a 16. 3.

Book XII. circle, draw GA at right angles to BD, and produce it to C, therefore AC touches the circle EFGH: Then, if the circumference BAD be bifected, and the half of it be again bifected, b Lemma, and fo on, there muft at length remain a circumference lefs 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 lefs fhall the ftraight lines. LD, DN meet the circle EFGH:

a 28. 3.

B

H

E

K GMD

F

N

So that if ftraight lines equal to LD be applied in the circle ABCD from the point L around to N, there fhall 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 centres 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 HG. The ftraight line KA from the centre of the circle in which the greater fides are, is greater than the ftraight line LE drawn from the centre to the circumference of the other circle.

a

If it be poffible, let KA be not greater than LE; then KA must be either equal to it, or lefs. Firft, let KA be equal to LE: Therefore, becaufe in two equal circles, AD, BC in the one are equal to EH, FG in the other, the circumferences AD, BC are equal to the circumferences EH, FG; but becaufe 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, which is

impoffible:

OF EUCLID.

285

impoffible: Therefore the ftraight line KA is not equal to Book XII.

LE.

But let KA be less than LE, and make LM equal to KA, and from the centre 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 to, and lefs than EF, FG, GH, HE: Then, be- a 2. 6. cause EH is greater than MP, AD is greater than MP; and

G

K

M

A

BE

F

the circles ABCD, MNOP are equal; therefore the circum. ference AD is greater than MP; for the fame 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 likewife equal to it, which is impoffible: Therefore KA is not lefs than LE; nor is it equal to it; the ftraight line KA muft 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 base less 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 from the centre to the circumference of the circle defcribed about the triangle.

PROP.

Book XII.

See N.

a 15. 3.

b 16. 12.

Tth

PROP. XVII. PRO B.

O defcribe in the greater of two fpheres which have the fame centre, a folid polyhedron, the fuperfi cies of which fhall not meet the leffer fphere.

Let there be two fpheres about the fame centre A; it is required to defcribe in the greater a folid polyhedron, the superficies of which fhall not meet the leffer sphere.

Let the spheres be cut by a plane paffing through the centre; the common fections of it with the fpheres fhall be circles; because the fphere is defcribed 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 fphere, because the diameter of the fphere, which is likewise the diameter of the circle, is greater than any straight line in the circle or fphere: Let then the circle made by the fection of the plane with the greater fphere 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 circles, defcribe! 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 fphere in the point X; and let planes pafs through 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 c 18. 11. paffes through 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 halves BE, BX, KX, are equal to one another: Therefore, as many fides of the polygon as are in BE, fo many there are in BX, KX equal to the fides BK, KL, LM, ME: Let thefe polygons be defcribed, and their fides be BO, OP, PR, RX, KS, ST, TY, YX, and join

OS,

OS, PT, RY; and from the points O, S draw OV, SQ perpen- Book XII. diculars to AB, AK: And because the plane BOXD is at right n angles to the plane BCDE, and in one of them BOXD, ÖV

a

IL

is drawn perpendicular to AB the common fection of the planes, therefore OV is perpendicular to the plane BCDE: For the a 4. def, fame reafon SQ is perpendicular to the fame plane, because the plane KSXN is at right angles to the plane BCDE. Join VQ; and, because in the equal femicircles BXD, KXN the

circumferences BO, KS are equal, and OV, SQ are perpendicular to their diameters, therefore d OV is equal to SQ, d 26. 1. and BV equal to KQ: But the 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 Q A, wherefore VQ is parallel to BK: And becaufe OV, SQ are each of e 2. 6. them at right angles to the plane of the circle BCDE, OV is parallelf to SQ; and it has been proved that it is alfo equal f 6. 11. to it; therefore QV, SO are equal and parallel : And because g 33. 8. QV is parallel to SO, and alfo to KB; OS is parallel to BK; and therefore BO, KS which join them are in the fame plane h

in

1

Book XII, 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 fhown to be parallel to the fame KB; wherefore TP is parallel to SO, and the quadrilateral figure SOPT is in one plane: For the fame reason, the quadrilateral TPRY is in one plane: And the figure YRX is alfo in one plane.

a 9. II.

2. 11.

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 poJyhedron between the circumferences BX, KX compofed of pyramids the bafes of which are the quadrilaterals KBOS, SOPT, TPRY, and the triangle YRX, and of which the com mon 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 fphere, compo.

fed