGH; for the cylinders AX, ES are of the same altitude; and as the cylinder EO to the cylinder ES, so (13. 12.) is the altitude MF to the altitude MP, because the cylinder EO is cut by the plane TYS parallel to its opposite planes. Therefore 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 base 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 (A. 5.) to KL, and therefore the cylinder AX is equal (11. 12.) 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 base EFGH, so is the altitude MN to the altitude KL; therefore MN is greater (A. 5.) 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 (11. 12.) 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 same ES; whence the cylinder AX is equal to the cylinder EO; and the same reasoning holds in cones. Q. E. D.

PROP. XVI. PROB.

To describe in the greater of the two circles that have the same centre, a polygon of an even number of equal sides, that shall not meet the lesser circle.

Let ABCD, EFGH be two given circles having the same centre 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 lesser circle.

Through the centre K draw the straight line BD, and from the point G, where it meets the circumference of the lesser circle, draw GA at right angles to BD, and produce it to C; therefore AC touches (16. 3.) the circle EFGH: then, if the circumference BAD be bisected, and the half of it be again bisected, and so on, there must at length remain a circumference less (Lemma.) than AD: let this be

LD; and from the point L draw LM perpendicular to BD, and produce it to N; and join LD, DN. Therefore LD is equal to DN; and because LN is parallel to AC, and that AC touches the circle EFGH, therefore LN does not meet the circle EFGH: and much less shall 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.

Ir two trapeziums ABCD, EFGH be inscribed in the circles, the centres 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 side AB greater than EF, and EC greater than HG; the straight line KA from the centre of the circle in which the greater sides are, is greater than the straight line LE drawn from the centre 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, DC in the one, are equal to EH, FG in the other, the circumferences AD, BC are equal (28. 3.) 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 to LE.

But let KA be less than LE, and make LM equal to KA, and from the centre L, and distance LM, describe 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 (2. 6.) to and 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 same reason, the circumference DC is greater than PO: therefore the whole circumference ABCD is greater than the

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