the plane GH; and these planes cut

P

off the cylinders, PR,

RB, DT, TQ: And be-O

P

NS

cause the axes LN, NE,

EK are all equal; there. R

NS

B

H

D

E B

RH

C

FD

X

Y

the plane GH, and the plane AEFC: And because the parallel 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 line from the centre of the circle AB: For the same reason, each of the straight lines drawn 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 circumfer ence, 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 it is to be shewn, 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: Therefore the common sections of these planes with the cylinder produced are circles the centres of which are the

BG are equal: And be- T cause the axes LN, NE, V EK are equal to one another, as also the cylinders PR, RB, BG, and that there are as many axes as cylinders; therefore, whatever multiple the 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 is of the axis KF, the same multiple is the cylinder QG of the cylinder GD: And if the axis 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 GQ; 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 KF, as the cylinder BG to the cylinder GD. Wherefore, if a cylinder, &c. Q. E. D.

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 to the axis GH:

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 same altitude being to one another as their bases, (11. 12.) therefore the

a

RO; therefore the common section of the plane TYS and the cylinder EO is a circle, and consequently ES is 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 (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 MN 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.

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 poly. gon of an even number of equal sides not meeting the lesser circle. Which was to be done.

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 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 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 centre to the circumference of the circle described about the triangle.

Let there be two spheres about the same centre A; it is required to describe in the greater a solid polyhedron, the superficies of which shall not meet the lesser sphere.

Let the spheres be cut by a plane passing through the centre; the common sections of it with the spheres shall be circles; because the sphere is described by the revolution of a semicircle about the diameter remaining unmoveable; so that in whatever position the semicircle be conceived, the common section of the plane in which it is with the superficies of the sphere is the circumference of a circle; and this is a great circle of the sphere, because the diameter of the sphere, which is likewise the diameter of the circle, is greater (15.3.) than any straight line in the circle or sphere: Let then the circle made by the section of the plane with the greater sphere be BCDE, and with the lesser sphere be FGH; and draw the two dianeters BD, CE, at right angles to

one another; and in BCDE the greater of the two circles, describe (16. 12.) a polygon of an even number of equal sides not meeting the lesser circle FGH; and let its sides, in BE the fourth part of the circle, be BK, KL, LM, ME; join KA, and produce it to N; and from A draw AX at right angles to the plane of the circle BCDE, meeting the superficies of the sphere in the point X; and let planes pass through AX, and each of the straight lines BD, KN, which, from what has been said, shall produce great circles on the superficies of the sphere, and let BXD, KXN be the semicircles thus made upon the diameters BD, KN: Therefore, because XA is at right angles to the plane of the circle BCDE, every plane which passes through XA is at right (18. 11.) angles to the plane of the circle BCDE; wherefore the semicircles BXD, KXN are at right angles to that plane: And because the semicircles BED, BXD, KXN, upon the equal diameters BD, KN, are e