multiples whatever, viz., the axis K M and cylinder G Q; and it has been demonstrated, that if the axis KL be greater than the axis K M, the cylinder PG is greater than the cylinder GQ; if equal, equal; and if less, less. Therefore (v. Def. 5), the axis EK is to the axis K F, as the cylinder B G is to the cylinder GD. Wherefore, if a cylinder, &c. Q. E. D.

Corollary.-Every section of a cylinder by a plane parallel to its base is a circle. and, every section by a plane through its axis is a rectangle.

PROP. XIV. THEOREM.

Cones and Cylinders upon equal bases are to one another as their altitudes. Let the cylinders E B and FD be upon the equal bases A B and C D. The cylinder E B is to the cylinder F D as the axis G H is to the axis K L. Produce the axis KL to the point N, make L N equal to the axis GH, and let C M be a cylinder of which the base is CD, and axis L N.

F K

IL D

Because the cylinders E B and C M have the same altitude, they are to one another as their bases (XII. 11). But their bases are equal. Therefore, also, the cylinders EB and CM are equal. Because the cylinder F M is cut by the plane CD parallel to its opposite planes, the cylinder CM is to the cylinder FD, as (XII. 13) E the axis LN is to the axis K L. But the cylinder C M is equal to the cylinder E B, and the axis LN to the axis G H. Therefore, the cylinder E B is to the cylinder FD as the axis GH is to the axis K L; and the cylinder E B is to the cylinder FD as (V. 15) the cone ABG is to the cone CD K, because the cylinders are triple (XII. 10) of the cones. Therefore, the axis G H is to the axis K L, as the cone A B G is to the cone CDK, and as the cylinder EB is to the cylinder FD. Wherefore cones, &c. Q. E. D.

PROP. XV. THEOREM.

B

N M

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

Let the circles ABCD and EFGH, the diameters of which are A C and E G, be the bases, and K L and MN the axes, or the altitudes, of the equal cones ALC and ENG, and of the equal cylinders AX and E O. The bases and altitudes of the cylinders AX and EO are reciprocally proportional; that is, the base ABCD is to the base EFGH as the altitude M N is to the altitude K L.

First, when the altitudes MN and KL are equal. Because the cylinders AX and EO are also equal, and cones and cylinders of the same altitude are to one another as their bases (XII. 11). Therefore the base ABCD is equal (A 5) to the base E F G H. Wherefore the base ABCD is to the base E F G H as the altitude MN is to the altitude KL. Next, when the altitudes K L and M N are unequal, and M N the greater of the two. From MN take MP equal to K L, and through the point P let the cylinder EO be cut by the plane T Y S, parallel to the opposite planes of the circles F H and R O. Because the common

R

section of the plane TY S and the cylinder EO is a circle (XII. 13 Cor.) ES is a cylinder, of which the base is the circle E F G H, and altitude M P. Because the cylinder A X is equal to the cylinder E O. The cylinder AX is to the cylinder ES as (V. 7) the cylinder EO is to the cylinder E S. But the cylinder A X is to the cylinder ES (XII. 11) as the base ABCD is to the base EF GH; for the cylinders A X and ES are of the same altitude; and the cylinder EO is to the cylinder ES (XII. 13) as the altitude MN is to the altitude M P, because the cylinder EO is cut by the plane TYS parallel to its opposite planes. Therefore the base ABCD is to the base EFGH as the altitude MN is to the altitude M P. But MP is equal to K L. Therefore the base A B CD is to the base EFGH as the altitude M N is to the altitude KL; that is, the bases and altitudes of the equal cylinders A X and E O are reciprocally proportional.

X

YPS

DI

C

E

G

B

F

Next, let the bases and altitudes of the cylinders AX and EO be reciprocally proportional; viz., the base A B C D to the base E F G H as the altitude M N to the altitude KL. The cylinder A X is equal to the cylinder EO.

First, let the base A B C D be equal to the base EFGH. Because the base A B CD is to the base EFGH as the altitude M N is to the altitude KL; MN is equal (V. A) to K L. Therefore the cylinder A X is equal (XII. 11) to the cylinder E O.

But let the bases ABCD and EFGH be unequal, and let ABCD be the greater. Because A B C D is to the base E F G H as the altitude MN is to the altitude KL. Therefore M N is greater (V. A) than KL. The same construction being made as before, because the base A B C D is to the base EFGH as the altitude M N is to the altitude KL; and, because the altitude K L is equal to the altitude MP. Therefore the base ABCD is (XII. 11) to the base E F G H as the cylinder A X is to the cylinder ES; and the altitude MN is to the altitude MP or K L, as the cylinder EO is to the cylinder ES. Therefore the cylinder A X is to the cylinder ES as the cylinder EO is to the cylinder E S. Therefore the cylinder A X is equal to the cylinder EO: and the same reasoning holds in cones. Wherefore the bases, &c. Q. E. D.

