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 bifccted, and the half of it be again bifceted, and fo on, there muft at length remain a circumference lefs b than AD. let this be LD; and from the point L draw

a. 16. 3.
b. Lemma.

LM perpendicular to BD, and pro

duce it to N; and join LD, DN.

therefore LD is equal to DN, and
because LN is parallel to AC, and B

that AC touches the circle EFGH;

therefore LN does not meet the

A

F

N

circle EFGH. and much lefs fhall
the ftraight lines LD, DN meet the
circle EFGH. fo that if ftraight lines equal to LD be applied in
the circle ABCD from the point L around to N, there shall be
defcribed in the circle a polygon of an even number of equal fides
not meeting the leffer circle. Which was to be done.

LEMMA II.

F two trapeziums ABCD, EFGH be infcribed in the circles the centers of which are the points K, L; and if the fides AB, DC be parallel, as alfo EF, HG; and the other four fides AD, BC, EH, FG be all equal to one another; but the fide AB greater than EF, and DC greater than IIG. the ftraight line KA from the center of the circle in which the greater fides are, is greater than the ftraight line LE drawn from the center to the circumference of the other circle.

If it be poffible, let KA be not greater than LE; then KA muft be either equal to it, or lefs. First, let KA be equal to LE. therefore because in two equal circles, AD, BC in the one are equal to 18. 3. EH, FG in the other, the circumferences AD, BC are equal * to the circumferences EH, FG; but because the ftraight lines AB, DC are refpectively 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 alfo equal to it,

277

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

LE.

But let KA be lefs than LE, and make LM equal to KA, and from the center L, and diftance LM defcribe the circle MNOP, meeting the ftraight lines LE, LF, LG, LH, in M, N, O, P ; and join MN, NO, OP, PM which are refpectively parallel b to, and b. 2... lefs than EF, FG, GH, HE. then, becaufe 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 fame reafon, the circumference BC is greater than NO; and be caufe the ftraight 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 reafon, the circumference DC is greater than PO. therefore the whole cir cumference ABCD is greater than the whole MNOP; but it is likewife equal to it, which is impoffible. therefore KA is not less than LE; nor is it equal to it; the ftraight line KA must therefore be greater than LE. Q. E. D.

COR. And if there be an Ifofceles triangle the fides of which are equal to AD, BC, but its bafe lefs than AB the greater of the two fides AB, DC; the ftraight line KA may, in the fame manner, be demonftrated to be greater than the ftraight line drawn rom the center to the circumference of the circle defcribed about he triangle.

Book XII.

Sec N.

T°

PROP. XVII. PRO B

To in the great, a
O defcribe in the greater of two fpheres which
have the fame center, a folid polyhedron, the
fuperficies of which fhall not meet the leffer sphere.

Let there be two spheres about the fame center A; it is required to defcribe in the greater a folid polyhedron the fuperficies of which fhall not meet the leffer sphere.

Let the fpheres be cut by a plane paffing thro' the center; the common fections of it with the fpheres fhall be circles; becaufe the sphere is described by the revolution of a femicircle about the diameter remaining unmoveable; fo that in whatever pofition the femicircle be conceived, the common fection of the plane in which it is with the fuperficies of the fphere is the circumference of a circle; and this is a great circle of the sphere, because the diameter of the a. 15. 3. fphere which is likewife the diameter of the circle, is greater 2 than any ftraight 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 leffer fphere be FGH; and draw the two diameters BD, CE at right angles to one another. and in BCDE the greater of the two b. 16. 12. circles defcribe b a polygon of an even number of equal fides not meeting the leffer circle FGH; and let its fides, 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 fuperficies of the sphere in the point X; and let planes pafs thro' AX and each of the ftraight lines BD, KN, which, from what has been faid, fhall produce great circles on the fuperficies of the sphere, and let BXD, KXN be the femicircles 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 thro' XA is at right angles to the plane of the circle BCDE; wherefore the femicircles BXD, KXN are at right angles to that plane. and because the femicircles BED, BXD, KXN, upon the equal diameters BD, KN are equal to one another, their fourth parts BE, BX, KX are equal to one another. therefore as many fides of the polygon as are in the fourth part BE, fo many there are in BX, KX equal to the fides BK, KL, LM, ME. let these polygons be defcribed, and their fides be BO, OP, PR, RX; KS, ST, TY,

C. 18. 11.

迄

YX, and join OS, PT, RY; and from the points O, S draw OV, Book XII.
SQ perpendiculars to AB, AK. and because the plane BOXD is at

right angles to the plane BCDE, and in one of them BOXD, OV

is drawn perpendicular to AB the common fection of the planes,

d

therefore OV is perpendicular to the plane BCDE. for the fame d.4 Def.12.
reafon SQ is perpendicular to the fame plane, because the plane
KSXN is at right angles to the plane BCDE. Join VQ, and be-
cause in the equal femicircles BXD, KXN the circumferences BO,

KS are equal, and OV, SQ_are perpendicular to their diameters,
therefore* OV is equal to SQ, and BV equal to KQ, but the • 26.1,
whole BA is equal to the whole KA, therefore the remainder VA is
equal to the remainder QA. as therefore BV is to VA, fo is KQ to
QA, wherefore VQ is parallel to to BK. and because OV,SQ are e, 2. 6,
each of them at right angles to the plane of the circle BCDE, OV is
parallel f to SQ; and it has been proved that it is also equal to it ; f, 6. 17,
therefore QV; SO are equal and parallel. and because QV is pa 8, 33. 1,
rallel to SO, and alfo to KB; OS is parallel to BK; and therefore h. g 11.

Book XII. BO, KS which join them are in the fame plane in which these parallels are, and the quadrilateral figure KBOS is in one plane. and if PB, TK be joined, and perpendiculars be drawn from the points P, T to the ftraight lines AB, AK, it may be demonftrated that TP is parallel to KB in the very fame way that SO was fhewn to be parallel to the fame KB; wherefore a TP is parallel to SO, and the quadrilateral figure SOPT is in one plane. for the fame reafon the quadrilateral TPRY is in one plane. and the figure YRX is also in one

h. 9. 11.

plane i. therefore, if from the points O, S, P, T, R, Y there be drawn ftraight lines to the point A, there fhall be formed a folid polyhedron between the circumferences BX, KX composed of pyramids the bafes of which are the quadrilaterals KBOS, SOPT, TPRY, and the triangle YRX, and of which the common vertex is the point A. and if the fame conftruction be made upon each of the fides KL, LM, ME, as has been done upon BK, and the like be done alfo in the other three quadrants, and in the other hemifphere; there fhall be formed a folid polyhedron defcribed in the sphere,