PROP. XVI. PROBLEM.

In the greater of two concentric circles to inscribe a polygon of an even number of equal sides, that shall not meet the less circle.

Let A B C D and E F G H 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 less circle E F G H.

Through the centre K draw the straight line BD, and from the point G, where it meets the circumference of the less circle, draw A C touching (III. 17) the circle E F G H. If the circumference BAD be bisected, and the half of it be again bisected, and so on, there will at length remain an arc less

B

A

H

F

(XII. Lemma 1) than A D: let this arc be L D. From the point L draw LN parallel to BD, and join LD and D N. Because LD is equal to DN (I. 4) L N is parallel to A C, and A C touches the circle E F G H. Therefore LN does not meet the circle EFGH; and much less do the straight lines LD and DN, meet the circle EFGH. Therefore, if straight lines equal to LD be applied in the circle ABCD from the point L all round to N, there will be inscribed in the circle a polygon of an even number of equal sides not meeting the less circle. Q. E. F.

LEMMA II.

If two trapeziums ABCD and E F G H be inscribed in the circles, the centres of which are the points K and L; and if the sides A B and D C be parallel, as also E F and HG; and the other four sides AD, BC, E H, and FG, be all equal to one another; but the side AB greater than the side EF, and the side D C greater than the side HG; the radius KA of the circle in which the greater sides are, is greater than the radius LE of the other circle.

If it be possible, let K A be not greater than 'LE; then K A must be either equal to LE, or less than it. First, let K A be equal to LE. Because in two equal circles, the two straight lines AD and B C, in the one, are equal to two straight lines EH and FG in the other, the arcs AD and BC are equal (III. 28) to the arcs E H and F G. But the straight lines A B and DC are respectively greater than the straight lines EF and G H. Therefore the arcs A B and D C are greater than the arcs EF and H G. Wherefore the circumference ABCD is greater than the circumference E F G H; but it is also equal to it; which is impossible. Therefore the radius K A is not equal to the radius L E.

But let K A be less than LE; make L M equal to K A, and from the centre L, and distance LM, describe the circle MNOP, meeting the straight lines LE, LF, LG, and LH, in M, N, O, and P. Join MN, NO, OP, and PM; these straight lines are respectively parallel (VI. 2) to and less than the straight lines EF, FG, GH, and HE. Because

K

EH is greater than MP, AD is greater than MP. But the circles A B CD, and M N O P are equal. Therefore the arc A D is greater than the arc MP. For the same reason, the arc B C is greater than the arc N O. Because A B is greater than EF, and EF is greater than MN; much more is A B greater than MN. Therefore the arc A B is greater than the arc M N. For the same reason, the arc DC is greater than the arc PO. Therefore the circumference A B C D is greater than the circumference MNOP; but they are likewise equal, which is impossible. Therefore K A is not less than LE; nor is K A equal to K E. Therefore the radius K A is greater than the radius LE. Wherefore if two, &c., Q. E. D.

Corollary. If there be an isosceles triangle, the sides of which are equal to A D and BC, but the base less than A B, the greater of the two sides A B and C D, it may be shown also that K A is greater than the radius of the circle described about the triangle.

In the greater of two spheres which have the same centre, to inscribe a polyhedron the superficies or faces of which shall not meet the less.

Let there be two spheres having the same centre A. It is required to inscribe in the greater sphere a solid polyhedron, the superficies of which shall not meet the less.

Let the spheres be cut by a plane passing through the centre; the common sections of it with the spheres are circles; for the sphere is described by the revolution of a semicircle about the diameter remaining fixed. Therefore, in whatever position the semicircle is supposed to be, the common section of the plane in which it is, with the superficies of the sphere, is the circumference of a circle; and this section is a great circle of the sphere, because the diameter of the sphere, which is also the diameter of the circle, is greater (III. 15) than any chord in the circle or the sphere. Let the circle made by the section of the plane with the greater sphere be, BCD E, and that with the less, FGH. Draw the two diameters B D and CE at right angles to one another; and in BCDE, the greater of the two circles, inscribe (XII. 16) a polygon of an even number of equal sides not meeting the less circle F G H. Let the sides of this polygon in B E, the fourth part of the circle, be B K, KL, LM, and M E. Join K A, and produce it to N. From A draw (XI. 12) A X at right angles to the plane of the circle B CD E, and let it meet the superficies of the sphere in the point X. Let planes pass through the straight line A X, and each of the straight lines BD and K N; their sections with the superficies of the sphere are great circles. Let BXD and K X N be their semicircles, and BD and K N their diameters. Because X A is at right angles to the circle B CD E, every plane which passes through X A is at right (XI. 18) angles to the circle BCDE. Therefore the semicircles BXD and KX N are at right angles to BCD E. Because the semicircles BED, BXD, and K X N upon the equal diameters BD and K N, are equal, their halves B E, BX, and K X, are equal. Therefore as many sides of the polygon as are in B E, so many are there in BX and K X, equal to the sides BK, K L, L M, and ME. Let these polygons be described, and let their sides be BO, OP, PR, and RX; KS, ST, TY, and Y X. Join O S, PT, and RY; and from the points O and S, draw O V and SQ perpendiculars to AB and A K.

Because the plane BOX D is at right angles to the plane B C D E, and in BOX D, OV is drawn perpendicular to A B, the common section of the planes, therefore O V is perpendicular (XI. Def. 4) to the plane BCDE. For the same reason SQ is perpendicular to the same plane, the plane KSX being at right angles to the plane B C D E. Join V Q. Because in the equal semicircles B X D and K X N, the arcs BO and KS are equal, and OV and SQ are perpendiculars to their diameters. Therefore (I. 26) O V is equal to SQ, and BV to KQ. But the whole BA is equal to the whole K A. Therefore the remainder V A is equal to the remainder QA and BV is to V A as K Q is to QA. Wherefore VQ is parallel (VI. 2) to B K. Because OV and SQ are each of them at right angles to the circle B C D E, O V is parallel (XI. 6) to SQ; and it has been proved that it is also equal to it. Therefore QV and SO are equal (I. 33) and parallel. Because QV is parallel to S O, and also to KB; OS is parallel (xi. 9) to BK. Therefore BO and KS, which join them, are in the same plane, and the quadrilateral figure K BOS is

P

I

T

H

in that plane. If P B and TK be joined and perpendiculars be drawn from the points P and T to the straight lines A B and A K, it may be proved, in the same manner, that TP is parallel to K B. Therefore (XÍ. 9) TP being parallel to SO, the quadrilateral figure SOPT is in one plane. For the same reason, the quadrilateral TPRY is in one plane, and the figure YRX (XI. 2) is also in one plane. Therefore, if from the points O, S, P, T, R, and Y, straight lines be drawn to the point A, a polyhedron will be formed between the arcs BX and K X, composed of pyramids, whose bases are the quadrilaterals KBOS, SOPT, TPRY, and the triangle YRP, and whose common vertex is the point A. If the same construction be made upon each of the sides K L, LM and ME, as upon B K, and the same be done also in the other three quadrants, and in the other hemisphere; a solid polyhedron will be inscribed in the sphere, composed of pyramids, whose bases are the

D

E

M

G

quadrilaterals K BOS, &c., and triangle Y R X, and those formed in the same manner in the rest of the sphere, and common vertex the point A. Also the superficies on this polyhedron does not meet the less sphere of which the circle FGH section. For, from the point A draw (XI. 11) AZ perpendicular to the piane of the quadrilateral K BOS, meeting it in Z, and join B Z and Z K. Because AZ is perpendicular to the plane K BOS, it makes right angles with every straight line meeting in that plane. Therefore AZ is perpendicular to B Z and KZ; and it may be shown, as in Prop. XIV, Book III, that the straight line B Z is equal to the straight line ZK. In the same manner it may be shown that the straight lines drawn from the point Z to the points O and S, are equal to BZ or ZK. Therefore the circle described from the centre Z, and distance Z B, will pass through the points K, O and Z, and K BO'S will be a quadrilateral inscribed in the circle. Because K B is greater than Q V, and QV equal to SO. Therefore K B is greater than SO. But KB is equal to each of the straight lines BO and K S. Therefore each of the arcs cut off by K B, BO and K S, is greater than that cut off by OS; and these three arcs together with a fourth equal to each, are greater than the same three together with that cut off by OS; that is, than the whole circumference of the circle. Therefore the arc subtended by KB is greater than the fourth part of the whole circumference of the circle about KBOS, and the angle BZK at the centre is greater than a right angle. From the point G draw G U at right angles to A G, and join A U. If then the circumference B E be bisected, and its half again bisected, and so on, there will at length be left an arc less than the arc subtended by a straight line equal to G Ü, incribed in the circle BCDE; let this be the circumference K B. Therefore the straight line K B is less than G U. Because the angle R ZK_is obtuse, BK is greater than B Z; in the triangles AGU and A B Z, it may be shown, as in Prop. XV, Book III, that A Zis